# Telescoping Cubes

As a practice for the telescoping tool, I'll evaluate the well known sum $\displaystyle\sum_{k=1}^{n}k^{3}=\bigg[\frac{n(n+1)}{2}\bigg]^{2}.$

The idea is to find terms $\{u_{k}\}$ such that $k^{3}=u_{k+1}-u_{k}.$ This done we would be able to write

\begin{align}\displaystyle\sum_{k=1}^{n}k^{3}&=\sum_{k=1}^{n}(u_{k+1}-u_{k})\\ &=\sum_{k=1}^{n}u_{k+1}-\sum_{k=1}^{n}u_{k}\\ &=\sum_{k=2}^{n+1}u_{k}-\sum_{k=1}^{n}u_{k}\\ &=u_{n+2}-u_{1}. \end{align}

Obviously, $u_{k}$ needs to be determined up to a constant and, knowing the answer, we can get the required formula. However, setting out to find $\{u_{k}\}$ as the altimate goal may be rather unnecessary. Cube terms appear in the difference $(k+1)^4-k^{4}=4k^{3}+6k^{2}+4k+1.$ The extra terms do not cause an unsurmountable hindrance:

\begin{align}\displaystyle 4\sum_{k=1}^{n}k^{3}&=\sum_{k=1}^{n}((k+1)^4-k^{4}-6k^{2}-4k-1)\\ &=\sum_{k=1}^{n}((k+1)^4-k^{4})-\sum_{k=1}^{n}(6k^{2}+4k+1)\\ &=((n+1)^4-1)-6\sum_{k=1}^{n}k^{2}-4\sum_{k=1}^{n}k-\sum_{k=1}^{n}1. \end{align}

We may now recollect that $\displaystyle\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6},$ $\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2},$ and $\displaystyle\sum_{k=1}^{n}1=n$ which leads to

\begin{align}\displaystyle 4\sum_{k=1}^{n}k^{3}&=((n+1)^4-1)-6\sum_{k=1}^{n}k^{2}-4\sum_{k=1}^{n}k-\sum_{k=1}^{n}1\\ &=(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n, \end{align}

which may not be as nice a formula as we might have hoped to obtain but it is verifiably correct. Now, let's simplify:

\begin{align}\displaystyle 4\sum_{k=1}^{n}k^{3}&=(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n\\ &=(n+1)[(n+1)^3-n(2n+1)-2n-1]\\ &=(n+1)[(n+1)^3-(2n+1)(n+1)]\\ &=(n+1)^{2}[(n+1)^2-(2n+1)]\\ &=(n+1)^{2}n^{2}. \end{align}

Not too difficult.

Note: there is a whole page of telescoping examples.