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?

We have:
$$(1+2+...+k+(k+1))^2=(1+2+...+k)^2+(k+1)^2+2(k+1)(1+2+...+k)$$
So by induction hypothesis:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+2(k+1)(1+2+...+k)$$
We know $1+2+...+k = \frac{k(k+1)}{2}$.(You can prove it by induction easily). Therefore:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+k(k+1)^2=1^3+2^3+...+k^3+(k+1)^3$$
