<s>[INST] Simplify $\sum n(n+1)$ According to WolframAlpha,
$$\sum_{n=1}^k n(n+1)=\frac{1}{3}k(k+1)(k+2)$$
and it is easy to verify this if we use induction.
However, I would like to know how one can actually come up with this, other than by thinking about how to force the terms to cancel out.
I tried: Since $k(k+1)/2=1+2+\cdots+k$,
$$\begin{align}
\sum_{n=1}^k n(n+1) &= 2\bigg(1(k)+2(k-1)+3(k-2)+\cdots+k(1)\bigg)\\
& = 2\sum_{n=1}^k n(k-n+1)\\
\sum_{n=1}^k n(n+1) &= 2\bigg(1(k)+2(k-1)+3(k-2)+\cdots+k(k-(k-1))\bigg)\\
& = 2\bigg( (1+2+3+\cdots+k)k-(0\cdot1+1\cdot2+2\cdot3+\cdots+(k-1)k) \bigg)\\
& = k^2(k+1)-2\sum_{n=1}^{k-1} n(n+1)\\
& = \sum_{n=1}^k (-2)^{k-n}n^2(n+1)
\end{align}$$
I feel like I'm only making it worse..
 [/INST] Note that $n(n+1) = 2\bigg(\frac{n(n+1)}{2!}\bigg)=2\binom{n+1}{2}$. Then the sum becomes ....
$$\begin{align}
S(k) &= \sum_{n=1}^k{n(n+1)}\\
&= 2\sum_{n=1}^k{\binom{n+1}{2}}\\
(1) &= 2\binom{k+2}{3}\\
&= 2\cdot\frac{k(k+1)(k+2)}{3!}\\
&= \frac{k(k+1)(k+2)}{3}\\
\end{align}$$
Where $(1)$ is just the hockey-stick identity.
</s>