Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.

Notice $$
\begin{align}
k^3 
&= \color{red}{(k^3 - k)} + \color{blue}{k}\\
&= \color{red}{(k-1)k(k+1)} + \color{blue}{k}\\
&= \frac{\color{red}{(k-1)k(k+1)}(k+2)-(k-2)\color{red}{(k-1)k(k+1)}}{4}
+ \frac{\color{blue}{k}(k+1) - (k-1)\color{blue}{k}}{2}
\end{align}$$
is telescoping. We have
$$\begin{align}
\sum_{k=0}^n k^3 
&= \frac{(n-1)n(n+1)(n+2)}{4} + \frac{n(n+1)}{2}\\
&= \frac{n(n+1)}{2}\left(\frac{(n-1)(n+2)}{2} + 1\right)\\
&= \left(\frac{n(n+1)}{2}\right)^2
\end{align}
$$
