How to proof that for all n $ (\sum_{i=1}^n i)^2 = \sum_{i=1}^n i^3 $

Question

What i had so far was this: claim: How to proof that for all n $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3 $$

Proof

To proof: How to proof that for all n $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3 $$

basis $n=1$:

$$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3\\ (1)^2 = (1^3)\\ 1 = 1$$

Induction $n = k$: $$ \left(\sum_{i=1}^k i\right)^2 = \sum_{i=1}^k i^3 $$ My assumption is that this is true

$n = k+1$: $$ \left(\sum_{i=1}^k+1 i\right)^2 = \sum_{i=1}^k+1 i^3\\ \left(\sum_{i=1}^k i +(k+1)\right)^2 = \sum_{i=1}^k i^3 +(k+1)^3 $$

I got stuck here and i dont know how to continue.

This has been discussed at MSE several times. Did you have a look here and the related questions to it? — Dietrich Burde, Oct 15 '17 at 18:12
I dont know what you mean by that but i guess so @Dr.SonnhardGraubner — bassie, Oct 15 '17 at 18:14
@bassie $$1^3+2^3+ \cdots + n^3 = \sum_{i=1}^n i^3$$ and $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ so it is indeed a duplicate and I suggest taking a look at the answers provided there. — Andrew Tawfeek, Oct 15 '17 at 18:19
You do realize that forcing the equal signs to be outside of the dollar signs makes it hard to read and confusing? Please don't do that. — Simply Beautiful Art, Oct 15 '17 at 20:23

score 0 · Answer 1 · answered Oct 15 '17 at 18:12

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Prove that $$\sum_{i=1}^k i^3=\frac{1}{4} k^2 (k+1)^2;\;\sum _{i=1}^k i=\frac{1}{2} k (k+1)$$

answered Oct 15 '17 at 18:12

Raffaele

How did you come to this equation from my equation – bassie Oct 15 '17 at 18:14
@bassie The other answerer has proved the formulae – Raffaele Oct 15 '17 at 18:54

1 Answers1