TITLE: Hard combinatorial identity: $\sum_{l=0}^p(-1)^l\binom{2l}{l}\binom{k}{p-l}\binom{2k+2l-2p}{k+l-p}^{-1}=4^p\binom{k-1}{p}\binom{2k}{k}^{-1}$
QUESTION [5 upvotes]: Does anyone have idea to prove that?
$$\sum_{\ell=0}^{p}(-1)^{\ell}\dfrac{\binom{2\ell}{\ell}\binom{k}{p-\ell}}{\binom{2k+2\ell-2p}{k+\ell-p}} = \dfrac{4^p\binom{k-1}{p}}{\binom{2k}{k}}$$ is true for all
$k \in \mathbb{N}, p \in \mathbb{N} \cup \{0\}$.
We use the convention that, if $p>k-1$, then $\binom{k-1}{p}=0$

REPLY [3 votes]: We obtain
  \begin{align*}
\color{blue}{\sum_{l=0}^p}&\color{blue}{(-1)^l\binom{2l}{l}\binom{k}{p-l}\binom{2k+2l-2p}{k+l-p}^{-1}}\\
&=\sum_{l=0}^p(-1)^{p-l}\binom{2p-2l}{p-l}\binom{k}{l}\binom{2k-2l}{k-l}^{-1}\tag{1}\\
&=4^{p-k}\sum_{l=0}^p\binom{-\frac{1}{2}}{p-l}\binom{k}{l}\binom{k-l-\frac{1}{2}}{k-l}^{-1}\tag{2}\\
&=4^{p-k}\binom{k-\frac{1}{2}}{k}^{-1}\sum_{l=0}^p\binom{-\frac{1}{2}}{p-l}\binom{k-\frac{1}{2}}{l}\tag{3}\\
&=4^{p-k}\binom{k-\frac{1}{2}}{k}^{-1}\binom{k-1}{p}\tag{4}\\
&\color{blue}{=4^p\binom{k-1}{p}\binom{2k}{k}^{-1}}\tag{5}
\end{align*}
  and the claim follows.

Comment:

In (1) we change   the order of summation $l\rightarrow  p-l$.
In (2) we use the binomial identities $$(-4)^n\binom{-\frac{1}{2}}{n}=\color{green}{\binom{2n}{n}}= 4^n\binom{n-\frac{1}{2}}{n}$$
In (3) we use the binomial identity (the essence)
\begin{align*}
\binom{k-\frac{1}{2}}{l}\binom{k-\frac{1}{2}}{k}^{-1}
&=\frac{\left(k-\frac{1}{2}\right)^{\underline{l}}}{l!}
\cdot\frac{k!}{\left(k-\frac{1}{2}\right)^{\underline{k}}}\\
&=\frac{k!}{l!}\cdot\frac{1}{\left(k-l-\frac{1}{2}\right)^{\underline{k-l}}}\\
&=\frac{k!}{l!(k-l)!}\cdot\frac{(k-l)!}{\left(k-l-\frac{1}{2}\right)^{\underline{k-l}}}\\
&=\binom{k}{l}\binom{k-l-\frac{1}{2}}{k-l}^{-1}\\
\end{align*}
with $n^{\underline{k}}=n(n-1)\cdots(n-k+1)$ the falling factorial.
In (4) we apply the Chu-Vandermonde identity.
In (5) we use the first identity stated in (2).