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I am having difficulty proving the inductive hypothesis $(k+1)$ for the following statement: $$\sum_{i=1}^{n-1} (i(i+1)) = \frac{(n)(n+1)(n-1)}{3}$$

This is what I have so far:

$$(Step \ 1) \sum_{i=1}^{k-1} (i(i+1)) + (k+1)(k+2) = \frac{(k+1)(k+2)(k)}{3}$$

$$(Step \ 2)\frac{(k)(k+1)(k-1)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k)}{3}$$

$$(Step \ 3) \frac{(k^3 + 3k^2 +8k + 6)}{3} \ != \frac{(k^3 + 3k^2 + 2k)}{3}$$

I do not know what I did wrong. Thank you.

Martin Sleziak
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gordon sung
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    I believe that the last term to add is $k(k+1)$ rather than $(k+1)(k+2)$. – Nathan Weckwerth May 06 '16 at 03:09
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    Yes. And also on the right side $\frac{k^3+3k^2+2k}{3}$. Check it. – jrs May 06 '16 at 03:12
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    I don´t know why you use the variable k, although in the statement only the variable n is involved. Beside this, in the second line the upper limit is still $n-1$. – callculus42 May 06 '16 at 03:18
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    In steps 1 and 2 you are assuming that which you are trying to prove. After you make the corrections which others have pointed out, you should set the left side of step 1 equal to the left side of step 2, then simplify that expression and hope it equals the right side of step 1. You cannot assume from the outset that it will equal what you want it to equal. – John Wayland Bales May 06 '16 at 03:49
  • I am sorry, but why is it not i = (k+1) and thus we should have (k+1)(k+1+1)?thank you – gordon sung May 06 '16 at 04:23
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    Since your question seems to be mainly about your specific attempt (as opposed to asking any proof of this claim) I think that the tag ([tag:proof-verification]) should be used. See the tag-info for more details. – Martin Sleziak May 06 '16 at 05:05
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    You can probably find some other posts on this site about the same (or at least very similar) sum. For example: http://math.stackexchange.com/questions/43317/evaluate-sum-limits-k-1n-k2-and-sum-limits-k-1n-kk1-combinatoria or http://math.stackexchange.com/questions/1642906/simplify-triangular-sum-of-triangular-numbers-sum-i-1n-frac12ii1 – Martin Sleziak May 06 '16 at 05:08

1 Answers1

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Inductive hypothesis (=the claim holds for $k$): $$\sum_{i=1}^{k-1} i(i+1) = \frac{k(k+1)(k-1)}3.$$

Using this you want to show that it is true for $k+1$ instead of $k$: $$\sum_{i=1}^{k} i(i+1) = \frac{(k+1)(k+2)k}3.$$ (This is the inductive step.)

You have $$\sum_{i=1}^{k} i(i+1) = \sum_{i=1}^{k-1} i(i+1) + k(k+1) \overset{(*)}= \frac{k(k+1)(k-1)}3 + k(k+1) = k(k+1) \left(\frac{k-1}3 +1\right) = k(k+1)\cdot \frac{k+2}3 = \frac{k(k+1)(k+2)}3. $$

The step marked $(*)$ is the step where we are using the inductive hypothesis. Notice that in the inductive step I was adding $k(k+1)$. You were attempting a proof along the same lines, but you were adding $(k+1)(k+2)$.

Martin Sleziak
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