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Mathematical induction example

  1. Mar 30, 2015 #1
    A step in this process of proving Sn: 1+4+7+...+(3n-2) = n(3n-1)/2
    confuses me. I hope someone can clarify this for me.
    I do not require the work done, I need clarification on a step only. Thanks!

    2. Relevant equations
    After assuming n=k, we say Sk: 1+4+7+...+(3k-2) = k(3k-1)/2
    When assuming n=k+1, we say Sk+1: 1+4+7+...+(3k-2) + (3k+1) = k(3k-1)/2 + (3k+1)
    The book states the reason for adding (3k+1) on both sides of the equation is because 3(k+1)-2 = 3k+1

    3. The attempt at a solution
    Why is this the case? What is 3(k+1)-2 ? I know it equals 3k+1 because 3 multiplied by (k+1) is 3k+3, then -2 makes it 3k+1. But where did 3(k+1)-2 suddenly come from? It seems arbitrary and is without explanation in the text book.
    Last edited: Mar 30, 2015
  2. jcsd
  3. Mar 30, 2015 #2
    When using induction you are trying to show that the kth case implies the k+1th case. So after adding 3k+1 on both sides you get Sk + 3k+1=Sk+3(k+1)-2, which if you are to assume the inductive hypothesis, holds with n=k+1.
  4. Mar 30, 2015 #3


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    I would have said "By the induction hypotheses, for n=k we assume that 1+4+7+...+(3k-2) = k(3k-1)/2."

    Here it is better to say what we are to prove, which is that ##S_{k+1}## is true, which is:$$

    1+4+7+...+ (3k-2)+(3(k+1)-2) = \frac{(k+1)(3(k+1)-1)} 2$$
    This is gotten by just writing ##S_n## when ##n=k+1##. Now if you simplify the last term on the left side, you will see why adding ##3k+1## to both sides of ##S_k## will make the left sides equal and, hopefully, the right side equal to what you want.
    Last edited: Mar 30, 2015
  5. Mar 31, 2015 #4
    I got it now.
    That term 3(k+1)-2 that I thought was totally random came from plugging in (k+1) into the k in (3k-2)

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