How to prove that $\forall n\in \mathbb{N}$, $\sum ^{n}_{i=1}i^{3}=\frac {n^{2}(n+1)^{2}}{4}$?

Question

Use mathematical induction to prove that $\forall n\in \mathbb{N}$, $$\sum ^{n}_{i=1}i^{3}=\dfrac {n^{2}(n+1)^{2}}{4}$$

$$\begin{align*} \sum_{k=1}^{n+1} k^3 &= \sum_{k=1}^{n} k^3 + (n+1)^2 \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^2 \\ &= \dfrac {n^{2}(n+1)^{2}+4(n+1)^2}{4} \end{align*}$$

Is it true? What to do next?

Answer 1

First, don't forget the base case, $n = 1$: It holds.

Your inductive hypothesis is fine, but your inductive step is off.

You want: $$\begin{align*} \sum_{k=1}^{n+1} k^3 &= \left(\sum_{k=1}^{n} k^3\right) + (n+1)^{\color{blue}{\bf 3}} \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^\color{blue}{\bf 3} \\ &= \dfrac {n^{2}(n+1)^{2}+4(n+1)^3}{4}\\ &= \frac{(n+1)^2(n^2 + 4(n+1))}{4}\\ &=\frac{(n+1)^2 (n^2 + 4n + 4)}{4}\\ & = \frac{(n+1)^2(n+2)^2}{4} \end{align*}$$