# Proof by induction help. I seem to be stuck and my algebra is a little rusty [duplicate]

Stuck on a homework question with mathematical induction, I just need some help factoring and am getting stuck.

$\displaystyle \sum_{1 \le j \le n} j^3 = \left[\frac{k(k+1)}{2}\right]^2$

The induction part is: $\displaystyle \left[\frac{k(k+1)}{2}\right]^2 +(k+1)^3$ is where I am having a problem.

If you could give me some hints as to where to go since I keep getting stuck or writing the wrong equation.

I'll get to $\displaystyle \left[{k^2+2k\over2}\right]^2 + 2{(k+1)^3\over2}$

Any push in the right direction will be appreciated.

$(\frac{k(k+1)}{2})^2+(k+1)^3$

$=\frac{k^2(k+1)^2}{4}+(k+1)(k+1)^2$

$=\frac{(k+1)^2}{4}(k^2+4k+4)$

$=\frac{(k+1)^2}{4}(k+2)^2$

I assume you mean that you're trying to show that $$1^3+\dots+k^3=\left[\frac{k(k+1)}2\right]^2,\tag{1}$$ as otherwise, it makes no sense. For the induction step, you're supposing that $(1)$ holds for some $k$, and wish to show that $$1^3+\dots+k^3+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2.$$ By $(1)$, we need only show that $$\left[\frac{k(k+1)}2\right]^2+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2,$$ or equivalently, that $$(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2-\left[\frac{k(k+1)}2\right]^2.\tag{2}$$

Hint: $a^2-b^2=(a+b)(a-b)$ will let you rewrite the right-hand side of $(2)$ in a nicer way, without having to go through nearly as much polynomial multiplication.

\begin{align} \left(\frac{k(k+1)}{2}\right)^2+(k+1)^3=\frac{k^2(k+1)^2+4(k+1)(k+1)^2}{4}\\ =\frac{(k+1)^2(k^2+4k+4)}{4}\\ =\frac{(k+1)^2(k+2)^2}{4}\\ =\left(\frac{(k+1)(k+2)}{2}\right)^2\\ \end{align}

You have: \begin{align*} \left[ \frac{n (n + 1)}{2} \right]^2 + (n + 1)^3 &= \frac{n^2 (n + 1)^2 + 4 (n + 1)^2 (n + 1)}{4} \\ &= \frac{(n^2 + 4 (n + 1)) (n + 1)^2}{4} \\ &= \frac{(n^2 + 4 n + 4) (n + 1)^2}{4} \\ &= \frac{(n + 2)^2 (n + 1)^2}{4} \\ &= \left[\frac{(n + 1) (n + 2)}{2} \right]^2 \end{align*}