Simplify $\sum_{j=0}^{m}\sum_{\ell=0}^{p}(-1)^{j+\ell}\binom{j+k}{j}\binom{2(k+m)-p}{2(k+j)}\binom{\ell}{m-j}\binom{2m+k}{p-\ell}$

Question

Let $m\in\mathbb{Z}_{>0}$ and $p\in\mathbb{Z}$ with $0\leqslant p\leqslant 2m.$ Then how could we simplify the following double sum? \begin{align*} \sum_{j=0}^{m}\sum_{\ell=0}^{p}(-1)^{j+\ell}\binom{j+k}{j}\binom{2(k+m)-p}{2(k+j)}\binom{\ell}{m-j}\binom{2m+k}{p-\ell} \end{align*}

Answer 1

Here is a starter. We can find a closed form for the inner sum.

We obtain \begin{align*} \color{blue}{\sum_{\ell=0}^p}&\color{blue}{\binom{l}{m-j}\binom{2m+k}{p-\ell}(-1)^{\ell}}\\ &=\sum_{\ell=m-j}^p\binom{\ell}{m-j}\binom{2m+k}{p-\ell}(-1)^{\ell}\tag{1}\\ &=\sum_{\ell=0}^{p-m+j}\binom{\ell+m-j}{m-j}\binom{2m+k}{p-\ell-m+j}(-1)^{\ell+m-j}\tag{2}\\ &=\sum_{\ell=0}^{p-m+j}\binom{\ell+m-j}{\ell}\binom{2m+k}{p-m+j-\ell}(-1)^{\ell+m-j}\tag{3}\\ &=\sum_{\ell=0}^{p-m+j}\binom{-m+j-1}{\ell}\binom{2m+k}{p-m+j-\ell}(-1)^{m-j}\tag{4}\\ &\,\,\color{blue}{=\binom{m+k+j-1}{p-m+j}(-1)^{m-j}}\tag{5} \end{align*}

Comment:

• In (1) we set the lower index to $m-j$ since $\binom{p}{q}=0$ if $0\leq p<q$.

• In (2) we shift the index to start with $\ell=0$.

• in (3) we use $\binom{p}{q}=\binom{p}{p-q}$.

• In (4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (5) we use Vandermonde's identity.

We conclude \begin{align*} \color{blue}{\sum_{j=0}^{m}}&\color{blue}{\sum_{\ell=0}^{p}(-1)^{j+\ell}\binom{j+k}{j}\binom{2(k+m)-p}{2(k+j)}\binom{\ell}{m-j}\binom{2m+k}{p-\ell}}\\ &\,\,\color{blue}{=(-1)^m\sum_{j=0}^{m}\binom{j+k}{j}\binom{2(k+m)-p}{2(k+j)}\binom{m+k+j-1}{p-m+j}} \end{align*}