The sum of the cubes of the first n natural numbers

Submitted by Structure on Wed, 12/12/2007 - 22:00.

We shall prove that

$\sum_{k=1}^{k=n}k^3=1^3+2^3+3^3+...+n^3=\frac{n^2(n+1)^2}{4}$

First proof is by induction.
For n=1 we have

$1^3=\frac{1^22^2}{4}=1$

We shall prove that

$\sum_{k=1}^{k=n+1}=1^3+2^3+3^3+...+n^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$

$\sum_{k=1}^{k=n}k^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$

But from induction supposition we have to prove

$\frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$

$(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}-\frac{n^2(n+1)^2}{4}=\frac{(n+1)^2}{4}((n+2)^2-n^2)=(n+1)^3$