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?
Induction on k.

Base Case: 1=1

Induction Hypothesis: $(1+2+{...}+k)^2$=$1^3 + {...} + k^3$

Based on the hypothesis: 
$(1+2+{...}+k+(k+1))^2$

=$(1+2+{...}+k)^2$+2$(1+2+{...}+k)(k+1)$+$(k+1)^2$

=$1^3 + {...} + k^3$+$k(k+1)^2$+$(k+1)^2$

=$1^3 + {...} + k^3 + (k+1)^3$

Done.
We have $(a+b)^2=a^2+b^2 + 2ab.$ Thus $$(1+2+...+(k+1))^2=(1+2+...+k)^2+(k+1)^2+ 2(1+2+...+k)(k+1)=1^3+...+k^3+(k+1)^2 + 2(1+2+...+k)(k+1)$$ 

Using $1+...+k=\dfrac{k(k+1)}{2}$ we get $$(1+2+...+k+1)^2=1^3+...+k^3+ (k+1)^2 + k(k+1)^2=1^3+...+k^3+ (k+1)^2(k+1)=1^3+...+(k+1)^3$$ and the inductive step is proven.