Induction With The Sum Of Cubes

Consider the set $$S=\left\{n\in\mathbb{N}\mid 1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2\right\}.$$ We need to prove that $S=\mathbb{N}$. We proceed by induction. We know $1\in S$ because $1^3 =1$ and $\left(\frac{1(1+1)}{2}\right)^2 = (1)^2=1$. Assume for some $k\in \mathbb{N}$ that $k\in S$, which means we've assumed that $$1^3+2^3+\cdots+k^3=\left(\frac{k(k+1)}{2}\right)^2.$$ We need to prove that $k+1\in S$, which means we need to prove that $$1^3+2^3+\cdots+k^3+(k+1)^3=\left(\frac{(k+1)((k+1)+1)}{2}\right)^2.$$ We'll start with the expression $1^3+2^3+\cdots+k^3+(k+1)^3$ and modify it to obtain the right hand side of the equation above. We now compute $$\begin{align} 1^3+2^3+\cdots+k^3+(k+1)^3 &= \left(1^3+2^3+\cdots+k^3\right)+(k+1)^3 && (\text{group the first $k$ terms})\\ &= \left(\frac{k(k+1)}{2}\right)^2+(k+1)^3 && (\text{substitute using our assumption})\\ &= \frac{k^2(k+1)^2}{4}+\frac{4(k+1)^3}{4} && (\text{prepare to combine fractions})\\ &= \frac{(k+1)^2(k^2+4(k+1))}{4}&& (\text{combine and factor})\\ &= \frac{(k+1)^2(k^2+4k+4)}{4}&& (\text{expand})\\ &= \frac{(k+1)^2(k+2)^2}{4}&& (\text{factor})\\ &= \left(\frac{(k+1)((k+1)+1)}{2}\right)^2&& (\text{note $k+2=(k+1)+1$}) .\end{align}$$ This shows that if $k\in S$, then $k+1\in S$. By mathematical induction, we know that $S=\mathbb{N}$. This prove that for every $n\in \mathbb{N}$, we have $$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$