Hard combinatorial identity: $\sum_{l=0}^p(-1)^l\binom{2l}{l}\binom{k}{p-l}\binom{2k+2l-2p}{k+l-p}^{-1}=4^p\binom{k-1}{p}\binom{2k}{k}^{-1}$ 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 $\binom{k-1}{p}=0$
 A: 
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).
A: Let's adopt the definition of the Binomial coefficient through the Falling Factorial.
$$
\left( \matrix{
  r \cr 
  m \cr}  \right) = \left\{ {\matrix{
   {{{r^{\,\underline {\,m\,} } } \over {m!}}} & {\left| {\;0 \le m \in \mathbb Z} \right.}  \cr 
   0 & {\left| {\;\neg \left( {0 \le m \in \mathbb Z} \right)} \right.}  \cr 
 } } \right.
$$
with $r$ real or even complex.
Now, the binomial coefficient at the denominator of the sum shall be not null, that is it shall be 
$0 \le k-p$ with the above definition.  So let's put $m=k-p$ and rewite the proposed identity as
$$ \bbox[lightyellow] {  
\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,p} \right)} {\left( { - 1} \right)^{\,l} {{\left( \matrix{
  2l \cr 
  l \cr}  \right)\left( \matrix{
  m + p \cr 
  p - l \cr}  \right)} \over {\left( \matrix{
  2l + 2m \cr 
  m + l \cr}  \right)}}}  = 4^{\,p} {{\left( \matrix{
  m + p - 1 \cr 
  p \cr}  \right)} \over {\left( \matrix{
  2m + 2p \cr 
  m + p \cr}  \right)}}\quad \left| {\;0 \le m,p \in \mathbb Z} \right.
} \tag{1}$$
Note that  the bounds of the sum can be omitted since they are implicit in the binomials.
With the adopted definition, the central binomial can be written as
$$
\left( \matrix{
  2n \cr 
  n \cr}  \right) = 2^{\,2\,n} \left( \matrix{
  n - 1/2 \cr 
  n \cr}  \right) = \left( { - 1} \right)^{\,n} 2^{\,2\,n} \left( \matrix{
   - 1/2 \cr 
  n \cr}  \right) = \left( { - 4} \right)^{\,n} \left( \matrix{
   - 1/2 \cr 
  n \cr}  \right)
$$
Let's then remind that, for the Rising and Falling Factorial we have
$$
\eqalign{
  & 1 = x^{\,\underline {\,0\,} }  = x^{\,\underline {\,m\,} } \left( {x - m} \right)^{\,\underline {\, - m\,} }   \cr 
  & \left( {x + 1} \right)^{\,\overline {\, - m\,} }  = \left( {x - m} \right)^{\,\underline {\, - m\,} }  = {1 \over {x^{\,\underline {\,m\,} } }}\quad
 x^{\,\underline {\, - m\,} }  = {1 \over {\left( {x + m} \right)^{\,\underline {\,m\,} } }} = {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }} \cr} 
$$
That premised, the summands at the LHS can be written as
$$ \bbox[lightyellow] {  
\eqalign{
  & \left( { - 1} \right)^{\,l} {{\left( \matrix{
  2l \cr 
  l \cr}  \right)\left( \matrix{
  m + p \cr 
  p - l \cr}  \right)} \over {\left( \matrix{
  2l + 2m \cr 
  m + l \cr}  \right)}} = {{4^{\,l} \left( \matrix{
   - 1/2 \cr 
  l \cr}  \right)\left( \matrix{
  m + p \cr 
  m + l \cr}  \right)} \over {4^{\,m + l} \left( \matrix{
  m + l - 1/2 \cr 
  m + l \cr}  \right)}} =   \cr 
  &  = {1 \over {4^{\,m} l!}}{{\left( { - 1/2} \right)^{\,\underline {\,l\,} } \left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} \cr} 
} \tag{2}$$
And we can rewrite the RHS as
$$ \bbox[lightyellow] {  
\eqalign{
  & RHS = 4^{\,p} {{\left( \matrix{
  m + p - 1 \cr 
  p \cr}  \right)} \over {\left( \matrix{
  2m + 2p \cr 
  m + p \cr}  \right)}} = 4^{\,p} {{\left( \matrix{
  m + p - 1 \cr 
  p \cr}  \right)} \over {4^{\,m + p} \left( \matrix{
  m + p - 1/2 \cr 
  m + p \cr}  \right)}} =   \cr 
  &  = {1 \over {4^{\,m} }}{{\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,p} \right)} {\left( \matrix{
   - 1/2 \cr 
  l \cr}  \right)\left( \matrix{
  m + p - 1/2 \cr 
  p - l \cr}  \right)} } \over {\left( \matrix{
  m + p - 1/2 \cr 
  m + p \cr}  \right)}} \cr} 
} \tag{3}$$
and we can demonstrate that each summand on the left corresponds to a summand on the right
$$ \bbox[lightyellow] {  
\eqalign{
  & {1 \over {4^{\,m} l!}}{{\left( { - 1/2} \right)^{\,\underline {\,l\,} } \left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {1 \over {4^{\,m} }}{{\left( \matrix{
   - 1/2 \cr 
  l \cr}  \right)\left( \matrix{
  m + p - 1/2 \cr 
  p - l \cr}  \right)} \over {\left( \matrix{
  m + p - 1/2 \cr 
  m + p \cr}  \right)}}  \cr 
  & \quad \quad  \Downarrow   \cr 
  & {{\left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {{\left( {m + p} \right)!\left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } } \over {\left( {p - l} \right)!\left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } }}  \cr 
  & \quad \quad  \Downarrow   \cr 
  & {{\left( {m + p} \right)^{\,\underline {\,m + l\,} } } \over {\left( {m + l} \right)!}}{1 \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {{\left( {m + p} \right)!} \over {\left( {p - l} \right)!\left( {m + l} \right)!}}{{\left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } } \over {\left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } }}  \cr 
  & \quad \quad  \Downarrow   \cr 
  & {1 \over {\left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} } }} = {{\left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } } \over {\left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} } }}  \cr 
  & \quad \quad  \Downarrow   \cr 
  & \left( {m + p - 1/2} \right)^{\,\underline {\,m + p\,} }  = \left( {m + p - 1/2} \right)^{\,\underline {\,\,p - l\, + m + l\,} }  =   \cr 
  &  = \left( {m + p - 1/2} \right)^{\,\underline {\,p - l\,} } \left( {m + l - 1/2} \right)^{\,\underline {\,m + l\,} }  \cr} 
} \tag{4}$$
