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!

Hint.
$$
\sum_{k=1}^{n-1}(n-k)^3 = n\sum_{k=1}^{n-1}(n-k)^2-\sum_{k=1}^{n-1}k(n-k)^2
$$
then
$$
\sum_{k=1}^{n-1}k(n-k)^2 = n\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k^3
$$
