Sum of cubes proof Prove that for any natural number n the following equality holds: 
$$ (1+2+ \ldots + n)^2 = 1^3 + 2^3 + \ldots + n^3 $$
I think it has something to do with induction?
 A: By induction:
for $n=1$ it works. Then, suppose it works for a $n$. Then,
$$(1+...+n+(n+1))^2=\underbrace{(1+...+n)^2}_{=1^3+...+n^3\ by\ hyp.}+2(1+...+n)(n+1)+(n+1)^2$$$$=1^3+...+n^3+(n+1)\big(2\underbrace{(1+...+n)}_{=\frac{n(n+1)}{2}}+(n+1)\big)$$$$=1^3+...+n^3+(n+1)\big(n(n+1)+(n+1)\big)$$$$=1^3+...+n^3+(n+1)(n+1)(n+1)$$$$=1^3+...+n^3+(n+1)^3.$$
Q.E.D.
A: 
Here an illustration. The surface is the square of the sum.
Prove that each added layer (in different color) is a the wanted cube and you will be done.
A: This solution assumes you are allowed to use 
$$
V_1 = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}\\
V_2 = \sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}
$$
Use the perturbation method (set the sum of cubes equal to $V_3$. Consider 
$$
S_n = \sum_{k=1}^{n}k^4
$$
then
$$
S_n + (n+1)^4 = 1 + \sum_{k=1}^{n}(k+1)^4 = 1+S_n + 4 \sum_{k=1}^{n} k^3 + 4 \sum_{k=1}^{n} k^2 +  \sum_{k=1}^{n} k +n 
$$
Obviously $S_n$ cancels out, you know $V_1$ and $V_2$, so you can get the value for $V_3$ and see that it's equal to $\frac{(n(n+1))^2}{4}$.
A: There is another way, if you like telescoping:
Let $V_k = (k-1)*k*(k+1)*(k+2)$  and $U_k$ = Vk+1 - Vk
$U_k$ = $k*(k+1)*(k+2)*[(k+3) - (k-1)]$ = $4*k*(k+1)*(k+2)$ 
= $4*k^3 + 12*k^2 + 8*k$
So if you know $1+4+9+..+n^2$ you can get your sum pretty easily by summing the $U_k$ from 1 to n-1, you will get: 
$V_n$ -0 = $4*S_n + 12*C_n + 8*D_n$ , where $S_n$ is the partial sum of square and $C_n$ the partial sum of the cubes, and $D_n$ the partial sum of integers
$S_n = \frac{n*(n+1)*(2*n+1)}{6}$ , $D_n = \frac{n*(n+1)}{2}$,  if you don't know their value
