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!
 A: Perhaps not the most elegant, but something which works with minimum "trick" factor.
So, you have
$$
S_n \stackrel{\rm def}{=} \sum_{k=1}^{n-1} k(n-k)^2 = \sum_{\ell=1}^{n-1} (n-\ell)\ell^2
$$
Expanding out, you get
$$S_n = n\sum_{\ell=1}^{n-1}\ell^2-\sum_{\ell=1}^{n-1}\ell^3
$$
Now, do you know the closed-form expression for $\sum_{\ell=1}^{n-1}\ell^2$ and $\sum_{\ell=1}^{n-1}\ell^3$?
A: This is probably not the cleverest way, but it should work. You're trying to evaluate $\sum_{k=1}^{n-1}k(n-k)^2$. $$\sum_{k=1}^{n-1}k(n-k)^2=\sum(kn^2-2nk^2+k^3)=n^2\sum k-2n\sum k^2+\sum k^3$$ I know you can work out the first of those three sums. The other two are not as easy, but they are well-known and I'm sure you'll have no trouble finding places where they are done, even on this website. 
A: You want to find 
$$\sum_{k=1}^{n-1} k \, (n-k)^2=\sum_{k=1}^{n-1}  (kn^2-2nk^2+k^3)= n^2\sum_{k=1}^{n-1} k-2n\sum_{k=1}^{n-1} k^2+\sum_{k=1}^{n-1} k^3$$
$$\sum_{k=1}^{n-1} k \, (n-k)^2= n^2\frac{(n-1)n}{2}-2n\frac{(n-1)n(2n-1)}{6}+\left(\frac{(n-1)n}{2}\right)^2$$
A: A convolution approach is missing so far, so I am willing to introduce it. We may consider that by stars and bars, for any $|z|<1$ we have
$$ f(z)=\sum_{n\geq 1} n z^n = \frac{z}{(1-z)^2}, \qquad g(z)=\sum_{n\geq 1} n^2 z^n = \frac{z(1+z)}{(1-z)^3}$$
hence
$$ \sum_{k=1}^{n}k(n-k)^2 = [z^n] f(z)g(z) = [z^n]\frac{z^2(1+z)}{(1-z)^5} =\color{red}{\frac{n^2(n^2-1)}{12}}.$$
With or without this convolution approach, it is pretty clear that the LHS is a polynomial in the $n$ variable with degree $4$, hence it can be computed by interpolation, through the values of the LHS at $n\in\{1,2,3,4,5\}$, for instance.
A: 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
$$\sum_{k=1}^{n-1} k \, (n-k)^{2} = \frac{n^2 \, (n^2 -1)}{12}.$$
A: 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
$$
\sum_{k=1}^{n-1}k(n-k)^2 = n\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k^3
$$
