Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?

After checking this for k=1, k=2 you need to follow this:
Let's say $A_1=(1+2+...+k)^2$; $B_1=1^3+...k^3$ and $A_2=(1+2+...+ (k+1))^2$
$B_2=1^3+...(k+1)^3$
Then to use induction it is needed to prove that $A_2-A_1=B_2-B_1$ or:
$$(1+2+...+ (k+1))^2-(1+2+...+k)^2=1^3+...(k+1)^3-1^3+...k^3$$
And it is equal to:
$$(k+1)(1+2+...+k)+(1+2+...+k+1)=(k+1)^3$$
$$(k+1)(\frac{k(k+1)}{2}+\frac{(k+1)(k+2)}{2})=(k+1)^3$$
$$(k+1)^2\frac{2k+2}{2}=(k+1)^3$$
$$(k+1)^3=(k+1)^3$$
This is the prove.
