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Prove that $(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3$ for every $n \in \mathbb{N}$.

I'm trying to use induction on this one, but I'm not sure how to. The base case is clearly true. But when I add $n+1$ to the right and left hand sides, I don't know what I'm supposed to get. For example, when I extend the right hand side's sequence by $n+1$, I get $n^3+(n+1)^3$ at the end of the sequence, which is $2n^3 +3n^2+3n+1$.

This doesn't seem that meaningful, so I don't know how to proceed.

Aditya Hase
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frustguy750
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  • @Integrator Use "flag" and put the URL of the previous. – Suzu Hirose Nov 21 '14 at 03:53
  • @Integrator That's not the same identity ... – Zubin Mukerjee Nov 21 '14 at 04:01
  • @Zubin Yes it is, since $1+2+\dots+n=\frac{n(n+1)}2$. – Mario Carneiro Nov 21 '14 at 04:02
  • It's enough of a rearrangement to justify another question, yes? – Zubin Mukerjee Nov 21 '14 at 04:02
  • @Zubin It's a straight composition of two mathematical facts which can already be found on MSE. If all these merited separate question, there would be way too much noise. – Mario Carneiro Nov 21 '14 at 04:04
  • @MarioCarneiro Some theorems directly follow from definitions. Perhaps we should just not write out the theorems because they're "too much noise"? – Zubin Mukerjee Nov 21 '14 at 04:05
  • @Integrator Could you link to it in a comment so it shows up in "Linked"? – Mario Carneiro Nov 21 '14 at 04:08
  • @Integrator Okay, then it'd make sense to link those instead of the one you did ... perhaps I'm being pedantic – Zubin Mukerjee Nov 21 '14 at 04:08
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    @Zubin I think it's bad form to link a duplicate to another duplicate – Mario Carneiro Nov 21 '14 at 04:09
  • @MarioCarneiro Not at all, if the duplicate has answers of its own. If A points to B and C points to B, the content of A remains unnoticed. If C points to A instead, all three are easily accessible. –  Nov 21 '14 at 04:19
  • @Rafflesiaarnoldii Actually, the contents of all duplicates are available from the "hub" question which links to these in the "linked" sidebar, but ideally all answers should be focused on the hub question so one doesn't have to run around to find a good answer. (FYI there are at least four exact duplicates of this question which can be found this way.) – Mario Carneiro Nov 21 '14 at 04:23

1 Answers1

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When you extend the right side you don't get $n+(n+1)^3$, you get $\sum_{i=1}^ni^3+(n+1)^3$ In the original question it was $n$, but in the edit it was $n^3$. Both are incorrect.

Ross Millikan
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  • I assumed he was just writing about the last two terms, which are $n^3$ and $(n+1)^3$; the reason I edited it is because he said whatever he was talking about was equal to $2n^3 + 3n^2 + 3n + 1$, which is only true if he was talking about $n^3 + (n+1)^3$ and not $n+(n+1)^3$. Also, shouldn't you have made that a comment and not an answer? – Zubin Mukerjee Nov 21 '14 at 03:59
  • @Integrator: I thought this would be useful to OP to see where the problem was. I was the second close vote. – Ross Millikan Nov 21 '14 at 03:59
  • @Integrator It is not a duplicate of the question you linked. That is a different identity. – Zubin Mukerjee Nov 21 '14 at 04:00
  • @ZubinMukerjee: it looks the same to me, once you make the left side inside the parentheses the triangular number. – Ross Millikan Nov 21 '14 at 04:02
  • @RossMillikan Is $20/2 = 10$ the same identity as $5+5=20/2$? Seems like quibbling over words ... EDIT: maybe a better example is the well-ordering principle and mathematical induction, which are equivalent but shouldn't be called the same thing – Zubin Mukerjee Nov 21 '14 at 04:07
  • @Integrator You seem very particular about certain rules of the site, but you just upvoted an answer that was not an answer and should be a comment. – Zubin Mukerjee Nov 21 '14 at 04:13