1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Integrator Use "flag" and put the URL of the previous. $\endgroup$ – Suzu Hirose Nov 21 '14 at 3:53
  • $\begingroup$ @Integrator That's not the same identity ... $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 4:01
  • $\begingroup$ @Zubin Yes it is, since $1+2+\dots+n=\frac{n(n+1)}2$. $\endgroup$ – Mario Carneiro Nov 21 '14 at 4:02
  • $\begingroup$ It's enough of a rearrangement to justify another question, yes? $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 4:02
  • 1
    $\begingroup$ @Zubin I think it's bad form to link a duplicate to another duplicate $\endgroup$ – Mario Carneiro Nov 21 '14 at 4:09
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 3:59
  • $\begingroup$ @Integrator: I thought this would be useful to OP to see where the problem was. I was the second close vote. $\endgroup$ – Ross Millikan Nov 21 '14 at 3:59
  • $\begingroup$ @Integrator It is not a duplicate of the question you linked. That is a different identity. $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 4:00
  • $\begingroup$ @ZubinMukerjee: it looks the same to me, once you make the left side inside the parentheses the triangular number. $\endgroup$ – Ross Millikan Nov 21 '14 at 4:02
  • $\begingroup$ @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 $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 4:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.