An expression for the sum $\sum\limits _{k=1}^{n-1} k \, (n-k)^2$

Question

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!

Answer 1

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.

Answer 2

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 $\sum_{\ell=1}^{n-1}\ell^3$?

Answer 3

You want to find

$$\sum_{k=1}^{n-1} k \, (n-k)^2=\sum_{k=1}^{n-1} (kn^2-2nk^2+k^3)= n^2\sum_{k=1}^{n-1} k-2n\sum_{k=1}^{n-1} k^2+\sum_{k=1}^{n-1} k^3$$

$$\sum_{k=1}^{n-1} k \, (n-k)^2= n^2\frac{(n-1)n}{2}-2n\frac{(n-1)n(2n-1)}{6}+\left(\frac{(n-1)n}{2}\right)^2$$

Answer 4

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 $$\sum_{k=1}^{n-1} k \, (n-k)^{2} = \frac{n^2 \, (n^2 -1)}{12}.$$

Answer 5

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

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

Answer 6

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.