chapter binomial theorem binomial theorem very easy chapter very very scoring chapter right so I think we should be able to cover this in two classes I'll just give you an overview of this chapter so what are we going to study under this chapter for your competitive level exam point of view in school it is not there it has been removed from your school exams right so we'll only be studying it from the competitive level exam point of view so the first thing that we are going to talk about is the derivation of binomial theorem so this is a theorem how does it work okay so what is the derivation of it okay how come how come we had one class on this in school but it was not there in your CBC syllabus that's what Raja Ji Nagar students told me ma'am said it is important okay then she is definitely a very concerned and sensitive teacher because she knows that it is useful unlike many teachers who are not teaching at all guys one thing I would like to bring to your notice board has never said dad you should not teach it right board has said you should not test it okay and she is very correct to say that it is very important okay even when I you know last batch your senior batch 12 batch we never omitted anything from our syllabus even though CBC omitted a lot of things and see it paid off now the board's JEMain syllabus says nothing is omitted okay but while many coaching institutes they omitted the syllabus and they taught so that their effort is less we could have also done it happily but in the long run it is going to harm you a lot my second name is you clearing an exam my first name is you understanding the concept enjoying it and knowing it very well exam is a by-product if you if you are studying from a firm exam point of view your knowledge will be very short-lived it will be very ephemeral okay yeah it's not I'm not saying all this thing to show how great is Centrum Academy but yes we definitely believe there is some non-negotiables that understanding is the utmost priority for us all right so derivation of the binomial theorem the next concept is the concept of the general term okay middle term terms they can be middle more than one middle term so middle term or terms okay then we'll talk about numerically greatest numerically greatest binomial coefficient term and coefficient they sound very similar but they are not we'll we'll talk about it when the right time comes okay next is next is the application of binomial theorem application of binomial theorem sometimes some cases divisibility problem remainder problem finding the last digit second last digit third like that etc of a very heavy number so those applications are there with us for us in this chapter and finally we are going to talk about not finally second last you are going to talk about binomial coefficient series binomial coefficient series okay and lastly a bit of multinomial theorem what again so we did this in pnc yes you definitely did it in pnc but we have to talk about it in more detail binomial theorem and finally you have a bit of idea about i mean 15 or 20 minutes idea on binomial theorem for for for non-whole number index okay so these are the seven things that we are going to study in this chapter even though it looks like seven things but i'm sure we'll be able to cover at least the first third in today's class on you okay and four five six seven can be done in the next class okay so are we ready to start ourselves with binomial theorem not a new concept you have been using this in our we have used this in our limits chapter when you were doing one of the standard limits called the standard algebraic limits so not a new concept so let's start first of all binomial theorem it was discovered or invented you should say by by newton one second aakash these are chapters topic names something which are not important like you can copy it after the session has comments sorry chat session is over okay front of which teacher is there for you in your school hsr all right so binomial theorem before we derive this binomial theorem there is certain observation let's let's do a certain observation okay important observation so i'll be introducing you to such terms where n is a whole number now n can be a negative integer or a fraction also that is something which will be taking in the end of the chapter remember the last part of the chapter was binomial theorem for non-whole number index so right now we'll talk about the whole number index which is basically 90 percent of your uh you know concepts in your binomial theorem last 10 percent 5 percent concept will be based on when the power is not a whole number now why it is called a binomial theorem because it is applied to a binomial expression like this right now why normally it was called binomial because we consider that these two terms are dissimilar so they're two dissimilar terms because they are two dissimilar terms we use the word binomial okay but that is like you know very very uh you can say rudimentary form of binomial theorem and it was you know uh invented by uh Newton we actually can apply it to similar terms also right we can also do two plus three whole square by use of binomial theorem doesn't make any difference to uh to the approach okay so don't be like too much into like okay they have to be dissimilar yes when it was invented when it was on the rudimentary phase raw form people use it for uh dissimilar terms a and x are dissimilar term now again my aim is not to tell you that a is a constant and x is a variable no think as if a is a first term x is a second term and they're dissimilar from each other kind of right but don't take it very you know in uh uh exact form don't think like they have to be dissimilar terms they can also be applied to similar looking terms as i already told you a certain observations i would like you to have let me ask this as a question what is a plus x to the power zero let's start with the lowest whole number which is zero what is a plus x to the power zero one okay everybody's saying one what if i write one as one a to the power zero x to the power zero any problem with that we'll say no sir this is one only no problem okay good what is a plus x to the power one guys just have the good observation because i'm writing it out for you to have a good observation on this so it's a a plus x correct a plus x absolutely guy three can i write this as one a to the power one x to the power zero and again one a to the power zero x to the power one right banana is trying to understand what is going on yes absolutely banana okay are you i should write this in yellow it will look beautiful no art sir for beauty and all you raised it let's have consistently white and yellow white and yellow okay next is a plus x whole square what is it you'll say i mean the formula that comes in your mind is a square two ax plus x square correct what if i write it like this one a to x not two a one x one and one a not x two okay okay so don't worry about these numbers that you see worry about this a and x combination how it is like written worry about that look a pattern in that numbers i will talk about it in some time okay next a plus x whole cube these all formulas you have already done in your junior classes so you say a cube three a square x three ax square and x cube right what if i write it like this one a cube x not three a square x to the power one three a to the power one x square and one a zero x cube correct now when you look at these terms only the ones which i'm showing from the right side with a bracketed under bracket okay what pattern do you see there how are they listed down any pattern we are able to observe here no no they are not falling past this triangle past this triangle is followed by these numbers which i'll talk about in some time guy three huh these these numbers a and x they are basically sharing that power or sharing this n among each other or between each other yes or no yes absolutely number so if you see zero is shared by these two one is shared by these two so one zero zero one okay so it starts with the entire power n being taken by a and then what a does it shares one power at a time with x sharing is caring right so a starts with let's say it starts with the whole power x starts with zero then what happens a gives one power to x and again one more power so like that a starts giving one one power to x and slowly it becomes zero and x takes the entire power okay this is a pattern i would like you to observe over here as of now right now i know most of you do not know the formula of a plus x to the power four okay but if you look observe this pattern at least we can write down these expressions so let's say there is some number i don't know which number but i will have a to the power four x to the power zero again some number a to the power three x to the power one then you have a to the power two x to the power two then you have a to the power one x to the power three and then you have some number a to the power zero x to the power four okay so this observation actually helps us to write these terms at least okay now what about these numbers which i have not mentioned what about these numbers now these numbers come from an observation made by a french mathematician blaze Pascal and he realized that it comes from a pattern like this so this one that you see this one that you see here okay write it like this this one and this one you write it like this okay one two one write it like this one three three one you write it like this and he realized that when he was listing down these numbers okay the first number in the last number is always a one and the number in between comes from the addition of the top two if you add these two you'll get a two similarly if you add these two you get a three you add these two you get a three again okay so with this pattern in his mind he was able to construct the further rules so the next so that will come is one one and here i'll get a four this will be six this will be four again okay and these numbers will fortunately become the coefficients of this question mark or they will replace these question marks which i have put over here so it would become one one four six four one and this triangle that blaze Pascal came up with was later on honored after his name and it's called the Pascal's triangle the Pascal's triangle okay not a new concept to you we have already seen it when we were doing our permutation combination chapter i gave you Pascal's identity i gave you hockey stick identity they were all based on this triangle itself so thanks to this observation of pattern and thanks to this Pascal's triangle that now we'll be able to write down a plus x to the power any power we want of course not a very heavy power right now because you have to make triangles to that extent so i would request all of you take a clue from this and just give me what is a plus uh seven sorry a plus uh six a plus x to the power of six everybody please write this out 30 seconds 35 seconds 40 seconds you might take yeah i'll talk about binomial coefficient coefficient everything he wants you don't worry and how they are similar and how they're different at the same time write down but it doesn't take a lot of time no need to type just say done no need to type please okay punnab is done good punnab just hold your answer whatever is also done done done done okay good yeah so what i'm going to do is first i will write down those ax combination something a6 x0 something a5 x1 something a4 x2 something a3 x3 something let me write here something a2 x4 something a1 x5 and something a0 x6 okay so six came up as a power on a first with x power being zero and slowly slowly a started losing power one by one and x started gaining power one by one now to know these bracketed terms you have to take the help of pascal's triangle so pascal's triangle this will be one five ten ten five one one six fifteen twenty fifteen six one okay so this will become one six fifteen twenty fifteen six one right so you don't have to literally you know do a plus x to the power six by you know multiplying it to itself six times okay so thanks to this theorem thanks to this observation of pattern and this pascal's triangle that we are able to figure this out so later on newton he basically formalized it in the form of a theorem so what theorem he brought up and let us link the theorem that he found out with this particular approach so i'm going to the next page permission is there or anybody is copying i'm sure most of you would have written this okay can i go to the next page okay so what is the theorem now so newton what he did a plus x to the power n when n is in a whole number it is what it is basically a plus x into a plus x into a plus x like that it'll go n times isn't it okay now read addition as or and read multiplication as and like the way we used to do it in permutation combination so see when you are trying to multiply it okay you're going to get several terms okay yes or no so when you're going to expand it you're going to get several terms right now each of these term that you see let's say this is your term one this is your term two this is your term three and i don't know how many of you know how many terms will come okay so let's say this term number i do not know okay does anybody know what till what terms how many terms will i get from this anybody has an idea n n plus one n plus two okay many people are saying end okay there's one thing let's keep p question mark okay some of you have studied it they know it's n plus one terms okay should i write it down okay let me write it down okay n plus one terms will come okay you could also observe the pattern which i gave you just now you can just see how is your exponent and number of terms related to each other see now what i'm trying to do is everybody please pay attention here any term that you're formulating whether it is t1 t2 t3 till tn plus one every term is basically made up of either a or x coming from the this factor and either your x coming from this factor and either your x coming from this factor so read this like this as if this is an army to be formed by picking up people from these households so these are households let's say house one house two house three correct till house number n now let's say an army has to be formed and this army will contain either a or x not both that means every household has to contribute one member only one member whether a or x that is our call and every household must contribute one right so think i normally give this example in every binomial session class so think that there is a country which has got n households okay and every household has a man and a woman so a is a man x is a woman okay and the country all of a sudden has to go for a war okay so war has been called by the head of the country so now as a responsibility of every citizen or every household every household every household has to contribute one person okay not both because if both the people are going for the this thing who will take care of the house okay so either a or x will participate from h1 either a or x will participate from h2 either a and h will participate from h3 and so on right and every household must contribute a particular person right now let us say every household gives a then the army will be formed of all a's so that will be a to the power n right that is the first term which you which we used to write okay you may also write it in a better way a to the power n and no x so x to the power 0 if you want to okay let's say there's another way to make the army so this is the first way to make the army another way to make the army would be take a from n minus one households and one x from the remaining household what did i say take a from n minus one households let me write it in white only to maintain the pattern and take one x from the remaining household but now this one household who is contributing x that itself can be chosen in nc one way so many people ask them why are you choosing that house which is contributing x let's choose the house which is contributing a so that can be chosen in ncn minus one way both of them are same as we have learned in our permutation combination chapter okay so it can happen that x come x can come from household number one and rest households will give you a right or it can happen that second household gives you x right so the female member of the second household says i will participate in the war and the other household h1 s3 etc they give a like that now in the same way just keep thinking how many ways can you make an army where two of the you know family contributes to x and the rest contributes to a so that will be nc to a to the power n minus two x square right if this tent continues the last will be ncn a to the power zero x to the power n means all women of the family have decided to go for a war okay and i'm sure they will win it yes or no in order to make it consistent we let's have nc zero right so this is what newton figured out right so let me write it like this okay so these terms that we used to see in the pascals triangle were actually nothing but they were the ncr terms let me show you how this one i mean just i'll write on few of the okay this someone that you see it is actually zero c0 this is one c0 this is one c1 this is two c0 this is two c1 this is two c2 this is three c0 this is three c1 this is three c2 this is three c3 okay i think i already told you this in your combination chapter but just to reiterate the same thing okay so what newton figured out was a theorem actually so he said that a plus x so what's the theorem so the theorem is a plus x to the power n n being a whole number this theorem is working only for whole number okay this can be written as summation of ncr a to the power n minus r x to the power r are going from zero to n this is called the binomial theorem okay or this formula can help you to expand any such binomial expression any such binomial expression okay again i would like you to understand this fact a here is the first term over here and x here is the second term don't be like be so rigid about its definition that it has to be a a means some constant or it has to be just x no i can also apply something like this let's say five x square plus one by three x to the power of 10 i can apply binomial theorem on this so this will behave like your a and this will behave like your x this is just written for you to you know keep this formula in mind don't be very very rigid on its way the way i've written it a is your first term x is your second okay one more thing i would like to highlight over here if let's say there is an expression where there's a minus sign in between okay let's say this then what do we do is we put a plus and we include the minus sign with the second term okay so same theorem will be applicable right so don't be like okay if this is plus some different theorem will be there and if there's a minus some different theorem will be there no theorem is the same is just that the minus sign is made a part of the second term okay so let's do some simple exercise before i know we talk about anything else any questions so far so far looks easy any kasta no kasta okay let's move on some simple simple question we will take and then i will complicate your life why sir all dangerous dangerous problems i'm seeing oh you easy ones okay let's do one thing uh let's ignore the question my question is just find the expansion of this term forget about the seventh term okay as of now ignore this just write the expansion of this term sir 14 terms you have to write okay let me reduce the power let me make this as uh six right i know 14 terms will be slightly too much and just say done okay you don't have to even write them on your chat box i will do it for you write down the expansion of four x minus one by two root x whole to the power of six since this is your first question take some time and write it properly done okay so first term would be by the way as i told you include negative sign with the second term okay so the first term will be six c zero four x to the power six no need to expand it please do not waste your time doing it my purpose was just to show you the expansion in general okay next term will be six c one four x to the power five minus one by two root x to the power of one okay next will be six c two four x to the power four minus one by two root x to the power of two then will be six c three four x to the power three minus one by two root x to the power of three then six c four four x to the power of two 2, minus 1 by 2 root x to the power of 4, then 6c5, 4x to the power of 1, minus 1 by 2 root x to the power of 5, and then 6c6, 4x to the power of 0, minus 1 by 2 root x to the power of 6. If you really expand it you would realize one small observation that there will be alternative plus minus terms coming up. There will be alternative plus minus terms coming up because of this negative sign. So whenever there is an odd power on this term, a negative sign will feature in between. So the presence of an alternative plus minus terms in an expansion tells you that one of the terms or the sign in between has to be negative. Is this fine? Any questions? If this is clear to you, I have a question for you. Are you second question only? This is difficult. I would like you to try this out. Even if you are not able to do it, don't worry too much about it. Find the sum of this series. Now, this is linked to whatever we had learnt a little while ago. That's why I thought I would give you these questions. So summation of minus 1 to the power r, ncr times 1 by 2r, 1 plus 3 to the power r by 2 to the power 2r, etc. Look at these terms very carefully. Up to n terms and this r is going from 0 to n. This r is going from 0 to n. Let's see how many of you are able to attempt this question. Take your time. One minute, one and a half minutes and even if you are not able to answer, don't worry. I know this is a difficult one to crack given that we just started the chapter today. Yes, any success stories to share? Nobody, not yet. Difficult to crack? Okay. See, guys, let me take this out. Let me take this out. 30 seconds, you want more. Okay, Aditya. Good. That's like a fighter. Yes, alternating sign and these terms are also written in a very peculiar way. Right, Aditya. Okay. Now, see, if you just open up this term, let's say I open up this term like this, summation r equal to 0 to n minus 1 to the power r ncr and let's take the first term 1 by 2 to the power r. Okay. Next, again, summation r equal to 0 to n minus 1 to the power r ncr. This I can write it as correct me if I'm wrong, 3 by 4 to the power r. Right, because 2 to the power 2 r is like 4 to the power r. Similarly, the third term would be summation r equal to 0 to n minus 1 to the power r ncr 7 by 8 to the power r. And this continues dot, dot, dot for m terms. So m terms are like this. Okay. Now, let us look into the first term, this term. Let's try to analyze this term a bit. If you write it down, let's say in expanded format, you would write nc0, half to the power 0. Okay. Then you'll write minus nc1, half to the power 2, oh, sorry, half to the power 1. Correct. Then again, nc2, half to the power 2. And this will go with an alternating plus minus sign till ncn, okay, minus 1 to the power ncn, half to the power n. Okay. A clear analysis of this will tell you that, sir, you're writing actually 1 minus half to the power of n expansion, isn't it? Am I not writing 1 minus or you can treat it like 1 plus minus half, take it in brackets, whole to the power n expansion? Yes or no? Does this convince you first of all? Guy 3 is convinced. Others convinced? How do you not sort the other way? Other way, is it? Oh, half minus 1, half minus 1 to the power 3. Okay. See, if it is half minus 1, of course, you would realize, see, doesn't make a difference because you're writing the same thing in a reverse way. But yes, it is very important. If you write half minus 1 to the power of n, the first term would have been nc0, half to the power n, right, which may not have a negative sign, right? This may not be the same term as this because it depends upon n actually, right? The next term would be minus nc1, half to the power n minus 1, right, which may not be the term which is just second last term. So, sign will get different over here. Aditya, sign will differ. Are you getting my point? So, till we are aware of that n, we cannot comment that it is same as, see, if n was even, it won't make a difference, whether you write 1 minus half to the power n or n minus half to the power n. Pranab is spamming something on this group. What is this Pranab? Half. What are you trying to say? I didn't get you. You're trying to say the answer. No, no, no. Answer is still to find out. Wait, don't worry. Answer is not that. Wait. So, if this is understood, can I also say that I know the second term also? The second term is actually 1 minus 3 by 4 to the power n. Third term will be 1 minus 7 by 8 to the power n and this continues till m terms. Yes or no? Since first term is, all of you are convinced that it is this, you should also be convinced that the second term is this, third term is this and so on and so forth, correct? Which means, which means you are convinced that we are trying to sum up something of this sort till m terms, okay? Not to panic looking at it because it is actually a geometric series, okay? This is actually a geometric series whose first term is 1 by 2 to the power n. Second term is the multiplication of the first with 1 by 2 to the power n. Third term is again you multiply 1 by 2 to the power n and so on. So, this is clearly a geometric series with your first term being 1 by 2 to the power n, common ratio also 1 by 2 to the power n and number of terms is m. Number of terms is m, right? So, for such a series, we already know the formula and the formula is what? What is the sum? First term, right? 1 minus the common ratio raised to the power total number of terms by 1 minus 1 by 2 to the power n, okay? This is going to be your answer. If you want to, once again, 2 to the power n, not n. If you want to simplify it, it is your call, go ahead and simplify it. But my purpose was to just convey the idea to you. Does it make sense? Was it a difficult question? Actually, no. It was based on your basic understanding of the binomial theorem structure, right? Pranav, did you get your answer now? Did you get your mistake also? Okay, right? All right. Now, we'll do some analysis on the binomial theorem that we have just now figured out, a small analysis. The first analysis is the number of terms. The number of terms in a binomial coefficient is n plus 1. But this is subject to the fact that the terms involved are dissimilar, okay? Why I'm saying you this is because, let's say, somebody says 2 plus 3 to the power of 2. You'll only get one answer. That is 25, right? So, according to this observation, it should have been 3, but the answer is only 1, right? So, this is when you are assuming that the terms involved are such that if you expand them, they don't get observed in in each other, or some of the terms do not get combined and give you a single term, okay? So, number of terms is n plus 1 under the assumption that there is no combination of like terms happening, okay? So, this is the maximum number of terms that you can get. That is what you can actually say, okay? Anyways, now, this is to be perceived as if you are trying to distribute, please note that this is actually the number of ways to distribute two identical objects, or you can say n identical objects, n identical objects between two persons, okay? If you recall this concept that we had done in our permutation combination chapter, so let's say you're trying to distribute n identical laddus to two people, okay? Let's say Sita and Geeta, and laddus have to be distributed, and these laddus are all identical. How many ways can you do that? Remember, we had learned a formula, and the formula was n plus r minus 1, cr minus 1, correct? n being the total number of objects which were identical, r being the total number of people among which we were distributing the objects, so it will be n plus 2 minus 1, c2 minus 1, which automatically comes to be n plus 1, c1, which automatically comes to be n plus 1. Now, why I'm telling you this formula is because you can scale this concept up to find out the total number of terms even if you are given a trinomial, or to be more generic even if you are given a multinomial, right? For example, let's say if somebody asks you how many terms are there in a plus b plus c whole square? I know all of you know six terms are there, right? But let's say as a person who would like to solve this from p and c point of view, I would read it as if I'm trying to distribute two laddus between three people, right? Two identical laddus, right? Or pastries, Christmas time, cakes and pastries, okay? Two identical slices of cake is to be distributed between three people. How many ways can I do that? Your answer will be n plus r minus 1, c, r minus 1, that's nothing but 4c to 6, that means there will be six terms in this expression, a square, b square, c square, 2ab, 2bc, 2ca, okay? So again, this is subject to the fact that there is no observation, what I'm saying?  Because if let's say I use this to find the expansion of this, 1x plus 1 by x plus 1. Here you will realize you only have x square, 1 by x square, 1. Then you'll have a 2, then you'll have a 2x, then you'll have a 2 by x. And now 1 and 2 can be combined to become a 3. Normally we combine the like terms, right? So there are only five terms here, 1, 2, 3, 4, 5. So this formula will fail, this formula will fail, right? So don't think that you have got a, you know, what do you call it? The all-income passing formula to solve these kind of questions. So questions may be involving some like terms also. There this method will not work or there this formula will fail, so as to say. Are you getting my point? So this is number one observation. Okay, let me ask you this question. Let me ask you this question. I think I would have a question on that. The number of terms in the expansion of this are 36. Find n. Who will tell me this? Yes, anybody? Anybody? Okay, good. Very good, Aditya. Okay, so r is 3 here. n is not known to us. But we know for the fact that n plus r minus 1, cr minus 1 is 36. We very well know the formula of ncr, which is n plus 2 factorial by 2 factorial n factorial. n factorial, this is 36. This is nothing but, this is nothing but n plus 2, n plus 1 is equal to 72, right? 72 can be written as 9 into 8. So either you compare n plus 1 with an 8 or you compare n plus 2 with a 9. Either ways you'll get n value as 7 only. Answer is 7. Okay, so if you expand x plus y to the power 7, x plus y plus d to the power 7 and x, y, z are all such terms which are dissimilar to each other such that they don't combine on expansion or they don't give like terms on expansion. Then total number of terms here will be 36 terms. Then that would be a question in itself, Aditya. So these formulas will change. So there are different ways to deal with it. Don't worry. When such problems will come, we'll basically tackle it. Okay, so yes, of course as you correctly stated that there would be certain cases given to you where combination will happen. We'll talk about it when the problem comes. Okay, don't worry about it. So now we'll continue with our analysis. So this was the first analysis. Second thing is, second thing that I would like to discuss is the general term. General term basically helps you to figure out or find out a term with a certain power of the variable. Okay, for example, there was one of the question. Remember in the first question, the question was find the coefficient of x to the power 7 in that expansion and I said, okay, ignore that x to the power 7 and just do a general expansion. Those type of questions can be solved by this concept. So listen to this. The general term is basically the one which I wrote it in the binomial theorem expansion ncr a to the power n minus r x to the power r. This is actually your r plus one-th term. Remember this. This is not your r-th term. Guys, let me tell you because r starts from zero. So r is a number which starts from zero to n. Let me write it like this. r is a whole number which belongs to zero to n. So we cannot say this is not the r-th term because if you're saying this is the r-th term means you're trying to say zero th term. Zero th term doesn't make sense. Your terms can only take natural numbers positions, right? First term, second term, third term, right? So this is an expression for the r plus one-th term. If you put r as zero, you automatically get the first term which is nc zero a to the power n x to the power zero. If you put r as one, you automatically get the second term which is nc one a to the power n minus one x to the power one. Okay. Now many people ask me, why you give this formula? Why can't you give me this formula? t r where r is your natural number from one to n plus one. See, I can give this formula to you also but you will see that this formula is much more ugly to remember.  Which is, which is the more easier formula to remember? Tell me. Isn't it this, right? So if I want, I can give you this formula also but you will not be able to remember this and a lot of mistakes will happen and in fact, you have to write more actually, right? So better to write it like this rather than writing it like this. Okay. So I will erase it in case somebody gets confused and starts using it. Is this fine? Any question? The third thing that I would like to discuss is the middle terms. Middle term bracket s because there can be either one middle term or more than one. Okay. Now, why there can be more than one middle terms? See, let me let me write some expansions for you. Let's say a plus x to the power of two. Okay. If you expand it, you get this to be more specific nc zero a to the power two x to the power zero two c one x to the power one a to the power one two c two x to the power a to the power zero x to the power two. Correct? How many middle terms do you see here? You say only one.  So only one middle term is there. However, if I write a plus x cube, you will write three c zero a cube x not three c one a square x one three c two a one x square three c three a not x three. There are two middle terms over here. See, whenever there are even number of objects, they cannot be a single middle term. There has to be multiple middle terms. Right? For example, if you have four fingers, I ask you, which is a middle finger? Right? There's no middle finger. They're two middle fingers. Okay. So these two will be middle fingers. They're two middle terms. But if they're three, then which is the middle finger this one? Okay. Sorry. Okay. Is it fine? Understood? Okay. Now, if you make a pattern out of it, tell me, first of all, whether there will be one middle term or multiple middle terms or two middle terms depends upon what you say depends upon n. Correct? So let's try to device a mechanism or a formalization of this concept. So if n is even, there will be one middle term because n plus one will be odd, total number of terms will be odd. If total number of terms are odd, they can be only one middle term. Okay. So if n is even, then there will be only one middle term. But my question to all of you is, can you tell me what is the position and the expression of that middle term? Which term will be the middle term in this case? Think and answer. You can take multiple examples, you can take a plus x square, you can take a plus x to the power four, you can take a plus x to the power six and give me a formula in terms of n as to which will be the what will be the position and the expansion or the expression for that middle term. Very good punners. Anybody else? Very good Hariharan. So yes, people are giving me absolutely correct response. It will be n by two plus one-th term that will be your middle term. Okay. Anybody who wants to contest this? Are you all convinced? You can check it out, you can see for yourself, do a bit of verification. And that expression will be nc n by two a to the power n by two x to the power n by two. Got the point. And just to remind you, this will be the greatest binomial coefficient of all the terms. This will have the greatest binomial coefficient. Guys, I'm not trying to say greatest coefficient. I will talk about it very soon in today's class only. This will have the greatest binomial coefficient. nc n by two is the greatest binomial coefficient. Okay. So the middle term or terms will always have the greatest binomial coefficient. Okay. Anyways, coming to the fact when n is odd. When n is odd, we already discussed that total number of terms would be even. If total number of terms will be even, there will be two middle terms. There will be two middle terms. Can somebody tell me what will be the position of those two middle terms? And hence what is their expression? Take example, you can start with a plus x to the power one, you can take a plus x to the power three, you can take a plus x to the power five. So take some, you know, easy to manage cases with lighter powers. n by two plus n plus half. That is three n by two plus half. Gathri didn't quite get that. n by two plus 1.5. That's a good way to state it. Okay. So here, the position of your middle terms would be n plus one by two and n plus three by two. First convince yourself. Take some examples. Okay, let's take a plus x to the power five. Right. I will not write it, but I know for sure that there are six terms into this. So t one, t two, t three, t four, t five, t six. Okay. These are the two middle terms. Now three comes from five plus one by two and four comes from five plus three by two. So this expressions are absolutely correct. Okay. Now when you have to expand it, remember we had learned an expression for r plus one is term. So you have to write it in the same way. You have to write it in a slightly lengthy fashion like this so that you know this is your r. So now you can say n c r a to the power of n minus r, which happens to be n plus one by two x to the power r, which happens to be this. Okay. This term can written as t n plus one by two plus one. So this will become n c n plus one by two a to the power n minus one by two x to the power n plus one by two. Okay. So these will be the two middle terms and also needless to say that these two will be the greatest binomial coefficient and they would be equal. They would be the greatest binomial coefficient and they would be equal. Yes, Hariharan, do you have any questions? Do you have any question to ask? Can you scroll up? Okay. How much up? This much up? Top right. Yeah, top right. Here. Any questions? Okay. Now we're ready to take up some questions. Let's start solving few questions. Let me know if everybody is done copying with this. Done. Okay. Let's have this question. Yeah, this question which we had distorted. Let's now solve this question. Find the seventh term in this expansion. Of course, you will not sit and write all the 14 terms. Can we find an easier and a convenient way to directly write down the seventh term? If yes, please do so. And just type done on your chat box so that I can discuss it. Write down the seventh term of this binomial expansion. Yes, done. Let's discuss it. Guys, this is very simple. If you want the seventh term, you have to write it like this six plus one, because then this will tell you what is the R to be used for that. So it'll be N, C, R. First term to the power of N minus R. Second term. Remember, second term will have minus inclusive. Okay. So this will give you 13 C is six, four to the power, sorry, four X to the power seven. And this you can write it as by two to the power six X to the power three. So I will not do a lot of simplification, but yes, I will do certain simplification, which will make our life easy. Two to the power eight X to the power four. Okay, no need to expand it further because you would need a calculator for it, which is not allowed anyways. Is it fine? So thanks to the general term that we are able to write such, you know, expansions without having to go to that term. Okay, the same question, the same question. If I ask you, find the seventh term from the end. Okay, let's take a second part of the question. Find the seventh term from the end, from the end in the expansion of whatever was the expansion, I think four X minus one by two root X to the power of 13. Now, I know you can do it, but the safest and the fastest way to do this problem is people start counting from backwards, which is time consuming just reverse the position of these two terms. And instead of finding seven terms from the end, find the seventh term from the beginning that will be your answer of the seven term from the end. So if you reverse the position of these two terms, then the seventh term from end in this will be the seventh term from beginning in this. Okay, and this is much easier to deal. So we can directly write it as T six plus one in this. So it'll be 13 C six minus one to the power seven, four X to the power of six. Is it fine? Okay, you can simplify this and get your job done. So if some question says find the seven term, find some nth term from the end, reverse the given binomial term positions and then write down the same position from the beginning. Okay, you don't have to reverse count it. Is it fine? Any questions? Let's take another one. Let's take this question. Find the coefficient of X to the power eight in this expansion. Done. Okay, now see, in this type of question, obviously you will not sit and write all the terms and try to consolidate the powers and see which power is giving you power of eight. Okay, so there has to be a shorter and faster way to do these kind of questions. Correct. So here, what do we do is we say, let t r plus one th term contain X to the power eight. Okay. So see, it is impossible for us to write 11 terms and then consolidate the powers of X and check which one is finally giving you X to the power eight. That would be two time consumed. Okay. So in order to do this question, you have to first assume that let r plus one th term contain X to the power eight. So r plus one th term would be what? r plus one th term will be 10 c r x square to the power of 10 minus r and minus one by X to the power r. Okay. So let's say this contains X to the power eight term. Correct. If you take all the constants together and all the powers of X together, this is what we see. Correct me if I'm wrong. Now if this contains X to the power eight term, that means it clearly implies that this power over here is actually an eight. That means three r is 12. So r is equal to four. Correct. That means your fifth term contains X to the power eight and that term itself would be 10 c four minus one to the power four. Okay. X to the power eight. So this will be called as the coefficient. Guys, again, get this clear. All the constants combined together with the variable, all the constants which is present with the variable that is called the coefficient of that variable.  By the way, 10 c four, I think I s 210. I have to just check it out by 24. Yeah, 210. So the answer is 210. Is this fine? Why three? Why three are here? See, we have 20 minus two are already and this X will also have an power of our right, right? Exponent rule. Exponent loss. See, I, okay. I think I have to write it in a much more easier way. See, this term 10 cr X to the power two into 10 into r is what? 20 minus two are this is minus one to the power r by X to the power r. So if you consolidate this to single power of X, what will you write? Okay, and minus one to the power r will be clubbed with the constant. This is what I wrote in a direct way. Got it. Okay. Now, one small thing I would like to add to this since we are doing these kind of problems. Let us say, let's say hypothetically, if you solve for r and r did not come out to be a positive integer. That means r came out to be negative integer or a fraction. So please note this. If r does not belong to a whole number, then the coefficient of such a term should be written as zero. That means such a term was actually not present only in the expansion. So if something is not present in the expansion, we have to write its coefficient as zero. Right. So it may happen that you're solving a question and you try to get an r and you got, let's say, a five by two or you get r as minus three hypothetically. Then basically, such a term is not there in your expansion. Okay. And if it is not there in the expansion, you can always write its coefficient as zero. Is that fine? Any questions? Any concerns? All right. We'll take more questions. Don't worry. Okay, let's take this term. If the coefficient of three consecutive terms in the expansion of one plus X to the power n is 165, 330 and 462 respectively, find the value of n, find the value of n. Coefficient of three consecutive terms. I'm not saying coefficient of first three terms, three consecutive terms. Very good, Pranav. Anybody else? Guys, do not leave the meeting. I'm taking attendance right now. Some of you have become very quiet, like Pratik. Pratik, how are you? How are you doing all this? And there is to be a gentleman by the name of Shashwat. He's not there in the class also. Who else? Rehan, Rishabh, Sauras Singh, Adansar. Yes, two people have answered so far. Others, otherwise keep participating, looks lively. Have you seen a cricket match? What is the role of a wicket keeper? His role is not only to keep the wickets. Sledging, huh? Yes, of course he sledges the opposite team, but also keeps on encouraging the team members. Isn't it? Come on, come on, you will get it. You will get it on his ball. Like that he says, isn't it? I remember they keep saying, oh, this is a weak batsman. This is a weak batsman. He'll go, he'll come and go. One ball only, he'll come and go. Rishabh Pant, I have not seen him playing, to be very, very sorry about that. Modern day cricket, I'm so much alienated from the modern day cricket. It was happening too much, IPL and all those. We used to wait for those series to happen, India versus Australia or India versus Pakistan or the Saja Cup, which is to wait for three months, four months. Now, cricket is overloaded. Some westernies, see to be very frank, I don't even know the entire team that plays for India. I know a few of them, Virat Kohli. I think I know only one person. Shikhar Dhawan, sometimes I hear his name. Not in 80s also, 80s was like beyond me. I was like, I belong to Tendulkar, Javed, sort of ganglis. Yeah, you're absolutely correct. It's not a commodity. It's no more a game. It's commodity. Yes, you can say it was an entertainment. It is always an entertainment, but not as more of a commodity. Yeah, all for commercials, yes. Some big scam will surface out after IPL and all. Same with me, yes. The first IPL I saw in 2008, and after that I never wasted my time. Because I thought it's like one of the occasions they have conducted such a gala event where all the states are participating. After that, it is every year. Who will waste time sitting in the front of TV watching 2-2 matches back-to-back? We are not that jobless also, please. Let's discuss this. See here, your coefficients will actually be the binomial coefficients. So, can I say, let's say the coefficient of the rth term, which is ncr-1 is 165. Coefficient of r plus 1th term is 330 and coefficient of r plus 2th term is 462. Now, the only way to figure out what are the values of n and r is the use of one of the properties which I discussed in the permutation combination chapter, which was this property. Ncr by ncr-1 is n minus r plus 1 by r. So, if I use it on this two, first of all, let's say, so 330 upon 165 is going to be n minus r plus 1 by r. By the way, this is 2. So, 2r is equal to n, or you can say 3r is equal to n plus 1. 3r is equal to n plus 1. This is the first formula that we get. Okay, we'll check. The next is, if you replace your r with an r plus 1, then ncr plus 1 by ncr, that will be 462 upon 330. If I'm not mistaken, this is divisible by 11. Yes or no? 11, 11, 11. Yes, 11. So, it will be 42 times 11 and this will be 30 times 11. And of course, we can go by 6 as well. So, 66 to be more, friend. So, this will be 5 and this will be 7, 7 by 5. But ncr plus 1 by ncr is just replacement of r with r plus 1. That's it. Okay. So, 7r plus 7. If you open this bracket over here, it will give you 5n minus 5r. Correct me if I'm wrong. Right? So, 12r is equal to 5n minus 7. So, clearly I can say 5n minus 7 is 4 times n plus 1. That means n is equal to 11. That's your answer. From 1 into. I mean, this was a question which was already done multiple number of times while we were doing PNC chapter, just a repetition of the same. Okay. I will take up more questions. But right now, I think most of you need a break. We started at 3.45. So, it's almost like 2 hours, 20 minutes kind of a thing. So, let's have a small break. Yeah, we definitely need a break. Okay. So, the next concept, in fact, the next problem that we're going to take is a slight complex question based on a similar concept. Oh, I think we have not done any question on middle terms. I'm so sorry about it. Let's take some middle terms question. Let's take this question. Find the middle term. In fact, they should write middle terms if at all there can be more than one. So, find the middle term or terms in the expansion of 3x minus x cube by 6 to the power 9. So, if n is odd, we have already discussed that there will be two middle terms. Correct? So, there will be two middle terms. So, what are those middle terms? Let's figure it out. Just write done once you're done. No need to simplify it. Just say done so that we can discuss it. Okay, enough. No issues. No issues. Okay. So, n is an odd number 9. Basically, your n. Okay, this is your n. If this is odd, there are two middle terms. They are two middle terms. And what are those middle terms? Tn plus 1 by 2th and Tn plus 3 by 2th. By the way, they are just one difference apart. So, if this is 5th, then this will be 6th. Okay. So, for T5, we have to use r value as 4. So, it will be 9c4 3x to the power of 5 and minus x cube by 6 to the power of 4. Okay. No need to waste time in the simplification process because this is a simple question to solve. And for 6, I have to keep my rs5. So, 9c5 3x to the power of 4 and minus x cube by 6 to the power of 5. Okay. So, these will be your two middle terms. Any questions? Easy. Okay. So, we'll start complicating it slightly. Okay. Next question for you is, which is more? 101 to the power of 50? 100 to the power of 50 plus 99 to the power of 50. Which is more? Okay.  So, please justify your answer. Which of them is more? And please justify it. Why? Any idea anybody? Okay. 101 to the power of 50. Can I write it as 100 plus 1 to the power of 50? Okay. Can I say it is 100 to the power of 50 plus 50c1 100 to the power 49 plus 50c200 to the power 48. Till 50c50. Okay. Now, 99 to the power 50, in a similar way, can I write it as 100 minus 1 to the power 50? Now, since there's a negative sign over here, when you expand it, you will get alternating plus minus signs. Okay. As you can see here on your screen. Okay. But this will also end in 50c50. Right? Now, subtract these two. Subtract these two. Okay. Subtract these two. When you subtract these two, on the right side, if you check, you will end up getting the second term twice up, fourth term twice up like that. So, you'll get two times 50c100 to the power 49, 50c300 to the power 47, and so on. And the last term, I mean, I don't care about the last term, by the way, I don't need it. Okay.  Correct. Yes or no? 50 into 2 will be 100. And this is only 100 to the power 49. And other terms will be twice 50c3. I will not waste time writing it. I'll just say dot, dot, dot. Okay. Now, pay attention. Everybody please pay attention. This is 100 to the power of 50. And this is a positive term. Isn't it? This all term over here, which I'm not writing, it's actually a positive term. Right? I don't think so. There's a negativity in that term in what form, whatsoever. Right? So, what do you figure out from here? We figure out that 101 to the power 50 minus 99 to the power 50 minus 100 to the power 50 is a positive. Oh, sorry. I should have kept this symbol. Yes or no? And this is a positive term. That means 101 to the power 50 minus 99 to the power 50 minus 100 to the power 50 is greater than zero because it's a positive term. Correct? So, if this is a positive term, this is greater than zero. And if this is greater than zero, it automatically implies that 101 to the power 50 is greater than 100 to the power 50 plus 99 to the power 50. Okay? So, this guy alone is bigger than the sum of the other two. So, this is the winner. This is the term which is greater than the sum of these two. Is it fine? This is a very famous type of question. Normally, it comes in comparative exam also, but it's there in Adi Sharma also. Is it fine? Any questions? Any concerns? Please do ask. Okay. Can I move on to the next question? Let's take some more questions. Find the term independent of X in this expansion. Okay. What are the meaning of independent of X? Independent of X means it's a constant term, Gayatri. That means there's no X term related in that. That is the constant term. For example, if I say which term is independent of X in this expansion. So, when you expand it, okay, this term is independent of X. I'm just giving an example here. This term is independent of X. Independent of X means no X in that term. Right. Right. Front of when they say find the term independent of X, they do not mean the position of course only. They mean the value of the term also. This is a misinterpreted term by many students. When they see this, find the term independent, they mention the position. No. They're asking you the value of the term or the expression of the term. Very good front of. Nice. Aditya has a different answer to give. Okay, Aditya. I'm not saying right or wrong to anybody. Okay. All right. Okay, let's check. See here, when you're trying to find out the term independent of X, of course, you're not going to write all the 10 terms to see which term is independent of X. Right? It's like not a feasible way to solve the question in that way. So we'll say, okay, let t r plus one of the term be independent of X. Be independent of X. Now, we all know t r plus one of the term expansion is what n c r. This term to the power of n minus r. This term to the power of r. Okay, now consolidate all the constants together. So what are the constants present over here? Three by two to the power n minus r will be present minus one by three to the power r will be present and consolidate all x terms together. Now x terms will have a power of 18 minus two r minus r, which is 18 minus three r getting the point. Now you are claiming that this term, this guy is independent of X. Okay, if this guy is independent of X, independent of X, it means 18 minus three r should have been zero because then only your x will vanish off, isn't it? That means r has to be six. Right? If r is six means seventh term is independent of X. Okay, and that term will be nine c six three by two to the power of three minus one by three to the power of six. Okay, let's simplify this nine c six. This will give you if I'm not mistaken one by eight and will have three to the power three. Right? Yes or no? Nine c six is not much nine into eight into seven by six. Okay, and one by eight into nine nine. Sorry, nine and nine will go off eight and it will go off. So it's seven by 18. So the term independent of X over here is seven by 18. I will give you a light demonstration of this. I will use my GeoGibra tool. Sir, GeoGibra can be used here also. Yes, why not? So let's name this function as G of X. Somebody will take the custa of telling me the term three by two X square, three by two three by two X square minus minus. This is what I don't like. Minus minus one upon three X. Hold to the power off. What are the power guys? I want to go back. Okay, as you can see, okay, now since there is a nine power on the top and nine power in the denominator, they have actually taken the LCM, I believe.  Right? So try doing 3919104 upon 10077696. This will give you, as per our calculation, it should come out to be seven by 18. Okay, I can quickly check on my phone, does it give it like that? So 3919104 upon 10077696. Okay, this gives you 0.388 into 18 gives you seven. Yes, this is seven by 18. Clear? Any questions? Let's move on to the next question. Find the sum of all rational terms in this expansion. Find the sum of all rational terms in this expansion. Excellent. Very good. Pranav has cracked this. Well done. Anybody else who would like to contribute? No, that's not correct. Okay, let's try to look into this problem. So, see, if you write down any R plus one-th term, you will be writing 15CR, three to the power one-fifth of 15 minus R and two to the power one-third of R, isn't it? Correct? Now, if you want this term to be a rational term, that means the powers on three should be integers, isn't it? So, 15 minus R by five should be an integer. Okay, and so should be one by three R. They should also be an integer, right? So, these two should be integers, right? So, what values of R will make both of them integers, those will be the R values permissible for which I will get an R, for which I'll get the TR plus one-th term as a rational quantity. Is this point clear, first of all? Then only it makes sense to proceed further. So, does this make sense? Unless until these powers are ration, integers, I will not get a rational quantity out of it because if any one of them happens to be a non-integer, okay? Now, many times you will ask me, sir, what if three has a, you know, non-integeral power and two has a non-integeral power but it combines to give you such a power for which we get an integer, it cannot happen because three and two are prime numbers, they do not share anything in common. Are you getting my point? Okay, so now, what is R here? R has to be a number from zero to 15, right? So, let us check for which R from zero to 15 both will be integers. Let us start with this first because this looks simpler to me. So, this says if R by three has to be an integer and R has to be zero to 15, R can only take multiples of three, zero, three, six, nine, 12 and 15, right? Out of these, which will make even these as integers, let us check that also because both of them should be happy. It's not about one of them being happy, both of them should be happy. So, if you put a zero, yes, zero works for this, okay, so zero is qualified. If you put a three, you get 12 by five. 12 by five is not an integer, so three is ejected, okay? Even though it is making R by three is an integer, it is failing to make 15 minus R by five as an integer. Six, no, six will also fail, nine will also fail, 12 will also fail, 15, yes, 15 will work, okay? So, the final values of R that is going to give you a rational term is R equal to zero and R equal to 15. Now, R equal to zero is basically your first term and R equal to 15 is your last term, so only the first and the last terms were rational, rest all the terms were irrational, so all 16, out of 16, 14 of them were irrational and only two of them, that is the first and the last, they were rational. So, let's check what is the first term, so first term would be 15 C zero, three to the power one-fifth of 15 minus zero into two to the power zero. I think that gives you three to the power three, 27, 27, okay? And for 15, I will get 15 C 15, three to the power zero, two to the power five, that gives me 32 into one, so the sum will be 27 plus 32, answer is 59. Only two of you could get it, Aditya and Pranav, well done guys. No Charan, that's unfortunately not correct, but well tried, happy to see everybody trying. Go, Taitu. Yes, got it. Okay, let's take more questions. Find the term independent of X in the product of four plus X plus seven X squared times X minus three by X to the power of 11. Okay, so in this product, okay, what will be the term which will not have X in it, that is what will be the constant term that will come from the product of this?  Let's have two, two and a half minutes for this, very good. Three of you have responded so far. Last 30 seconds I can give you. Okay, five, four, three, two, one, go. Nine of you have voted so far, what happened to the others? Is it so difficult? Okay, let's check. Out of nine of you who voted, five of you have said option B. Okay, so B and C had close calls. I see which terms will give you constants, right? So if this four multiplies with a constant over here, correct, so I need a constant in this, this X multiplies with some term containing one by X over here and the seven X squared multiplies with some term containing one by X squared over here. Okay, so what am I looking for? I'm looking for constants somewhere. Okay, I'm looking for a term containing one by X, correct, and I'm looking for a term containing one by X square because only these terms when multiplied to four X and seven X square respectively can generate a constant, right? So I will not solve this as three independent questions. I will say let R plus one term be the one which individually satisfies these criteria. So I will take first the generic version. So R plus one term of this guy will be what? 11 CR X to the power of 11 minus R minus 3 by X to the power R, right? So that gives you 11 CR minus 3 to the power R X to the power 11 minus 2R, correct me if I'm wrong, correct?  Why not possible? Because R is going to give you 11 by 2. R cannot be a negative integer. R cannot be a fraction. R has to be a whole number, right? So there will be no constant term in the expansion of X minus 3 by X to the power 11, okay? For term containing one by X, for term containing one by X, you just have to equate this to minus one. That means 2R is equal to 12. So R is equal to 6 now. This is possible, okay? And that term would be and the term would be the seventh term of that expansion, okay? Which will be 11 C6, yeah, 11 C6 and this will be minus 3 to the power 6, okay? X to the power automatically minus one will come. You don't have to put the values, okay? So this will be your, this will be your term over here. So I would write it down over here. It's 11 C6 into 3 to the power 6, okay? Similarly, one by X square, this guy has to be minus 2. Again, this is not possible, okay? So the only term which is constant will be when your one multiplies with this. So the answer will be 11 C6 into 3 to the power 6. This is going to be your answer. Is it there in the option? Yes, option B is correct. Option B is correct. So Janta was correct. Most of you said B, however, by one more vote. Is it fine? Any questions here? Any questions? Any concerns? Please note, the constant term was not obtained and the power of 1 by X, the power of X to the power minus 2 was not obtained, right? Because our values corresponding to that was coming out to be non-whole numbers, non-whole numbers have to be rejected. Or you can say coefficient will be zero for those cases. Is that okay? Any question? Any concerns? Please do let me know. Done? Okay. Next question. Yeah. So if P to the power 4 plus Q cube is 2, P is positive, Q is positive, find the maximum value of the term independent of X in this expansion. Yes, any idea how to do this question? I think this word maximum is creating a lot of issues, right? Okay. But don't worry, there are some hints also given to you in that question. Hints. Let's see who is able to exploit that or use that. I'm launching the poll now. In case you're done with it, you can press on the poll. Good, good, good. Okay. At the count of five, we can stop the poll. Five, four, three, two, one, go. Please vote, please vote. Okay. Most of you have voted for B. That is 14 C6. Let's check. See, first of all, the process is very similar. Let TR plus 1 at term be independent of X. Okay. Be independent of X. Okay.  So if you want it to be independent of X, the consolidated power on, let's write it like this, and consolidated power on X will be 14 minus R by 12 minus R by 9. Correct. And this guy must be zero. This guy must be zero. Correct. Then only I'll get a term independent of X. That means 14 minus R by 12 should be equal to R by 9. Okay. So let's simplify this. 42 minus 3R is equal to 4R. R comes out to be 6. Correct. R comes out to be 6. So this term will be 14 C6, P to the power 8, Q to the power 6. Correct. Now the problem is I have to find the max value of this subject to the fact that P 4 X, Q, Q is 2 and P and Q are positive terms. Guys, this is a big indication to you that you have to use AMGM inequality. You have to use AMGM inequality. I'm getting the point. So can I say if P and Q both are positive, so will be P to the power 4 and Q to the power 3. Correct. So can I say P to the power 4 plus Q to the power 3 by 2 would be greater than equal to P to the power 4, Q to the power 3 under root. Correct. Yes or no. That means 2 by 2 is greater than equal to P to the power 4, Q to the power 3. That means P to the power 4, Q to the power 3 is less than equal to 1 and so is P to the power 8, Q to the power 6. That means max value of this fellow is going to be 1. So max value of this is going to be 1. That means max value of this entire thing is going to be 14 C6. Answer is option number B. Janta is absolutely correct. Now people have solved it. I'm sure you are not aware of this fact that you can use AMGM inequality. Is it fine Pranav? Do you want to copy from here, this page? Then question. Do let me know once you're done. Correct. Next concept that we are going to talk about.


Hey guys, it's Nate, AKA The Foot Accountant. Welcome back to the channel. EA has one mission right now in FIFA 23 for you and me, and that is they are trying to release the content that can possibly drain our clubs and make us spend coins doing SBCs like this one to get us to spend our coins and get rid of cards in our club, basically mud our clubs before Team of the Year, which is a huge pack-based promo. Now, we talk about this a lot, but this is something that EA do and we have to just be smart. This week, we even saw it start last week with a lot of these SBCs coming out already. I think it's gonna take place again this week and maybe even to a more higher degree, right? More SBCs, more upgrade packs this coming week. We have to talk about that today because those upgrade packs could continue even today on Sunday. So I wanna take a look at that and how it's gonna impact the fodder side of the market because some of your 83s, 84s, and especially 85s once again are going up in price earlier than expected because that pack, the 100 players upgrade, dropped yesterday. Now, we also talked a lot yesterday about the market crash, right? You guys might be like, Nate, where's this market crash? I was a little surprised today to see that there was not much crashing on the market but then again, I thought about it and really, it just comes down to nobody has a reason to panic yet. I mean, where's there really some imminent fear about Team of the Year? The market's done pretty well over the past week and a half. Since prices really started to drop about 10 days ago, we really haven't seen too much of a crash since then. It's still gonna go down. I really do believe that. It's more of a when, not an if type of scenario. But until we see some sort of catalyst to make people really panic sell cards in this market, it might just be, again, the stagnant market that we've been seeing for the past couple of days. But I do wanna talk about that Golden Bopay because it's not like cards aren't dropping. They're just not dropping really fast in price and there's not a lot of panic because that Bopay is down 40,000 coins and that's a card we have to keep a close eye on with Team of the Year and his card probably being out of packs this coming Friday. So there's a lot to talk about today and of course, what to expect today. It's content on Sunday at FIFA 23. We're gonna cover all that and more. If you're excited for it, hit the thumbs up. Subscribe if you're new. Let's start by talking about the bit of a surprise yesterday and the beginning just like last week of how EA is gonna try to drain our clubs and drain our coins by having us tempted to do these upgrade packs. Now, if you don't have a lot of coins or let's say your club is in the mud, especially after last week's maybe 83 times 10 or maybe you've been crafting stuff. Yes, there is the opportunity to get some extra fodder right now, right? We've got the best week in league rewards we've ever had. We've got these cups objectives in here and plenty of other objectives like the Centurion of matches, that upgrade SPC Chase, the league SPC, extra objective packs that you get in here too. It's actually not that hard to get fodder right now. But again, EA's dropping a lot of SPCs that people are putting their fodder into and they're doing that to try to get that stuff out of your club before the big pack promo of team of the year. They do this all the time. We've talked about it before. It's something very common because they want you to feel like you need to open packs this weekend and in the next couple of weeks as team of the year is coming. And again, like I said, this 100 players pack, which I really was not expecting this yesterday. I mean, it makes sense, right? The hindsight is always 20-20. When you look back, you're like, oh yeah, that did drop a week ago on Saturday. But honestly, I thought that was a one-off SPC. Here it is again, right? It's available for six days, 83, 84, 85 squads, one team of the week required. And of course, just like last week, this made SPC fodder bounce, especially on the low tier. Take a look at 85s from about seven and a half or basically 8,000 coins, what they were earlier in the day on Saturday. They spiked to almost 10,000 coins. Now they are back down a little bit, especially your 84s and your 83s have risen as well, right? 83s all at 1.3K, 84s at 4,000 coins. And you look inside some of the prices here on these 85s, some of these guys were actually 10,000 coins, right? 9.9 to 10K is what a lot of these guys were after they were in the low to mid eighths, late Friday night, early Saturday morning. Now, we talked in yesterday's video about potentially investing in these cards in the Saturday to Sunday timeframe with the way that it looks now, a lot of these have risen up on the low tier, right? The 83s to the 85s have gone up, honestly, a bit too much for me to want to invest. Now, there are more SPCs that are incoming as well and today's content, which we'll talk about in a little bit, will really impact this side of the fodder market because if we don't get anything today, if today is a really quiet L content day, then those fodder prices, just like they did last Sunday, will probably drop off some more and will create an absolutely incredible opportunity for another short-term investment and quick flip, right? But the other side of this story is, what if EA go out and today release another icon pack? Now, this would not follow the same schedule that they did last week. Last week, we had the 88 plus mid or World Cup icon on Monday and last week on Sunday, we saw fodder prices drop back down. Remember last week, 85s like Serge Nabri went from about 8,000 coins Sunday morning because of that 80 or that 100 players pack, the same one that we had yesterday. This pack was dropped last week. These 85 rated cards went from the high 8K range down into the low to mid 7Ks and then they boomed back up on Monday with the 88 plus icon that was released when that pack was out. So that's the same scenario that we're maybe looking at this week, but again, like I said, if EA go out and drop a icon upgrade SPC or some sort of other SPC today that would be a gamble pack, right? We're talking hero player picks, icon player picks, anything along those lines. If they'd go back to back days with SPCs like that, if they're in full out, try to drain our clubs mode, that would make fodder go even higher today. So on the types of the market right now that I think are maybe still investable, I think the 86s are low enough. They're not that low, but they're low enough to maybe get a couple on bid and club stock. If you think EA is gonna drop an icon pack today, that's where I would go. If you think EA are gonna follow the same schedule as they did last week, then I would wait to buy these cards until late Sunday night tonight as people are getting their weekend rewards and putting those cards up on the market. People are finishing their games, getting tradable packs, listing the stuff up. It'd be a great time to stock then if we don't get any content today, right? It kind of seems like a 50-50 to me, but we'll see, right? EA are creatures of habit and they might be following the same schedule as last week. So I'm probably not gonna buy or invest in any fodder unless I see it go back down. I remember perfectly last week buying 85s in the low 7K range. I'm not saying they're gonna go that low today. If we don't get any content today, hopefully 85s can go back to like 7,500 coins, right? Or maybe even just below 8,000 coins will be our buy price for these before they start to go up into the week. And especially you think about high-rated fodder, right? Look at 91 de Bruins, 61,000 coins. Yeah, these guys were 67K last week on Monday. I mean, it would take a big icon SBC or a more expensive hero player pick type of option to see these cards go up. Now, if I had to guess what type of upgrade SBC we're getting next between today on Sunday or tomorrow on Monday, maybe Tuesday in the next couple of days, I do feel like it will be either another icon upgrade or a hero upgrade. Now, we have not had a hero player pick yet. If I had to choose the opportunity of an SBC to drop that would maybe create some panic and also just be like a hype upgrade pack SBC. I think a world cup hero player pick, even a player pick that's between regular heroes and world cup heroes. I don't think we've had a world cup or just a hero in general player pick yet. That would be something that would be really hyped that a lot of people would want to do and try to gamble on, right? So we had an icon player pick that was dropped two weeks ago, of course, that could also be the base icon player pick which would be a bit of an L in my opinion since base icons are just kind of behind the curve at this point. But I feel like that would be the other SBC that could potentially be dropping either today on Sunday, tomorrow on Monday, or maybe Tuesday at the latest. But again, one of those types of upgrade SBCs is gonna come so keep a close eye on the fodder market. And then, even as we look later into the week, what do we have last week, right? We had the 83 times 10 SBC. I think we will either see the return of that or an 85 times three. We also had an 82 plus upgrade that was added to the code. Maybe just like last week on Sunday, they'll drop another upgrade pack out in the game. I think since this 80 plus upgrade is kind of tied to the objective, maybe that's a little bit less likely today, but watch out for another sort of upgrade pack like that. Even if it's like a simple 83 plus double that is repeatable a couple of times, those sorts of packs right now, everybody is gonna go and do and save for a team of the year with that promo coming so soon, just honestly, five, six days from now, right? So just be careful with that stuff today and watch out for EA trying to drain coins from your club. Just be careful is what I would say. Some of those packs look really fun to do and really interesting and maybe even craftable, but especially when you get into team of the year, some of the stuff you're gonna look back and be like, man, I wish I didn't do that or I'm glad I didn't do that. So just be very careful. That's the main point of the video today is I'm just trying to warn you guys to be careful. And a lot of you guys know this and stuff, but this stuff right here, just be very careful with these gamble packs as we head into team of the year. Now, let's talk about the market because of course prices yesterday, they went down a little bit, but they didn't drop that much. Of course, I could look at this insurance team and say like, hey guys, look, prices went down. Now, this is always expected, right? Zlatan from 4.2 million coins down to three mil flat dropped literally a million coins yesterday. He was just three million coins and kind of bounced back up to about 3.3, but he didn't sell and he's kind of back down. Now, I was expecting Zlatan to be three mil tomorrow, he dropped a little bit more and a little bit faster than I honestly thought he would, but that's a big price drop off for him. Royce, 250, I sold him last night at 3.80 right after I got off the video. Verate's down over 100,000 coins. A lot of your mid to high tier players in this promo's, Salah as well, 900K, they have dropped off a lot. I think they'll drop down a little bit more today. We do have squad battle rewards today that maybe some people are playing squad battles to get some of those objectives done like for Berardi and stuff. There will be a few tradeable packs hitting the market for those today. I do expect to see these cars dropping down a bit more, but then I also think they could rebound a bit into tomorrow and Monday. So watch for low prices today. Here, we will still get landing rounds. We will still get packs in the store. That is a given on the weekend, but the rest of the market really didn't drop off too much apart from this guy. Mbappe, 770 like all day yesterday and is now kind of down at like 720,000 coins. This is not low enough in my opinion to be buying right now, but as I've said, we're keeping a close watch on this card specifically because he will be going out of packs for a team of the year as he is expected to get a team of the year that will be in packs. So I still don't think this is low enough. In my opinion, I think there's gonna be further panic later on this week, but keep a close eye on this card as that was one card that actually had a bit of a drop off yesterday. Some cards, and it's really the midfielders. I know we looked at Conte last night. Conte was way too low last night. He was a really good buy at 800K. He went back up, but a lot of your Premier League midfielders and just midfielders in general are down. I think it's a combination of the Fred, right? I overlooked that a bit in yesterday's video. Fred's price and Fred's card definitely impacting Conte a little bit too, but all midfielders in the game, specifically Prem midfielders, seemed to be down at the moment, right? We looked at the Conte yesterday, the Yaya, and then that card, Bruno Gamerich that we just looked at right there. All these Prem midfielders are down because of kind of the trifecta of midfielders that were dropped, even Virati, that trifecta of midfielders in the Centurions promo this weekend. So I think that's why some of your mid to high tier and even some of your lower tier mids are down because people want to try those cards. But in total, the rest of the market, like I said in the intro of the video, it's gonna take another sort of catalyst type situation or something else has to create panic on this game because right now just the fact that Team of the Year is coming in five days has not created enough panic on this market to make prices drop a ton, right? Really not a true market crash type situation like we have seen in previous years. Now granted, you cannot look at the market right now and say that the market is up in general because look at Mbappe, right? Literally, last weekend he was 840,000 coins and today he stands at 730. So the market is dropping but it's just not in the fashion that we usually see for a Team of the Year market crash. So you take a look at other cards in the market as well. It's not the exact same story but card prices have just slowly dropped instead of it being like a sharp drop off, right? That last weekend timeframe, Yaya was two million coins last week on the seventh and is now down to 1.7. But that's just kind of the market that we're in. We need something else to happen whether it's the end of the voting which the voting I think ends tomorrow and Monday like really late. So it's like early Tuesday when the voting ends if it's leaks that start getting stuff out there if it's a loading screen maybe on Tuesday it's really just gonna be kind of a waiting period until we see the switch flip and we see some cards dropping price. I still do believe that there will be price drops and I know and I say this and I have to mention in every video because a lot of people get in the comments they say like, you're just making this video to create a market crash. No, honestly, I mean we talked about selling cards from our teams literally a week and a half ago around even before the market started to drop off on those graphs that we were just looking at. So a lot of you guys have known that if we have tradable cards in the club it's very more risky to be holding on to them. We've been saying that literally for like a week and a half, two weeks now. So we are obviously knowing the market is headed down. We are just kind of waiting for that kind of rock bottom or timing the low, right? It's really tough to get the absolute rock bottom price on a lot of these cards. I look at Comma Vinga right now 415K for him. That's really though for a Comma Vinga card for where he has been and just I didn't really watch the market personally. I didn't trade yesterday. I try to take Saturdays as a true day off and not try to trade too much. But I just noticed yesterday that there wasn't a ton of crazy price drop offs like we originally were expecting. So we continue to wait. We continue to see what it takes to make the market really drop off because there's still gameplay demand as well. And we've seen that recently just a little bit more on this market really fast. I mean, yes, we're waiting for some sort of catalyst to cause panic, but after the winter wildcards promo the new player base of players that have been on this FIFA game since Christmas. There's a lot of new players I really do believe. I wish I could like ask EA or see the stats or something of how many new players have gotten FIFA or been playing the game since Christmas because that's when the market really started to go up and that's when the demand kind of changed. And there was a lot more demand as people get coins, buy cards for teams and stuff like that. So I think it's all those factors and just the fact that the team of the year panic is not set in yet for most people. That's why we haven't seen a market crash. So we'll keep watching the market and seeing where it goes here over the next day or two. Now, what I wanna talk about here at the end of the video is just other content that we maybe see today. There's one sole leaked SBC and it's barely even worth mentioning. Alex Fernandez has added to come as an SBC during Foot Centurions. These are his expected stats. I think it kind of reminds me of the Raul Garcia except don't slay the Raul Garcia. That man is a beast. If you've watched the streams, you know. So if that comes out, hopefully it's cheap because that should be pretty cheap. And the last thing I wanna talk about here is I wanna look up a tweet that I saw yesterday and talk to you guys about this. Deena Talley. And this idea of like Foot Hero Team of the Years has been going around on Twitter, on Instagram because I think a lot of people saw Deena Talley post this. I think chat, guys, call me chat again, man. You're not the YouTube chat. You're the YouTube comments, right? Anyways, I really feel like this is just a concept. And I think if you actually look below this tweet, this was just a concept post by somebody who was like making a cool image and saying like, wow, wouldn't it be sick if we got Team of the Year Heroes? I don't actually think that we are actually gonna get the Team of the Year Heroes, right? I saw a couple people tweeting about it. Heroes and icons both are getting Team of the Year cards. I don't, guys, be careful with the stuff that you see on the Twitter machine because this guy, this is also the same guy that said we were gonna get wins if your opponent quit on a draw in footchamps. So that obviously does not happen, but I saw Deena Talley and I saw a couple people tweeting about this Deena Talley post yesterday. I think Deena Talley just literally posts concept graphics that are maybe tweeted at him or these tagged him because at the top of this picture right here, it literally looks like two people's profile pictures that made a concept and then whipped this card up and is like, oh yeah, this would be sick, right? Well, it would be sick. And a Team of the Year Hero would be insane until we see some really concrete evidence around that. Just take that with a big time pinch of salt because it feels to me like it might not be legit, but that's just something I saw on Twitter yesterday that I wanted to bring to your guys' attention. I'm sure that you guys have been seeing that as well and been talking about it because that would be pretty sick, right? The concept is amazing, but it doesn't look super real to me. Now, we did get the mini release yesterday. We had Rafa, Coleman and Paloise and I think these guys are worth covering just for a quick second because, I mean, this Paloise card is not bad and let me talk about the showdown as well. Paloise is about what, 20,000 coins? I mean, I'm not gonna compare him to the Winter Wild Card center back that we had last year that took the game by storm, a causa, but I think this guy is actually a really solid card. Two-two is not good, but left foot is kind of nice, medium high, his pace, you definitely need a shadow, 97 strength, 90 composure, 95 aggression, decent card for 20,000 coins. That's one that you don't have to worry about losing too many coins about because he's already so cheap, so I'm a fan of that. The Rafa super sub card is exactly that, a nice super sub with 94 pace, 90 dribbling, so a nice card that if you wanna throw it in for some extra pace is a super sub late in the game. That's a cheap card there and then our other mini release card, I don't even remember who it was, Coleman mid, but if you're an Everton fan, then boom, you have a nice right back, you can try out for your team and then I wanna talk about these showdowns really fast at the end of the video.  I think I still stand by that, but that was maybe a slight overreaction. I think it's just a W. The reason why I think that is, these are both, especially Meketarian, a very meta FIFA player, right? You guys maybe remember for birthday Meketarian and previous special cards from this guy, he's always very meta for the five-throw weak foot, finesse shot, outside foot trade. Does he definitely need a plus two to be usable? Absolutely, if Inter win this game, we're talking 80 physical, 86 passing, 86 shooting, 89 pace, 90 dribble. That looks like a really, really good card in a certain tacky mid. So if they go ahead and win this game, 60K for this card, not that bad. And then Origi, this is a dub for me. It's just a straight up fun card for like 30, 40,000 coins, one squad, five-star skills, 99 heading accuracy. Like this to me is just cool. So I'm a fan of this, GGs to EA. I love it when they make players that are not that usable, incredibly usable and like surprisingly usable with the five-star skills. And there's so much Syria hype right now in the game. I love it. So GGs to EA Sports for that showdown. Game is on Wednesday. And I honestly just saw this as well. The EA Sports Super Cup. Of course, EA's got to release a showdown for a cup that they sponsor. Of course, there's a tie there. So anyways, that is going to be the video for today. Hopefully you guys enjoy Sunday. Good luck with your Champs games, of course. If you guys enjoy the video today again, leave a like, comment down below if you have any questions and subscribe if you're new. It's been Nathan Foot Account. I'll catch you guys later. Peace out.