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..

Your solution is very close. Your last calculation shows
\begin{equation}
\sum_{n=1}^{k}n(n+1)=k^2(k+1)-2\sum_{n=1}^{k-1}n(n+1).
\end{equation}
The sums on the left and right side are almost the same, except for the coefficient and the limits of summation. If you note
$$\sum_{n=1}^{k-1}n(n+1)=\sum_{n=1}^{k}n(n+1)-k(k+1),$$
then you can immediately write the first identity as
\begin{align}
\sum_{n=1}^{k}n(n+1)&=k^2(k+1)-2\sum_{n=1}^{k}n(n+1)+2k(k+1)\\
&=k(k+1)(k+2)-2\sum_{n=1}^{k}n(n+1).
\end{align}
Now, both sides have your desired sum $\sum_{n=1}^{k}n(n+1)$, which you can solve for and you have your result.
