Computing $\sum\limits_{r=1}^{n} r^{4}\binom{n}{r}^{2}$

Question

I want to prove the following identity: $$\sum\limits_{r=1}^{n} r^{4}\binom{n}{r}^{2}=\frac{4^{n-2}n^{2}(n^{3}+n^{2}-3n-1)\Gamma\left(n-\frac{3}{2}\right)}{\sqrt{π}\Gamma(n)}.$$ I saw a similar question here, but this is completely different. I know that $$\binom{n}{r}^{2}=\binom{n}{r}\binom{n}{n-r},$$ Also $$\binom{n}{n-k}=\oint_{|{z}|\ =\ 1}\frac{(1+z)^{n}}{z^{n+1-r}}\,\frac{\mathrm dz}{2πi},$$ I have no idea to started.

Answer 1

In fact, an "explicit" formula for $\sum_{r=0}^{n}r^k\binom{n}{r}^2$ (assuming $n,k$ are nonegative integers with $k$ fixed, and under the agreement of $0^0=1$) can be obtained using Stirling numbers of the second kind. We use the identity from the linked article, written as $$r^k=\sum_{s=0}^{k}{k\brace s}(r)_s=\sum_{s=0}^{k}s!{k\brace s}\binom{r}{s},$$ which gives $$\sum_{r=0}^{n}r^k\binom{n}{r}^2=\sum_{s=0}^{k}s!{k\brace s}\sum_{r=s}^{n}\binom{r}{s}\binom{n}{r}^2$$ (the lower limit of $r=\color{red}{s}$ justified by $\binom{r}{s}=0$ if $r<s$); now $\binom{r}{s}\binom{n}{r}=\binom{n}{s}\binom{n-s}{r-s}$ hence $$\sum_{r=0}^{n}r^k\binom{n}{r}^2=\sum_{s=0}^{k}s!{k\brace s}\binom{n}{s}\sum_{r=s}^{n}\binom{n}{r}\binom{n-s}{r-s}.$$ And the inner sum is computed using Vandermonde identity: it is $$\sum_{r=s}^{n}\binom{n}{n-r}\binom{n-s}{r-s}=\sum_{r=0}^{n-s}\binom{n-s}{r}\binom{n}{n-s-r}=\binom{2n-s}{n-s},$$ giving finally $$\color{blue}{\sum_{r=0}^{n}r^k\binom{n}{r}^2=}\sum_{s=0}^{k}s!{k\brace s}\binom{n}{s}\binom{2n-s}{n-s}\color{blue}{=\sum_{s=0}^{\min\{k,n\}}{k\brace s}\frac{(2n-s)!}{(n-s)!^2}.}$$

For $k=4$ and $n>1$, taking out $\frac{(2n-4)!}{(n-1)!(n-2)!}$, we're left with $n^2(n^3+n^2-3n-1)$ as claimed.