Prove by induction $\sum_{k=1}^n k^3 =( \sum_{k=1}^n k )^2 $

Question

Can anyone show me how to prove this example by induction? I can't figure it out. $$\sum_{k=1}^n k^3 =\left( \sum_{k=1}^n k \right)^2 $$

Answer 1

Hint You can first pose $S_n = \sum_{k=1}^n k $ and apply the identity $(a+b)^2 = a^2 + 2ab + b^2$ with $S_{n+1}^2 = (S_n + (n+1))^2$. Using the recurrence hypothesis, you will find the result.

Answer 2

Let $u_n=\sum_{k=1}^n k^3$ and $v_n=(\sum_{k=1}^n k)^2$. You have to prove $u_n=v_n$ for all $n\ge1$. The base case is clear. Now, $$u_{n+1}=u_n+(n+1)^3$$ and $$v_{n+1}=v_n^2+2(n+1)\left(\sum_{k=1}^n k\right) + \left(\sum_{k=1}^n k\right)^2$$ So all you have to do is compare those two induction steps.

Answer 3

Inductive step: $$\sum_{k=1}^{n+1} k^3=\left(\sum_{k=1}^n k\right)^2+(n+1)^3=\left(\frac{n(n+1)}{2}\right)^2+(n+1)^3=(n+1)^2\left(\frac{n^2}{4}+n+1\right)=\\ \frac{(n+1)^2(n+2)^2}{4}=\left(\frac{(n+1)(n+2)}{2}\right)^2=\left(\sum_{k=1}^{n+1}k\right)^2$$