# Calculus: Simplifying a sum of products of binomials

Let $j$ be an integer. Let $r$ be an even integer such that $r\in 1,\ldots,2j$.

For a given integer $u$, I want to prove the following identity:

$$\sum_{n=0}^u \sum_{k=0}^{2j}\sum_{p=0}^r \frac{\binom{k}{p}\binom{2j-k}{r-p}\binom{2j}{k}\binom{j-1-n}{u-n}}{(2n+2)!}(-1)^{r-p+u-n}2^{2(n-j+1)}(k-j)^2\prod_{l=1}^{n}((k-j)^2-l^2)=\begin{cases}\binom{2j}{r} &\text{ if }r=2(u+1)\\\\ 0 &\text{ else}\end{cases}$$

I tried with different values of $j$, $u$, and $r$ and this was always true. Does someone have an idea on how to prove it or prove it wrong?

Thank you!

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