Hello all welcome to Centrum Academy YouTube live session on binomial theorem. This is session one of this chapter so those students have just Join us on YouTube live I would request you to type your name in the chat box so that I know who all are attending this session Hi guys, I am Sugosh Ashutosh Bharat Hope you guys are well prepared for your tomorrow's physics exams So let's give a minute or so to others. It's still one minute left to 11 o'clock. No, it's exactly 11 o'clock now. Good morning, Sheram Hello, Niranjan, Andrew, Omkar, Samir, Sushant All right, so let's begin with this session So binomial theorem as you have already studied permutation combination It's basically a follow-up chapter which will use a lot of things from permutation combination Especially your binomial coefficients, you know NCR everything would be governed by your understanding of permutation and combination chapter Okay, now let's start this chapter. First of all, let us understand. What is a binomial expression? So any algebraic expression which contains Two dissimilar terms right so any algebraic expression Okay, which contains two dissimilar terms Okay having two dissimilar terms is called a Binomial expression and these terms may or may not be raised to a power So a generic structure of a binomial expression would look like x plus a to the power of N Where N is the exponent here But guys remember that when I say x plus a it is just to show you that these are two dissimilar terms Okay, and both can be you know numbers both can be Different variables one can be a variable one can be a constant and so on and so forth So we will be talking about Binomial expressions of two types one where you have your N as whole numbers and Another would be covering any real number. So you can say non whole numbers You know, we'll be covering those powers where N is not a whole number This will do later in this chapter towards the end But our prime focus would be on when N is a whole number Okay So I hope all of you have joined in the sessions away Gaurav. Hello, okay so we'll start with That binomial expansion where we have N as whole numbers Okay, or you can say we will talk about positive integral index, right? This is called the index of the binomial theorem Binomial expression. So this is called the index So we'll be talking about where our index is a whole number Now, let us start with our very very basic understanding where we have come across binomial expressions in our past For example, I'll start with X plus a to the power of zero Okay, X plus a to the power of zero. We all know that this answer is actually one Okay When we talk about X plus a to the power of one the answer is X plus a correct When we have X plus a square the answer is X square plus two a X Plus a square When you have X plus a cube basically these are identities which we have already studied in our junior classes That is X cube three a square X plus three a X square Plus a cube. I'm sorry You can just Write three X square a Plus three X a square plus a cube. Is that clear so far? Okay, now if you see there is a pattern in which you have actually written these expressions Okay, the pattern is if you look at if you look at the these terms X and a you would see that these terms are basically You know having a maximum power given by this index for example, if you look at this term X and a both have the power of one and one If you look at This second term you would see that this index is two So X has index of two here and in this term you can you can actually write it like this You can write it like X to the power one a to the power one right Here will give you again to Okay, and again a has a power of two Now the trend is Whenever you're writing this expression, you know, and I'll talk about the numbers little later on I will talk about these coefficients little later on But as per the variables, I can see a pattern that it starts with the maximum power So X takes the maximum power on itself Okay, and this power slowly decreases so from three it became two and Slowly a has made an appearance Okay, so a has come now again the power to has further decreased to one and a has further gained in one more power and Finally the last term you can say X has become zero and a has taken the entire power on itself Okay, so as far as the X and a powers are concerned We can make the following assumptions or we can see the following patterns. So if I talk about Let's say Let me go out to the next page If I talk about X plus a to the power of four Okay, I don't know what is the number here, but it'll start with X to the power four Correct. Again, I don't know what is the number here? But X will decrease by one a will make its appearance now. So a a to the power one will come Again, there will be some number here X square a square will be there Again some number would be there and X to the power one a cube would be there and this trend will continue till X has completely lost all the power and a has gained all the power That means you can assume it to be going in this fashion correct now This was the pattern that we observe with respect to the powers of X and a Yes, you can say that but that a is taking away the powers of X Okay, so if I say the next expression, let's say X plus a to the power five So it would start with some number X to the power five a to the power zero Okay, then some number X to the power four a to the power one Then some number X cube a square Then some number X square a cube Okay, then again some number X to the power one a to the power four again some number X to the power zero a to the power five So at least we know what is the pattern with respect to the power on these two dissimilar terms of this binomial expression Now what about these numbers which I have not mentioned in these two expansions? Okay, so what about No, what about these numbers or coefficients with with these variables? So example here is one one This is also one two one This is one three three one like that So if you see this observation is basically Following a triangle It is following a triangle Which we call call the Pascal's triangle Okay We have already talked about this in Pascal's identity in our permutation combination chapter now officially I'm dealing with this Pascal's triangle This Pascal triangles tells you Tells the coefficients coefficients of any General term Okay in the binomial expansion Okay, how it tells and how is this Pascal triangle formed? Pascal triangle is basically A triangle which starts with one And ends with one except for the zero-width row. This is called the zero-width Row Okay, so and this will be zero-width column So This would be called the zero-width column And this will be called the first column Okay, and this is your first row So basically Pascal's triangle has several Rows in it. This is zero-width row. This is first row The second row is basically obtained see it starts with one ends with one no doubt about it So every row will start with every row Will start and end with A one So in the in the zero-width row, there's only one number. So it is starting with one and one is the only thing First row is basically starting with one ending with one Now this is your second row Let me write it on this side. This is your second row Second row is also starting with one ending with one and in middle basically the number would be The sum of these two numbers. So if you add them, this will be your term in the middle now you can see a comparison here When you wrote x plus a to the power zero your answer was one. It was actually this number When you wrote x plus a to the power of one, it is basically x plus a And x plus a both will have One one in front of them. So these are these two numbers which you have in front of x and a Okay When you wrote x plus a square You got x square plus two x a plus a square Now the number in front of these expressions that is one two and one. These are these three numbers one two and one right Now continuing with the third row of pascal's triangle again, it'll start with one end with one And the number coming in between these two would be the sum of these two which is three Again the number coming here would be three So that's how we get the coefficients of the expansion x plus a cube x plus a cube we all know is x cube plus three x square a Plus three x a square plus a cube right So these coefficients that is one three three one This is what we are getting over here one three three one Are you getting it? So now With all this information at our hand We are ready to fill in the blanks in this particular expansion. So what would be the number in these round? Boxes, let's figure it out So in x plus a to the power four I will have to go and write down the fourth fourth row elements Or fourth row entries of the pascal triangle. So again it starts with one ends with one And what will be the number here? It would be one plus three, which is four The number here will be three plus three, which is six The number here will be three plus one, which is four Now what we have to do is we have to take these numbers and start filling here one four six four one So now this is the expansion that you would get for x plus a to the power of five Right correct Bharat. You are correct in a similar way guys Uh, take a clue from pascal triangle and let us complete x plus a to the power five expansion Which is right now incomplete over here So it's very simple fifth row would start with Fifth row would start with a one This two would be Five so I'll write a one. I'll write a five This two will be 10 as Bharat has already pointed out So there will be a 10 over here Then six plus four will be again a 10 So the 10 will come over here again Now this will be five So a five will come over here And finally it will end with a one. So this will be one Okay Yes, see about the higher powers will will generalize this rule in some time right now. We are just You know taking a basic clue from our Past observations of these expansions So when it becomes a higher power like 20 and all So shushant will be soon giving you a general theorem on that Okay We have will yeah, we'll use the generalization right now. We are not generalizing right now. We are If we are giving these answers from our basic observation of the patterns Okay So if I have to write Somebody can tell me how would I write X plus a to the power of seven Just tell me The coefficients I would write down the X and the a terms without the coefficients. So there would be some term And start with x to the power seven a to the power zero Again, there would be some term x to the power six a to the power one c A is losing the power One at a time and sorry x is losing the power one at a time and a is gaining the power one at a time So it'll start with seven zero. That means x will take the entire power of seven A will be on zero power Then x will share one power with a so it'll become six And a will become power of one Next term will be x to the power five a square Again, we'll have x to the power four a cube Next term will be some x to the power three a to the power four Then I'll have x to the power two a to the power five Again, next term will be x to the power one a to the power six Again, finally last term would be x to the power zero a to the power seven So we'll stop here When the entire power of x has been transformed to a So see it starts with seven and zero and now it has become zero and seven Now what about these boxes which I have left empty? Right, we'll follow Pascal's triangle in the Pascal's triangle. I'll quickly Continue from where I left off Okay, so this was the fifth row which I had written Sixth row would be one six 15 20 15 six one Seventh row would be one seven 21 35 35 21 71 Is there any person who is not able to understand how these powers are how these Terms in the Pascal triangle are generated Please feel free to ask me on the chat box. Yeah, we'll talk about it. Adwet will do a lot of observation on it We have just started the chapter as of now So we'll make a lot of observation on this chapter. Don't worry about it So basically one seven 21 35 35 21 Seven and one will come in these places So now we are all set with respect to How the powers change in these two terms and what are the coefficients that You know come in the front of these terms. So we are fine with that Now I'm going to generalize this I'm going to actually generalize this Okay So when we talk about any such terms x plus a to the power of n Where My n is a whole number Okay Now, what is the meaning of x plus a to the power of n? The meaning of x plus a to the power of n is We are multiplying x plus a x plus a x plus a Basically we are doing it n number of times Right Yes, for non-whole number powers it will go to infinity. We are going to talk about it towards the end of this chapter Okay As of now when we say x plus a to the power of n Is it clear that it means the multiplication of x plus a x plus a x plus a n times Okay Now guys try to recall The fundamental principle of counting plus means or Right Multiplication means and Okay Now keep these two things in mind plus signifies or Multiplication signifies and correct Now when you're trying to generate a term from this multiplication What does fundamental principle of counting say it says you have to pick Either x or a from these two terms So one of them will be chosen Right either x or a will be chosen This means and and one of the terms from x and a from here would be chosen Again this multiplication means and One of the terms from these two would be chosen And so on so forth till you reach the nth expression of x plus a So what I'm trying to say is that From each of these terms You will either choose You'll either choose x or a And this and means From every such term which has been shown by a square bracket You will have to choose at least one of x or one of a Okay now see I will use permutation combination over here If let's say from the first term I choose x From the first term I choose x second term Also, I choose x and third term also I choose x and I keep on choosing x from all the terms Okay What will what will be my answer be if I choose x from all the terms my answer would be x to the power of n Yes or no Okay So this is how we generally get the very first term which is actually x to the power seven in this case This is x to the power n only right Now if I choose x from any n minus one such terms And a from the remaining term That means let's say I choose x from here x from here x from here like that From all the terms Just before let's say this last this is your n minus one-th term Okay, so I'm choosing x from all the n minus one expressions and I'm choosing a from this term Then what term would it result into you will say it would result into a term like this x to the power n minus one into a correct But now this a which you have chosen from the last term you could have chosen that a from any one of these n terms right For example, I could have chosen I could have chosen a from the first term And x x x from all the remaining terms Then still I would have got the same expression correct So how many ways are there in which I can Pick up a term from where I am picking up a only so you would say there are nc one ways because we know permutation So if I have n such terms and I have to pick up One of the terms which will contribute an a Then that would be done in nc one way correct in a similar way If let's say I pick up x x x from all the terms And I pick a a from the last two terms Then I would end up getting an expression like this correct And this term itself can be generated By picking up any two of these n terms from where you want to pick up an a so basically n minus two Terms are contributing x And any two terms are contributing a Okay, so how many ways can I pick up Two objects from n objects. So you'll say nc two So in nc two ways, I can generate Such an expression Is that clear? And if you continue with this procedure You can generate x to the power n minus three a cube In nc three ways And this continues Till you reach the last expression which is ncn x to the power n minus n a to the power n that means There's only one way in which you can Multiply all the a's when you pick up a from here a from here And so on till a from here a from here Then only you will be generating this term Is this general way of expanding this clear to everybody? Please type clr if it is clear Just to generalize this I can write nc zero over here a to the power zero over here Is everybody clear about how did I get this general expansion? Great So I have used my permutation combination. So p and c is used over here Okay Now when you get this expansion Then you can start relating the Pascal's triangle with these numbers that you see over here nc zero nc one nc two nc three so I'll quickly link Pascal's triangle with this However, you can also see the relation in the powers of x and a so x started with n a started with zero Then x lost one power a gained one power Again x lost one more power a gained one more power and so on So whatever I was writing in my you know previous expansions Surely by my observation is actually coming true So what I will do next is I will quickly relate Pascal's triangle and the binomial coefficients So when we write x plus a to the power n Okay, we basically Write something like this. We just now learned nc zero x to the power n a to the power zero nc one x to the power n minus one a to the power one nc two x to the power n minus two a square and this continues till I reach nc n x to the power n minus n a to the power n Right or you can write it becomes x to the power zero a to the power n Okay Now Pascal's triangle when you wrote a Pascal's triangle, let me tell you That when you were writing Pascal's triangle This term is actually zero c zero Right So basically you write the column number Sorry row number c column position of that number. So this comes in This comes in zero with row And zero with column so it becomes zero c zero This number that you see This number that you see it's actually one c zero I'm sorry zero c zero zero c zero because Oh, I'm sorry. I'm sorry it comes in One c zero because one is its row number And zero is its column number Okay, similarly this would be one c one Okay, we all know one c zero is one one c one is one Again, this one is what this one is two c zero This is two c one This is two c two Similarly, this is three c zero This is three c one This is three c two. This is three c three Similarly, this is four c zero Four c one Four c two Four c three Four c four and it keeps on going like that So basically these numbers that we write in the expansion. Let's say I'm writing x plus a to the power of four These numbers one four six four one is basically nothing but four c zero x to the power four a to the power zero Then four c one x to the power three a to the power one Then four c two x to the power two a to the power two And so on till we reach Four c four x to the power zero a to the power four So that's why when we were talking about ncr in permutation combination, we used to call it binomial coefficients And these binomial coefficients actually make your pascal's triangle. They make pascals Pascal's triangle Okay, and also I would like to reiterate pascal's identity So while we are constructing this number we constructed six by adding these two, right? So basically three c one plus three c two adds up to give you four c two Remember we did a pascal's identity That ncr minus one plus ncr is equal to n plus one cr so basically every Our element that we are filling in the pascal's triangle Is being derived from this identity which we call as the pascal's identity So now are we all able to understand how is Pascal's triangle generated and how does it contribute towards the coefficient of these binomial expression correct And this generalization is actually written as a theorem which we will know as the binomial theorem So this is just one part of the binomial theorem Where we have taken such indices which are whole numbers Is this clear, please type clr if it is clear And now we'll start with the analysis of binomial expression Any question with respect to this? Great So before I start analyzing let's take a quick question Question is Expand a minus three by b to the power of five Now please note before you start expanding it When we talked about x plus a When we talked about x plus a to the power of n This x is actually the first term in the expansion and a is the second term in the expansion Right treat it like this. So think like this Okay, so this two a is your first term in the expansion and let me tell you students Minus three b is yours minus three by b is your second term So always accompany the sign along with the term don't miss out the sign over here Are you getting this so treat treat two a minus three by b to the power five as Two a plus Minus three by b to the power of five that means This sign is taken along with three by b. So it is clubbed. So we club with the sign along with the number Now give me the expansion quickly. I'm giving you one minute Yeah, if x or a had a constant coefficient, it will be a part of it. For example, two a two a is taken together So this is taken together as your first term This is taken together as your second term So let the coefficient be with the term itself Okay, so in this case your first term would be what your first term would be five c zero Two a to the power of five Minus three b to the power of zero Now normally to explain you I'm writing this zero. Otherwise I will never write this zero Okay, just to maintain consistency and follow the formula I'm writing this zero Next term would be five c one Two a to the power of decrease the power of five by one and increase the power of minus three by b By one Okay Next term would be five c two Two a to the power of three Minus three by b to the power of two Okay, again next term would be five c three two a to the power of two minus three by b to the power of three again Five c four two a to the power of one minus three b to the power of four And finally five c five two a to the power of zero I would not have written this but just to explain you I'm writing to the power of zero And then you'll have minus three b to the power of five Okay, if you open this up you would realize that the first term is 32 a to the power five Okay, second term would be minus 240 a to the power four by b Okay Why minus has come because you can see there's a minus three by b and it is raised to a power of one Which is an odd number so a minus comes out in front Next term would be plus 720 a to the power three b cube Next again will be minus 1080 a square By b cube This will be I think b square. I'm sorry Yeah Next term would be Plus 810 a by b to the power of four And finally minus 243 by b to the power three This is your final answer So I hope the basic expansion is clear in your mind One thing I wanted to clear from this example is whatever is the sign coming between the first and the second term The sign will be taken along with the second term Okay Yeah, I'm sorry b to the power of five Does it make sense now? Okay, now I'll give you a generic a very targeted question. All of you please solve this Hope there is no problem with this. Okay. My question is find coefficient of x to the power four In x plus Under root x square minus one to the power of six Plus x minus Under root x square minus one to the power of six So hope everybody understands the meaning of coefficient coefficient means The term or number which accompanies this particular x to the power four term So for example, if it comes out to be k x four while you are expanding it, okay Let's say I'm assuming this is my answer then k will be called the coefficient of x to the power four So the number which accompanies or the term which accompanies that x to the power four will be called its coefficient So all of you please work this out I'm giving you around one minute Just type in the answer in the chat box So it will be a number for sure What is that number just type in in the chat box the no 30 is not the right answer Shavan Ashutosh, please check your calculation 30 is not the right answer No, even 15 is not the right answer No 42 also is not correct Okay, all of you are I don't know why you are making a mistake in this. Let's just solve this see, uh You know this expression is huge So I would call it a for the time being because it is occurring at both the places So if you look at the structure of the problem the structure of the problem says You want to write down the expansion of these two terms, okay Now little while ago we made an observation that The terms of or the expansion of such terms with a minus sign in between will lead to alternate negative positive terms So this will have an alternate Alternate positive negative terms right So if I write this expansion, let me write this expansion first x plus a to the power six It will be x to the power six plus Sixy one, which is I'm writing the direct result Six x to the power five a Then sixty two, which is 15 x to the power of four a square Next term would be 20 x to the power three a to the power three Again, it would be 15 x square a four Six x a to the power five and then finally a to the power six correct Even without expanding this I can say for sure it will be x to the power six minus six x to the power five a plus 15 x to the power four a square minus 20 x to the power three a cube Again plus 15 x to the part two a to the power four Again minus six x a to the power five plus a to the power six When you add these two that's what I've been expected to do in this I have to add these two Okay, so what do I get I get x plus a to the power six plus x minus a to the power six as twice of x to the power six Correct 15 x to the power four a square now a square. Who is what this is my a so a square will be x square minus one Right, so these two terms get cancelled These two gets cancelled these two gets cancelled. So next term would be 15 x square a to the power four a to the power four is like square of this And finally we are left with a to the power six a to the power six would be x square minus one to the power of three So this is what we get finally Now which terms will contribute x to the power four. Let's figure it out Can I say When this 15 x to the power four multiplies with this minus one I will get a 15 x to the power four In fact, I'll get a minus 15 x to the power four Correct In a similar fashion if you see this term That is 15 x square x to the power four minus two x square plus one This into this will generate an x to the power four, which is nothing but minus 30 x to the power four Okay Finally from this term If you expand it it will become x to the power six minus three x to the power four And whatever is the rest term I don't care because I've already got x to the power four So which will be minus three x to the power four And this a factor of two is waiting outside. So I'll get so many terms And a factor of two is waiting outside So finally my answer would be two times minus 48 x to the power four So my answer is going to become minus 96 Yes, uh gotta be wanted to ask something So guys, this is the uh, you know answer for this question any doubt regarding this Bharat, Gaurav, Ashutosh, Niranjan Please don't stop anybody from asking any question, right? You have whole right to ask me any question you want Okay, fine. Yeah Before I move on one more question so that you know, we can go and analyze the binomial theorem, which we just now discuss So let's take one more question. Okay, a very interesting one Probably this can come in your school exams as well So mark it as start and come in your school exam as well The question is Which is greater? Which is greater? 101 to the power 50 or 100 to the power 50 plus 99 to the power 50 Which is greater of these two is this greater or these two are these two come minor greater Again, think about this for some time. Then we'll solve this and please feel free to type in your answer on the chat box Whether expression a is greater or b is greater type in a if you think a is greater or type in b if you think b is greater Okay, Ashutosh. Good. Then you found your mistake Just type in a if you think a is the greater of the two or type in b if you feel b is the greater of the two That's fine. Shushant But what about 99 to the power 50 So Niranjan things a even shushant things a Bharat also thinks a Okay It's made for others to give give out their answers. Okay. Gaurav things b shavan also thinks b So i'm getting a mixed response Okay, Ashutosh also thinks a Sugosh Sugosh Advat omkar What do you think guys? Sugosh also thinks a okay. Omkar thinks we fine. Okay. Let's solve this now So guys, uh, what I'll do first I will first, uh, do this operation 101 to the power 50 Minus 99 to the power 50. I'll first do this operation. Okay. Let's see. What do we obtain from this operation? now As shushant rightly pointed out we could write 101 to the power 50 as this And 99 could also be written as 100 minus 1 100 minus 1 to the power of 50. Okay Let's see what comes out from this expression. Then we'll take, uh, you know our calls based on the result of this particular operation Now 101 100 plus 1 to the power 50 can be written as 100 to the power 50 plus 50 c1 100 to the power 49 And then of course one and all will come but it doesn't make any difference because one Raise to any power will still remain a one and when it is multiplied to this it doesn't make any difference to the expression so Without wasting much time I could write this expansion like this All the way till 50 c 50 Okay So this is the first term which is 100 plus 1 to the power 50 Now let's write the second term Again, it will be 100 to the power 50, but it will alternate in sign So next term will be this Then again plus again minus and so on Okay, by the way, tell me the last term will it be positive or negative here? 50 c 50 will this be a having a positive sign or a negative sign? Just type positive if you think positive just type negative if you think negative What would be the sign coming here? positive correct Okay, now let's subtract this When you subtract this you would realize you would be left with two times 50 c1 Okay 100 to the power of 100 to the power of 49 Okay Then these two terms will get cancelled this two will get cancelled these two will get cancelled the next term would be 50 c 3 100 to the power of 47 Right And so on Okay, the last term would be if I'm not wrong 50 c 49 Okay 100 to the power of 1 So if you open this up You would realize that 101 to the power 50 minus 99 to the power 50 is equal to Let's multiply 2 to all the terms. So 2 into 50 c1 will be 100 Into 100 to the power 49 Then we'll have 2 into 50 c 3 100 to the power 47 And so on I would like to bring this to your notice Can you see the screen guys? Hello Is am I audible? Oh, okay. You can see the screen. You can can you hear me also? Oh, okay. Fine. Fine. So I was thinking that you're getting disconnected Okay So guys 101 to the power 50 minus 99 to the power 50 I can write it as now 100 into 100 to the power 49 will become 100 to the power 50 And all these terms here on there are all positive terms They are all positive terms Okay, so what does this mean it means that It means that 101 to the power 50 minus 99 to the power 50 Minus 100 to the power 50 is made up of positive terms Right. This is a positive term Correct. So if 101 to the power 50 minus 99 to the power 50 Minus 100 to the power 50 is positive That is greater than zero Then I can safely comment that 101 to the power 50 would be greater than 100 to the power 50 plus 99 to the power of 50 Is that clear and hence This term which is your a is greater than this term which is your b Is that okay with this we are going to start analyzing Let's talk about the properties of binomial expansion of x plus a to the power n Let's talk about the properties of The binomial expansion x plus a to the power n The first property that we are going to talk about If you see the total number of terms in the expansion The total number of terms in this expansion was n plus one. So it had a total of n plus one terms so somebody I think adhavath Uh, you know pointed this out That the total number of terms is one more than the Index So if I say x plus a to the power of three The total number of terms that I would get would be Four which is actually which are these terms Okay, so you can see there are One two three four terms. They will be four terms Okay, x plus a square. There are three terms x plus a to the power four. There will be five terms and so on okay The second observation that you can make The second observation that we can make here is In any expansion the terms are having the same coefficient from the beginning and the end For example, if you see This is one. This is one Then this is three. This is three So what is happening is n c n is same as n c zero n c one is same as n c n minus one n c two is same as n c n minus two and so on Okay, this is very obvious from the uh properties that we learned in permutation combination as well That n c r is equal to n c n minus r Okay, it is because of this property that we are seeing this observation. Okay Another observation that you would see is the sum of the indices sum of the indices of x And a Would always be n Right, so two plus one is three Then here three plus zero is three Here one plus two is three Again here x to the power zero a to the power three is three. So some of the indices will always be three next property is next property that you would observe in this case is The general term In fact, we are going to spend a lot of time on the general term Okay The general term in the expansion of x plus a to the power Let me write it here The general term in the expansion of x plus a to the power n Normally we write it as the tr plus one-eth term Okay, normally we don't take tr. We write we take tr plus one and why we take tr plus one It will be very evident to you in some time So guys, I would like to know from you that when we were expanding x plus a to the power n Okay, the first term was n c zero x to the power n a to the power zero Second term was n c one x to the power n minus one a to the power one Correct. So this is your first term. This is your second term Okay, so if I go to That's a third term.  It will always have n c and this number would be one less than this number as you can see When it is one here, it is zero here When it is two here, it is one here So if I go to r plus one-eth term, this term will become n c r Okay And now look at the power also of x x is always raised to a power of n minus one less than this Correct. So it'll be n minus one less than r plus one which is r And a is also raised to a power of one less than the terms position So a will be also raised to a power of r So normally this is what we call as The general term in the expansion of x plus a to the power Okay So this term is much more convenient to write Visibly writing tr. So let me ask you somebody tell me what will be tr Then you would say it will be n c r minus one x to the power n minus r minus one a to the power r minus one So see it is much more complex to write It is much more complex to write And remember Okay, that's why we never take this as our General term we always prefer taking r plus one as our general term Right So the answer to why we don't take tr as our general term We take our tr plus one as our general term because This expression is much more complex to write and remember So if I have to find the first term that means if I have to find t one I will put r value as zero If I have to put find second term, I'll put r value as one like that If I have to find n plus one-th term, I will write my r as n Simple as that Is that okay? Why we have chosen this term as our general term if this is understood Can we start solving problems on the general term? Any doubt regarding this? What is the general term? third point Third point means You're asking about this This is very simple At any term if you add the powers of x and y it should give you n for example, if I say x plus a to the power four This is the basic of what we discussed in the derivation part also So it is this four x cube a six x square a square four x a cube a to the power four at any stage If you see the power will add up to give you four only So four plus zero is four three plus one is four two plus two is four one plus three is four Zero plus four is four. Okay. Simple. Now. Let's take problems which are based on the general term first question find the seventh Find the seventh term in the expansion four x minus one by two root x Raise to the power of 13 Okay I'll start with the very basic question find the seventh term In the expansion of four x minus one by two root x to the power of 13 now guys first of all never try to solve these problems by Completely expanding this please. Don't do that. It will lead to a wastage of time Right if you somebody's trying to you know open up this entire thing it will lead to 14 terms Right, so don't try to expand and do it So expansion should not be done because it will lead to 14 terms which will be huge Okay So if I want the seventh term and we already know the expression for tr plus one tr plus one is ncr First term that is four x to the power n minus r Second term is this To the power of r Now tell me what is n here and what should I put my r value as to get t7 Correct. So n is going to be 13 n is going to be 13 Okay, and r is going to be six Right, so this is my n for sure. No doubt about it And since I want seventh term r plus one should be seven that means r is six Okay, so t6 plus one which is nothing but t7 would be 13 c 6 4 x to the power 13 minus six Minus one by two root x to the power of six which is going to be 13 c6 4 x to the power of seven By the way, this minus will have no meaning because this is an even number Okay, so it'll become one by two root x to the power of six If you simplify this further You know, you'll note your answer to be 13 c6 two to the power of eight x to the power of four So that's what Bharath wrote 13 c6 times 256 Times x to the power four. Leave it here. No need to simplify Okay, so let's leave it here probably I think Andrew has simplified 13 c6 that is fine and so this is the answer Next question based on this Anybody having any doubt, please Please feel free to type in on the screen next question is independent of x Find the term independent of x In the expansion three by two x square Minus one by three x to the power of nine Okay, so read the question carefully find the term independent of x Independent of x means It should not contain any x-terminate it should just have a pure number Now again, if you are trying to expand it and figure out which term is independent of x Please do not do that. It will take a lot of time and effort from your end So I want to see how do you do this problem without expanding this particular Binomial term that's correct. That's correct. Niranjan. It's the seventh term. How did you figure that out by the way? Now when you say find the term it doesn't mean you tell me the position Term means you need to give me the full expression of the term So give me the value give me the expression of the term find the term doesn't mean the position of the term So position you are correct position is the seventh term But give me the value also. What is the term actually seven by 18? That's absolutely correct Bharat So guys again see the approach here This approach is important these type of questions are going to come a lot in your school exams as well Okay, the approach is we'll say let tr plus one be the term Independent of x Let tr plus one be the term independent of x Okay So first of all the expression for tr plus one. We all know is n c r first term to the power of n minus r Second term to the power of r correct So we know this is our expression for the r plus one th term Now start consolidating The powers of x together and the constants together So what I'll do is I will keep all my constants One side so basically these are my constants Okay And powers of x will be x to the power 2 to the power n minus r and divided by x to the power r correct So what I did I separated the powers of I separated constants and variables Okay Now all I need to do now All I need to do now. I need to consolidate the power of x Guys, uh, am I audible and visible to you? But guys is a screen visible to you all Am I audible guys to you all? Yes, can you see me? Can you see the screen? Okay, so guys what are we going to do now is I am going to Consolidate the power of x which comes out to be this number 18 minus 2 r minus r which is nothing but 18 minus 3 r Now since you don't want any x we don't want any x in the term Can I say for this x term to be absent the power should be zero If it is zero means r is going to be six That means your t six plus one which is nothing but your t seventh term would be independent of x And what would that term be that term would be nine c six three by two to the power of nine minus six which is three times minus One by three to the power of six Okay, let us simplify this nine c six will be nine into eight into seven by six And we'll have three to the power three by two to the power three into one by three to the power three Sorry one by three to the power six Okay, so this will get cancelled off three to the power three here and We can cancel off a factor of 12 so it will be 12 by 8 into 27 which is going to become Let's cancel this off by a factor of three Sorry four. We can have a three here and a two here Okay, and we'll have a seven as well. Okay, and this will be nine So it'll become seven by 18 as your final answer Okay, so this becomes our final answer Yeah Okay, so this is how we deal with the The general term now. Let us take different varieties of problems which are Centred around this particular concept my next question is find the fourth term from the end from the end in the expansion x cube by two minus Two by x square whole raise to the power seven So we have to find fourth term from the end again ashutosh I mean we sir we can do it normally right by the seventh term coefficient Yeah, you can write the seventh term but we have already found out the seventh term So I'm replacing my r as R as six there. Oh, you're asking about this problem. Which seventh term coefficient you are talking about? Oh, yeah, see you can do it But see what will happen is you will end up writing all the terms in the expansion And then you'll start seeing simplifying them to see which leaves your x or not Here the power was actually simple So you could figure it out by you know looking at some of the terms But when it becomes you know large power and complicated expressions It is always advisable to up take the approach of the general term. That is tr plus one approach 70 x is absolutely correct Bharat Sushant It is 70 x not 70 by x Bharat check your calculation. It is 70 x Okay, guys when it is asked From the end okay from the end Don't break your head figuring out which term is my fourth term from the end Anyways, you could write it down actually but it makes no sense because I'll tell you a short way for example, let's say you have seven right So you can write one two three four five six seven eight. So eight terms would be there So fourth term from the end will be your fifth term Correct. So you can find a t five by writing at as t four plus one Okay, and t four plus one will be seven c three correct seven c three x cube by two to the power of four minus two by x square to the power of Of three So you have seven terms So seven means there will be eight terms So one two three four five six seven eight fourth term from the end would be the fifth term, right? Yeah, 70 x is the correct answer So I once again once again, it's going to be three and this is going to be four once again I'll just rewrite this expression So we are finding the fourth term from the end So what we'll do is we'll switch the position of We'll switch the position of these two terms Okay So this will come here and this term will come here Now guys the fourth term from the end would be The fourth term from the beginning of this particular expansion Okay, it would be the fourth term from the beginning of this particular expansion Please note while you are shuffling the terms Take the sign along with it So read this as x cube by two plus Minus two by x square to the power of seven And then shuffle the position now find the Find the Fourth term from the beginning of this expression find t four In this so when you do t four which means you have to write t three plus one Okay, and that would be nothing but seven c three Minus two by x square to the power of Four Correct, which is seven minus three And x cube by two to the power of three Correct If you simplify seven c four or seven c three both mean the same thing seven c three is going to be Seven into six into five by six which is going to be 35 This is going to be 16 by x to the power eight This is going to be x to the power nine by eight Which is going to give you 70 x Is that fine? So whenever You are trying to find out any term from the end. It is advisable You flip the terms the first and the second term And just find out that position from the beginning it will give you the same result Next question is Let's take the next question Find the this is a very good question. Please think about it very carefully before answering Find the number of in the expansion of in the expansion of Nine to the power one fourth Plus eight to the power one sixth Whole race to the power hundred five hundred Whole race to the power five hundred Which are integers? Which are integers? Is the question clear? So basically There would be some terms While you are expanding this which will not be integers, which will be you can say They would be irrational in character Okay So we want to find The terms which are actually integers in the expansion of This term so how many won't find the number of terms so how many terms are integers in the expansion of Nine to the power one by four plus eight to the power one by six Whole race to the power Five hundred Forty is not correct Bharat No, 41 is also not correct No 188 is also wrong Is a very good question. It keeps on coming in many exams Is absolutely correct well done shawan very good. So the right answer is 251 guys Excellent. See guys how it works First of all this I can write it as three to the power half Okay Nine to the power one by four is as good as three to the power half This is as good as two to the power three to the power one by six, which is nothing but Uh two to the power one by two Okay, so this is two to the power one by two Whole race to the power five hundred Whole race to the power five hundred, okay Now Let's write the general term for this Answer is 251. Let's check how let's write the general term for this which will be 500 C r Three to the power half race to the power 500 minus r Into two to the power half Race to the power r Okay So it becomes 500 cr Three to the power 500 minus r by two two to the power r by two, please note r here can go from Zero to 500 now I will get integers only when These two powers These two powers are whole numbers Okay So this term Tr plus one will be integer when 500 minus r by two Right and r by two both of them belong to whole numbers now Let's start with Let's start with r by two term when can r by two be whole numbers? When r is zero two four six All the way till you reach 500 correct, correct And for all these values you would realize that 500 minus r by two will also be a whole number So for the same value let's say zero it will be 250 When you put two it will be 249 and so on till we reach 500 So basically from zero to 500 with a gap of two How many terms would be there very simple? Use your arithmetic progression for this Use a plus n minus one d as your the last term Let's say last term is 500 a is zero Common difference is two Okay, so this will give you n minus one as 250 So n will become 251 so this will become your Final answer for this question Let us take few more of this type because this type comes a lot in the entrance exams This was an easy problem to start with Let me take a slightly more complicated one any question with respect to this Please type clear on your screen if you have understood this So let's take another question next question is the number of irrational terms in the expansion of 5 to the power 1 by 8 plus 2 to the power 1 by 6 Whole raise to the power 100 Okay, read the question find the number of irrational terms Please take a minute or so to solve this Basically irrational terms means certs. So how many terms would be Serves in the expansion of this term Whoa Shushant You're correct 97 is the answer well done Shushant What about the others so far I've got the response only from Shushant Okay, so let us try to solve this problem correct Bharat 97 is correct So in the number of irrational terms number of irrational terms Would be nothing but total minus number of rational terms correct Now what would be the total number of terms total number of terms? We know it will be n plus 1. So it is 100 plus 1 which is 101 terms Okay, so altogether there will be 101 terms in this expansion Now let us write the expression for the general term that is the r plus 1th term So that will be 100 c r okay 5 to the power 1 by 8 to the power 100 minus r 2 to the power 1 by 6 to the power of r Which gives you 5 to the power 100 minus r by 8 And 2 to the power r by 6 Okay Now The term would be rational only when please note that r will range here from 0 to 100 The term would be rational r plus 1 would be a rational term When 100 minus r by 8 is a whole number As well as r by 6 is also a whole number Okay Now when is r by 6 a whole number very simple For what values of r you would say r can be 0 6 12 18 all the way till you reach 96 Correct Okay, what would make 100 minus r A whole number Can 0 make it a whole number You will say no 100 by 8 100 by 8 is not a whole number, right? It's not a whole number. Okay now can What are the what are the very first number that comes in your mind which is divisible by 8? So I think if I make this as 8 itself, which means r is going to be 92 Correct it can be 100 first of all Then 92 Okay, then again decrease a factor decrease a 8 value from this so which is going to be 84 correct And this will tend will go on till I reach till I reach something like Something like 4 Correct. Yes. So if you see in a reverse fashion So Bharat, you are correct. So it's 4 12 20 and so on till you reach 100 Now I have to see what are the common values of r in these two I have to see for the common values Okay, so the This is moving at a rate of 6 This has an increment of 6 this has an increment of 8 So basically the LCM of 6 and 8 is what I'm looking for which is actually 12 Sorry LCM of 6 and 8 is 24 I'm sorry. Yeah Okay So the first term common will be 12 Next term common would be after a gap of 24, which is going to be 36 Let me show you how So after this I'll have 24 30 36 42 like that dot dot dot After this I'll have 28 36 44 like this dot dot dot. So the next term common is 36 As you can see, there would be a gap of 24 Okay So the common values of r that is common r in both of these series would be 12 Correct 36 Then again a gap of 24, which is 60 Then 84 And that's it after that it will exceed 100. Okay So only for these four terms I would get Rational values So the number of rational terms would be Just four in number number of rational terms would be Just four in number So altogether there are 101 terms. So your answer is going to be 101 minus four, which is 97 terms Okay, is that clear? Great. So most of you are have been able to solve this problem. Well done So one last question we'll take on this and then we'll move on to other type of problems Question is find the sum of all rational In the expansion 3 to the power 1 by 5 plus 2 to the power 1 by 3 Whole raise to the power of 15 Very easy question as compared to whatever you have solved this question is easy You should be able to get all of you should be able to get the answer The question has a change now find the sum of all rational terms In the expansion of this So how do you solve such kind of problems without actually opening all the terms? 15 9 that's absolutely correct Sushant Bingo second time in a row 59 is correct Bharat. Also. You're correct again the same approach. There is no difference in the approach What about the others? Onkar you are correct. So tr plus one. Let's write it down. So 15 cr 3 to the power 1 by 5 to the power of 15 minus r 2 to the power 1 third to the power of r Okay So let's find out what will be the rational terms. I will obtain rational terms Only when This power and this power both are whole numbers. So both should be whole numbers Okay So when is r by 3 a whole number remember r can not go beyond 15 r can be 0 to 15 So this will be whole number only when r is 0 3 6 9 12 15 So when will be 15 minus r by 5 a whole number When this be will be a whole number Yeah, this will be a whole number when again r is 0 Okay So r will be 0 it will be r 3 then again when r is 5 r is 5 Then 10 Then 15 Now I have to see the common of these two So the common is 0 And 15 only So when r is 0 means t 1 Correct that would be simple t 1 is going to be 3 by 3 to the power 1 by 5 to the power of 15 Which is going to be 3 to the power 3 which is 27 And t 15 would be Sorry t 16 would be your when r is equal to 10. So t 15 plus 1 That would be your last term And the last term would be 2 to the power 1 third whole to the power 15, which is 2 to the power 5, which is 32 At these two your answer is going to be in front of you. That is 59 great now guys Moving on we were talking about the properties. So We were doing an analysis. We we were stuck with the general term so far Next important thing is you should be always be Knowing this formula very very well. These two formulas are very very heavily used nc r by nc r minus n minus r plus 1 by r Very important formula Okay, because you'll be using this in problem solving And of course nc r is equal to n by r n minus 1 c r minus 1 Okay, I already talked about these two formulas in p and c now. We'll see some questions based on these formulas Let's take some questions my question is the second third and fourth terms in the expansion of x plus y to the power n r<PHONE_NUMBER> respectively respectively then find the value of x y n Such type of problems come a come a lot in school exams as well And this property is a hint that will help you in solving this problem. Let's see how many of you get the answer I'll give you around two minutes to solve this and then we can take a small break Now, can you see the screen? Okay, great So t2 is given to you as 240 T3 is given to you as 720 And t4 is given to you as 1080 x is 2 correct y is 3 that is also correct ashutosh What about n? Absolutely correct shiram. So bang on n is 5. That's correct So you guys again very simple to solve this t2 means t1 plus 1 right And we know t1 plus 1 will be nc1 x to the power n minus 1 y to the power 1 okay 3 means t2 plus 1 which is nothing but nc2 x to the power n minus 2 y square And t4 is nothing but t3 plus 1 which is going to be nc3 x to the power n minus 3 y cube Okay, and these are the values given for this Let me call it as 1 2 and 3 correct Now let's do 2 divided by 1 So 2 divided by 1 will give me nc2 y by nc1 x And this is going to be 720 by 240 which is going to be 3 Now this expression seems familiar to me because I know this formula correct So here n is n r is 2 so my expansion here would be simply n minus 2 plus 1 by times y by x which is equal to 3 right which gives us n minus 1 by 2 times y by x equal to 3 Let me call this as 4 In a similar way if I do 3 divided by 2 right shiram n is 5 you are correct When you do 3 divided by 2 you get nc3 by nc2 again y by x And 1080 by 720 is 3 by 2 you can cancel it off by a factor of 360 Again the similar expression ncr by ncr minus 1 so it will be n minus r plus 1 by r Times y by x is equal to 3 by 2 Let me call this as 5 Okay, in fact I can further simplify this n minus 2 by 3 y by x Is equal to 3 by 2 okay Instead of calling this as 5 let me call this as 5 Let me call this as 5 Okay Now let us do Let us do 5 divided by 4 Okay, 5 divided by 4 you will get n minus 2 by n minus 1 Okay 3 here 2 up Is equal to half Is equal to half Which means 4 times n minus 2 is equal to 3 times n minus 1 which means 4 n minus 8 Is equal to 3 n minus 3 which means n is equal to 5 Right, so most of you have got this answer n is 5 well done Now once you've got n is 5 we can use any one of these equations. So let's put n is 5 In this expression. So when you put n is 5 you get 4 by 2 y by x is 3 So y by x is going to be 6 by 4 which is going to be 3 by 2 correct So we can say y is equal to 3 x by 2 Now let's use in the first expression itself Let me make some space it is getting too cluttered So t2 which is 240 is going to be 5 c1 x to the power n minus 1 which is 4 And instead of y I can put 3 x by 2 Which means 240 into 2 by 3 divided by 5 is x to the power 5 Let's cancel 80 16 32 so x becomes 2 So y is nothing but 3 x by 2 which is nothing but 3 into 2 by 2 that means y is 3 Okay, so this is the values that we get x as 2 y as 3 and n as 5 Is that clear? So the formula which I discussed with you proves very very handy while solving such kind of problems Now next concept that we are going to talk about will be after the break So all of you please take a break Of let's say seven minutes We'll meet at 110 p.m. Sharp Okay, and then we'll talk about further more concepts. There are a lot of concepts left So see you at 110. All right guys, let's resume back Hope everybody is back after the break Okay, so now we are going to Discuss about the middle terms in the binomial expansion Right So this is a very very important concept for a school exam as well Okay, a lot of questions are directly asked on middle terms So middle term can be found based on the value of n So when we say x plus a to the power of n this n plays a very very vital role in Finding or deciding the middle terms of the expansion So I'll take two cases the first being when n is even Okay It's very clear that the total number of terms we always have is n plus 1 So when n is even I will have odd number of terms I'll have odd number of terms in the expansion Okay, which means there will be only one middle term Which implies there will be Only one middle term would be there So when the number of terms are odd we'll only have one middle term now. Can somebody tell me what would be the What would be the Value of what will be the position of What will be the position of The middle term What will be the position of the middle term can somebody tell me You can take an example. In fact, you can take multiple examples. Let's say you take x plus a square Right, then you know that This is your second term is your middle term Okay, when you have x plus a to the power of four So you have x to the power four Then you have four x cube a Six x square a square then four x a cube then a to the power four like that, correct? So the third term is your middle term Okay Right so Bharat is correct. So it's basically nothing but if I go to any n where n is even Where n is even you would realize that the middle term would be n by two plus one-th This term would be your position of the middle term Okay So your answer here would be n by two plus one-th term would be your middle term correct And what would be the value of that term or what would be the expression for that term? So basically it is nothing but t n by two plus one-th we need, right? So our r is basically n by two So that will become n c r first term to the power of n minus r Second term to the power of r So this would be your expression for the middle term. This is the position of the middle term Okay. Yes, Andrew you all correct Omkar is also correct. Okay Now let us move on to case number two when the number of terms are odd So let's take the second case when n is odd When n is odd it implies the total number of terms Total number of terms would be even That means I will have two middle terms I will have two middle terms Now can somebody tell me what would be the position of the two middle terms? What will be the position of the two middle terms? You may take some examples and try to figure it out For example, let's say I start with x plus a to the power of three So I know it's going to be x cube 3x square a 3x a square plus a cube So my second and third terms are my middle term Okay, when I say x plus a to the power five I get x to the power five 5x to the power four a Then 10x cube a square Then again 10x square a cube Then 5x a to the power four plus a to the power five and so on So in this case my third And my fourth terms are my middle terms So how are these numbers two three related to this number? That's what we have to figure out Okay So the pattern here is that If you are going till x plus a to the power n x plus a to the power n Your middle terms would be n plus one by two For example, if you take five plus one by two you get a three Correct And the next will be n plus three by two That is five plus three by two, which is four And the same is applicable here also Three plus one by two is two three plus three by two is three Okay So basically the position of the middle terms would be n plus one by two-th term n plus three by two-th term Okay Now what about the values? So for n plus one by two-th term, I need to write it first of all like this n minus one by two plus one Correct Then this will start behaving as my r So my answer will be nc n minus one by two x to the power n minus r will be n plus one by two and a to the power of n minus one by two Correct In a similar way Tn plus three by two can be written as t n plus one by two plus one It's actually the very next term after this, isn't it? n plus one by two and n plus three by two are consecutive So if you add a one to this you get n plus one n plus three by two So it is going this is going to play the role of r now So it is going to be nc n plus one by two x to the power n minus one by two a to the power n plus one by two Guys one thing that you should remember is that Both the middle terms will have the same binomial coefficients Both the middle terms Okay will have the same binomial coefficients Okay, that is nc n minus one by two will be same as nc n plus one by two Is that clear? guys One small thing I wanted to clarify here I have seen a lot of confusion this about this particular thing, which I'm going to discuss in the past There are three terms which are very confusing One is Binomial coefficient One is coefficient and third is the term Okay Let me clarify this to you Let me just clarify this because I have seen a lot of ambiguity or you know misconception or wrong notion about the meaning of these terms What is binomial coefficient? Let me write a simple actually series. Let's say Let's say I write two x plus three to the power of Let's say four, okay When you expand this you write something like this four c zero Okay, two x to the power four Then you write four c one two x to the power of three into three Then you write four c two two x square three square Then four c three Two x to the power one three to the power four Okay, and finally Sorry And finally four c four Two x to the power zero Three to the power four, okay Now let's pick up any one of the terms. Let's say I pick up I pick up this term When I ask you what is the binomial coefficient for this term Your answer will be just four c two nothing else So this term is your binomial coefficient Are you getting this point? Okay, so if I ask you what is the binomial coefficient of the third term you would say four c two Correct If I say what is the coefficient of let's say x to the power two Then you would say four c two into two to the power two into three to the power two because All the thing which is present along with x to the power two will become the coefficient So this entire thing becomes coefficient Are you getting it? So in that you should not write this When I ask you what is the third term? Then you would mention everything that means this entire thing will be your third term So listen to or read the question very very carefully. Are they asking you only binomial coefficient? Are they asking you coefficient of some particular expression? Or are they asking you the term? Are you getting this? So if I say what is the binomial coefficients of the middle terms? So here you would say the binomial coefficient of the middle term would be four c two Getting the point if I ask you what is the middle term? Then you would say the entire thing that is the middle term Okay, so watch out for the question asked because After this particular session in fact concept I will be talking about Terms which will be sounding very very similar Okay Any question regarding These three terms it is very clear. What are the difference between binomial coefficient coefficient of an expression and the term itself? Now, let's take problems based on middle terms middle term Very simple problem to start with find the middle term In the expansion of 3x minus x cube by 6 To the power of 9 find the middle term You can have terms as well that depends upon the so since n is odd Okay, middle terms would be n plus 1 by 2th and n plus 3 by 2 th terms Which is going to be your fifth term And sixth term So t five will be nothing but t four plus one correct So n c r because r is going to be four This term to the power five This term to the power of four Okay If you're going to simplify this It's just going to give you 189 by 8 x to the power 17 Okay, don't waste time doing it because Uh, we can You can just see what is what has been done on your screen. I know you know how to do it So similarly t six will be t five plus one So that will be n c five 3x to the power four minus x cube by six to the power of five That would give you minus two hundred and sixteen x to the power 19 Okay, so these two terms would be the middle terms Is that clear? Please type in yes if it is clear guys, please type in yes if it is clear on your chat box Now guys, uh, we are going to start with a very very important concept here The concept of numerically greatest term Now again, I want you to first understand the meaning of numerically greatest term n gt Again, this is a very important concept which keeps on coming In uh regional entrance exams as well What are the meaning of numerically greatest term numerically greatest term means the term with greatest magnitude When I say magnitude means ignore the sign ignore the sign so minus hundred And 99 Which is numerically greatest? Type in you type it in your chat box, which is numerically greatest minus hundred or 99 correct So this is numerically greatest Of these two. Okay, so don't take the sign into your consideration only the magnitude is what we are looking at Okay Now if you're given If you are given any binomial expression like this Okay Then If I ask you What is the greatest? Binomial coefficient How will you find this out? What value among x a and n are required? To know the greatest binomial coefficient Or all the three are required which one of the three is required to know Or which ones are required to know the greatest binomial coefficient Right. So middle term Bharat. How do you get the middle term? What decides the middle term? Only n decides the middle term. Correct. So this becomes important while you are Finding the greatest binomial coefficient Okay, so greatest binomial coefficient only depends upon n For example, if I say n is 10 Then you would say the greatest binomial coefficient will be 10 c5 If I say n is 9 Then you would say greatest binomial coefficient would be what? Would be what? You will say 9 c4 and 9 c5 Correct. These two would be the greatest binomial coefficient So only n is required So we need only n is required To get the greatest binomial coefficient Okay for Greatest binomial Coefficient, but if I say I need Greatest coefficients Then what all do you require? out of x a and n What all information should be given to us to find the greatest coefficient or let's say greatest coefficient of The variable x We need n and we need a over here. That's it Because x anyways will remain in terms of x So whatever is along with x that would become the coefficient of that variable correct or variable x So I don't need the value of x. I need only n and a Okay, so it is clear in this case greatest binomial coefficient only n is required When you need greatest coefficient You need both n and a Okay But when you require numerically greatest term numerically greatest Focus on term We need all the information. We need x also a also n also Okay, so for numerically greatest term we should know We must know x a And n so in short we should know All the terms in this expression, of course, I mean You should know 3x plus 4 you should know 3 you should know what value of x you are putting in for example, if I say uh x 3x plus 4 to the power of 10 find the numerically greatest term when x is let's say 1 by 4 So indirectly I know x also I know I already I know 3 4 and 10 Okay, so this information is required This is required Are you getting this point shushant? Is that clear? No, let's not solve it. I am going to just uh, you know discuss the theory first Let us discuss the theory now Between any r plus 1th term and rth term Okay, let's talk about their modulus Only three types of relation can exist Either this can be greater than one Or equal to one Or less than one Correct Okay So let's say Tr plus 1 and tr the ratio of these two terms can only have The mod of these three terms can only have three relations Either this ratio can be greater than one That means either I can have mod tr plus 1 greater than mod tr Or I can have Mod tr plus 1 equal to mod tr Or I can have mod tr plus 1 less than mod tr There's no fourth relation between these two correct Okay So now what I will do is just watch out the theory in this case If you write tr plus 1 by mod tr For this expression, what will you write?  Now we all know that ncr by ncr minus 1 this term is mod n minus r plus 1 by r And from here I will get a on the numerator And x in the denominator Right so far so good Now this I can write it as n minus r plus 1 by r mod a by x That means this I can bring it out of the mod Why I can bring it out of the mod? Because we know that n minus r we all know that We know that n minus r plus 1 by r will always be positive It's because n is greater than r in fact greater than equal to r So this will always be positive This will always be positive. So the entire thing is positive. So I can just bring it out of the mod sign correct Yes or no Okay, now let's take this situation and move on That means let's say take this this situation and let's move on So now I'm trying to say is I'm trying to say n minus r plus 1 by r Mod a by x. Let's say I'm assuming that this is greater than 1 That is I'm trying to say that tr plus 1 Is greater than tr Guys, I know you would have a lot of questions in your mind So just hold on to your question till I have completed this process so that I can explain it to you in a better way Okay So I can say n minus r plus 1 by r is greater than mod x by So from this I can write this correct And I can write here as n plus 1 by r minus 1 greater than Mod of x by That means n plus 1 by r is greater than 1 plus mod x by a that means n plus 1 by 1 plus mod x by a Is greater than r or you can say r is less than n plus 1 by 1 plus mod x by a Okay So guys now listen to this very very carefully because If you miss out on this concept Then you would not understand what's going on Okay Now when I started this conversation, I told you to know the numerically greatest term In this expansion You should know x you should know a and you should know n Correct that means each of these terms here is known to me correct So I can evaluate this value very easily Let's say I evaluated this value n plus 1 by 1 plus mod x by a And I got a value m Okay, I got a value m right Now what I'm trying to say here is that When r is less than m I forgot a model here then Mod of tr plus 1 would be greater than mod of tr Is this statement clear? When r is less than m Mod of tr plus 1 would be greater than mod of tr. Is this statement clear? Please type clear on your screen if it is clear then I'll then only I will move forward So niranjan, shiram, andri, umkar clear What about Bharat? Shant Now see very very carefully. Okay There are two possibilities now this m This m could be an integer Or it could not be an integer right Okay, this m could be an integer Or it may not be an integer. There are only two possibilities Now let's take Situation number one if m is an integer Let's say if m is an integer m is an integer means it has some value which is an integer. Okay Now what is this expression trying to say When r is less than that integer This term will always be greater than this term. That means mod of tr plus 1 would be greater than mod of tr For example, let's say I put r as 1 Let's say I put r as 1 Okay Now and let's say m is some value which is higher than 1 Okay, so can I say mod of t2 Would be greater than mod of t1 Which means The numerical value of t2 is greater than Numerical value of t1 Okay, similarly when I put r as 2 I can say t3 is greater than mod t2 When I put r as 3 then mod t4 would be greater than mod t3 Let's say I reach a stage of m Right, let's say m minus 1 for the time being okay m minus 1 Where m is some integer Please note r can only take integer values Okay r can be 1 2 3 For only integer values because r is the position it represents the position Latest coefficient I will come little later on ashutosh. It is linked to this to this particular derivation So follow the derivation of The numerically greatest term then I will come to the derivation of greatest coefficient Right now. I'm talking about numerically greatest term Is that okay ashutosh I'll come later on to this question. What is the greatest coefficient? Okay We were clear about Greatest binomial coefficient now. We are doing greatest term and then I'll come to greatest coefficient So that would be our order of dealing with the things Okay Now guys pay very very good attention over here because this may be this is not so easy to understand in first go Now when r is equal to m minus 1 I am still less than m right Then in that case, can I say t m mod? Would be greater than mod t m minus 1 The moment I reach r equal to m Then this condition is violated because I have now surpassed I have become equal to m Okay, so when r is equal to m Please note that There will be an equality sign coming in between here So this condition was only valid when I was saying mod t r plus 1 is greater than mod t r When r becomes equal to m Then mod t m plus 1 will become equal to mod t m correct And when I Take r as m plus 1 that means I have exceeded m r has exceeded m That means this inequality This inequality has flipped now. So this inequality will also flip like this That means t m plus 1 T m plus 1 Would now become greater than Or you can say t m plus 2 would now become less than t m plus 1 Yeah, among these two, this is the largest among these two This is the largest like that. I'm actually drawing a comparison Let's see what will be my conclusion be Okay Next when I start putting more values like m plus 2 I would get t m plus 3 Less than mod t m plus 2 I can go all the way till I reach n Now see what is happening see the moral of the story here The moral of the story is The moral of the story is When when my value of r was 1 I'm writing it in a new pace so that it is clear to all of you When r is equal to 1 mod t 2 was greater than mod t 1 Similarly when r is 2 mod t 3 is greater than mod t 2 And this goes on till r is m minus 1 in that case t m would be greater than mod t m minus 1 The moment I reach m Then what will happen t m plus 1 would be equal to t m And the moment I start putting higher values Mod t m plus 2 will start becoming lesser than mod t m plus 1 okay, so You can start seeing the decrement in the value happening Okay, all the way till I reach n okay, that is T n plus 1 would be lesser than mod t n If you summarize this in a single inequality relation You would realize that things goes like this mod t 1 is less than mod t 2 is less than mod t 3 And so on till you reach mod t m Which is actually equal to mod t m plus 1 And again the value starts decreasing Okay Yes, now Bharat you have concluded it correctly. You would see that When I'm writing this all I realize that these two are the numerically greatest terms By the way, you will not find these proofs in many book. They just directly write the result because You know making this explain is becomes very tedious work for the book also You cannot just understand this by reading some statements Okay, so this is a scenario where your m was an integer So my conclusion here is that most of the books will just write conclusion conclusion is If your m which is nothing but n plus 1 by 1 plus mod x by a Is an integer Then there would be two numerically greatest terms And what are those terms that would be t m and t m plus 1 Okay, so this is the conclusion which I drove from This particular derivation as of now The next question that would arise in our mind is what will happen if m was Not an integer If m was not an integer then what will happen?  Okay This is my m and this is my right If m is not an integer, can I say this case will never happen? Duranjan, can you hear me now? Duranjan, can you hear me? Okay, great So till this stage is it clear that if m is not an integer I can never have r equal to m because r is related to the position Right r is related to the position It's related to the position So r can only take values like zero one two three So it will always be an integer And if this case is this expression m if it is not an integer r can never become equal to m clear So only these two possibilities will be there Okay Now again, let me start with the same thing If r is one, let's say m is a value which is higher than one It is let's say something like 10.24. Okay So when r is one, can I say mod t2 would be greater than mod t1? When r is two mod t3 would be greater than mod t2 Let's say I reach a value m minus one Till that's this term would be greater than This Okay The moment r becomes equal to m Which cannot happen.  Okay What will happen in this case? We know that greatest integer is also less than m See, let's say m is 10.24 Greatest integer m is 10 So technically 10 is less than 10.24 That means your r is less than m You are still following this relation r is less than m Then in this case also mod greatest integer m Plus one would be greater than Let me write here. Let me write like this It makes more sense. Yeah Greatest integer m minus one Sorry, uh Is that okay? Now the moment r becomes greatest integer m plus one Yeah, yeah, that's what I corrected Bharat. It would become t greatest integer m minus one. Is that clear Bharat? So I I immediately corrected it would become t greatest integer m and greatest integer m minus one. Okay Now the moment r becomes greatest integer m plus one it means it has exceeded m So if r has exceeded m then we are going to follow this relation That is mod mod this plus two Would be less than mod t greatest integer m plus one Okay And I can continue all the way till I reach n. So in that case this will be less than t n Now see the moral of the story here t2 is greater than t1 numerically t3 is greater than t2 and so on till I reach a point where t greatest integer m plus one is greater than t greatest integer m After that the value starts falling down So something like this is happening over here mod t1 is less than mod t2 mod t2 is less than mod t3 And it is going all the way till mod t greatest integer m is less than t greatest integer m plus one and after that it starts falling Let me just write it in a clear way After that it starts falling from here Let me write it in a new page so that So mod t1 was less than mod t2 mod t2 was less than mod t3 and so on Till I reached This expression and after that the value starts dropping I reach this So which is the numerically greatest term the numerically greatest term would be This fellow Right, so this fellow would be the numerically greatest term numerically greatest term So conclusion number two if your r n plus 1 by 1 plus mod x by a equal to m is not an integer Then there will be only one numerically greatest term and that would be t greatest integer m plus 1 is this clear? So I'll write the final judgment over here The final result is So from this exercise what we learned We learned in this entire discussion was If I have to find the numerically greatest term of this expansion Step number one is I will first find m m is nothing but n plus 1 by 1 plus mod x by a Then from here two results can happen My m could be integer My m may not be integer If n is an integer then they will be Two numerically greatest terms and what would be they they would be t m and t m plus 1 When m is not an integer there will be only one numerically greatest term And what would be that that term would be t Greatest integer m plus 1 So please remember this very very important What I will do is I'll quickly take a problem and then we'll close the session because I don't want to You know leave this without solving a problem on this Okay, I'll just take a simple example So just see this example carefully Oh, I'm sorry question is find the numerically greatest term in the expansion of 3 minus 5 x to the power of 11 when when x is 1 by 5 When x is 1 by 5 Okay Now be very very careful Right now if you compare this With x plus a to the power n Whatever we derived was for x plus a to the power n Then who's playing the role of x over here 3 is playing the role and the role of a has been played by minus 5x So remember step number 1 We'll find m m would be what n plus 1 by 1 plus mod x by a Okay So right now in this problem. This is my n So 11 plus 1 by 1 plus mod x x is 3 a is minus 5 x And put the value of x as 1 by 5 over here Okay This would give you 12 by 1 plus mod 3 by minus 1 Which is 12 by 4 which is equal to 3 Now we know that 3 is an integer Correct So if 3 is an integer We are going to follow This particular scenario Correct, so there would be two answers that I would get Okay So I'll have two numerically greatest terms two numerically greatest terms would be there That would be t m And t m plus 1 Okay Now what I will do I'll just show you why they're why they're numerically greatest and Why both of them are the answers because both of them will have the same numerical value. Let's check T 3 is nothing but t 2 plus 1 Correct t 2 plus 1 is 11 c 2 3 to the power of 9 minus 5 x to the power of 2 Okay Okay, so let's see what is the value here So it's 11 c 2 3 to the power of 9 Okay, and minus 1 by 5 will become minus 5 into 1 by 5 Square which is equal to 11 c 2 into 3 to the power 9 Now let's find t 4 t 4 is t 3 plus 1 So the answer will be 11 c 3 correct 3 to the power 8 Okay and minus 5 into 1 by 5 to the power of Now I'm trying to say That we got the same answer from both 65 into 3 to the power 8 this minus 165 into 3 to the power 8 So new both are but yes numerically both are equal So guys we'll continue with this numerically greatest term and yes I have to do numerically greatest coefficient also in fact greatest We're doing the next class But this concept is very very important. Let me tell you The understanding of this concept is very very important I don't understood anything. Please Watch this video again, especially the last part And then let me know if I need to start x-class Thank you so much and all the best for your physics exam for tomorrow Thank you so much for joining in youtube live Bye bye signing out from here centrum academy