Nested sum of binomial coefficient in closed form

Question

Let $\ell$ be an even positive integer. Express $$\sum_{k=0}^n\sum_{i=0}^\ell (-1)^i\binom{n}{k}^2\binom{2k}{i}\binom{2n-2k}{\ell-i}$$in closed form.

Is anybody to give a insight, please? [A Path to Combinatorics for Undergraduates] p.192 problem 8.9

Answer 1

Let $\ell=2m$, then $$S_{n,m}:=\sum_{k=0}^n\sum_{i=0}^{2m} (-1)^i\binom{n}{k}^2\binom{2k}{i}\binom{2n-2k}{2m-i}=\binom{2m}{m}\binom{2(n-m)}{n-m}.$$

Note that $$\sum_{i=0}^{2m} (-1)^i\binom{2k}{i}\binom{2n-2k}{2m-i}=[x^{2m}](1-x)^{2k}(1+x)^{2n-2k},$$ Hence $$\begin{align} S_{n,m}&=[x^{2m}]\sum_{k=0}^n\binom{n}{k}^2(1-x)^{2k}(1+x)^{2n-2k}\\ &=[x^{2m}]4^nx^n P_n\left(\frac{1}{2}\left(x+\frac{1}{x}\right)\right)\\ &=[x^{2m}]\sum_{m=0}^n \binom{2m}{m}\binom{2(n-m)}{n-m}x^{2m} =\binom{2m}{m}\binom{2(n-m)}{n-m} \end{align}$$ where $P_{n}$ is the $n$-th Legendre Polynomial.