An expression for the sum $\sum\limits _{k=1}^{n-1} k \, (n-k)^2$ I really don't know how to find the sum of the series: $$\sum\limits _{k=1}^{n-1} k \, (n-k)^2 = 1(n-1)^2+2(n-2)^2+3(n-3)^2+\dots+(n-1)1^2.$$ 
My attempt:
I tried to approach the old school approach of how we find the sum of arithmetic-co geometric progression but unable to do so.
Any help is appreciated!

For the series
$$\sum_{k=1}^{n-1} k \, (n-k)^{2}$$
consider 
$$\sum_{k=1}^{n-1} k \, (n-k)^{2} = n^{2} \, \sum_{k=1}^{n-1} k - 2n \, \sum_{k=1}^{n-1} k^{2} + \sum_{k=1}^{n-1} k^{3}$$
and use 
\begin{align}
\sum_{k=1}^{n} k &= \frac{n(n+1)}{2} \\
\sum_{k=1}^{n} k^{2} &= \frac{n(n+1)(2n+1)}{6} \\
\sum_{k=1}^{n} k^{3} &= \frac{n^2 (n+1)^2}{4}
\end{align}
to obtain
$$\sum_{k=1}^{n-1} k \, (n-k)^{2} = \frac{n^2 \, (n^2 -1)}{12}.$$
