Prove that $\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}$. In an attempt to answer this thread, I discovered an identity involving binomial coefficients.  However, I am not able to find a proof.  All tricks are welcome.

Let $n$ and $k$ be nonnegative integers with $k\leq n$.  Prove that $$\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}\,.$$


It's convenient to use the coefficient of operator $[z^j]$ to denote the coefficient of $z^j$ in a series. This way we can write e.g.
\begin{align*}
[z^j](1+z)^n=\binom{n}{j}
\end{align*}

We recall the Euler transformation formula of a series
  \begin{align*}
A(z)=\sum_{n=  0}^\infty a_nz^n\qquad\qquad 
\frac{1}{1-z}A\left(\frac{z}{1-z}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^n\binom{n}{j}a_j\right)z^n
\end{align*}
  This transformation formula together with a proof can be found e.g.  in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev.
This  formula is useful, but it can be considerably generalized
\begin{align*}
\text{from}&\qquad&\sum_{j=0}^n\binom{n}{j}a_j&=[z^n]\frac{1}{1-z}A\left(\frac{z}{1-z}\right)\\
\text{to}&\qquad&\sum_{j=0}^n\binom{n+pj}{m+qj}a_j&=[z^n]\frac{z^m}{(1-z)^{m+1}}A\left(\frac{z^{q-p}}{(1-z)^q}\right)\qquad q>p\tag{1}
\end{align*}
A proof of (1) can be found in Theorem 2.1 in Riordan arrays and combinatorial sums by R. Sprugnoli.

We can now show:

The following is valid
  \begin{align*}
\sum_{j=k}^n(-1)^{j-k}\binom{j}{k}\binom{2n-j}{j}2^{2(n-j)}=\binom{2n+1}{2k+1}\tag{2}
\end{align*}

$$ $$

We start by rearranging the left hand side of (2)
  \begin{align*}
\sum_{j=k}^n&(-1)^{j-k}\binom{j}{k}\binom{2n-j}{j}2^{2(n-j)}\\
&=(-1)^k4^n\sum_{j=k}^n\binom{2n-j}{j}\binom{j}{k}\left(-\frac{1}{4}\right)^j\tag{3}
\end{align*}
In order to use (1) we derive a generating function $A(z)$ for 
  \begin{align*}
a_j=\binom{j}{k}\left(-\frac{1}{4}\right)^j
\end{align*}
  We obtain
  \begin{align*}
A(z)&=\sum_{j= 0}^\infty a_jz^j\\
&=\sum_{j=k}^\infty\binom{j}{k}\left(-\frac{1}{4}\right)^jz^j\\
&=\sum_{j=0}^\infty\binom{j+k}{j}\left(-\frac{1}{4}\right)^{j+k}z^{j+k}\tag{4}\\
&=\left(-\frac{z}{4}\right)^k\sum_{j=0}^\infty\binom{-k-1}{j}\left(\frac{z}{4}\right)^{j}\tag{5}\\
&=\frac{\left(-\frac{z}{4}\right)^k}{\left(1+\frac{z}{4}\right)^{k+1}}\tag{6}
\end{align*}

Comment:


*

*In (4) we shift the index so that $j$ starts from $0$.

*In (5) we apply the binomial identity $\binom{-u}{v}=\binom{u+v-1}{v}(-1)^v$.

*In (6) we apply the  binomial series expansion.
Since we have derived $A(z)$ we can proceed with the main calculation.

We obtain from (3) and (6)
  \begin{align*}
(-1)^k4^n&\sum_{j=k}^n\binom{2n-j}{j}\binom{j}{k}\left(-\frac{1}{4}\right)^{j}\\
&=(-1)^k4^n[z^{2n}]\frac{1}{1-z}A\left(\frac{z^2}{1-z}\right)\tag{7}\\
&=(-1)^k4^n[z^{2n}]\frac{1}{1-z}\left(-\frac{z^2}{4(1-z)}\right)^k\frac{1}{\left(1+\frac{z^2}{4(1-z)}\right)^{k+1}}\tag{8}\\
&=4^{n-k}[z^{2n-2k}]\frac{1}{\left(1-z+\frac{z^2}{4}\right)^{k+1}}\tag{9}\\
&=4^{n-k}[z^{2n-2k}]\frac{1}{\left(1-\frac{z}{2}\right)^{2(k+1)}}\\
&=4^{n-k}[z^{2n-2k}]\sum_{j=0}^\infty\binom{-2k-2}{j}\left(-\frac{z}{2}\right)^j\tag{10}\\
&=4^{n-k}[z^{2n-2k}]\sum_{j=0}^\infty\binom{2k+j-1}{j}\left(\frac{z}{2}\right)^j\tag{11}\\
&=4^{n-k}\binom{2n+1}{2n-2k}\frac{1}{4^{n-k}}\tag{12}\\
&=\binom{2n+1}{2k+1}
\end{align*}
  and the claim follows.

Comment:


*

*In (7) we use the representation (1) with the parameter settings
$
(n,m,p,q)\rightarrow (2n,0,-1,1)
$.

*In (8) we use the representation (6) of $A(z)$.

*In (9) we do some simplifications and use the rule $[z^{u+v}]A(z)=[z^u]z^{-v}A(z)$.

*In (10) we apply the binomial series expansion.

*In (11) we use the binomial identity $\binom{-u}{v}=\binom{u+v-1}{v}(-1)^v$.

*In (12) we select the coefficient of $z^{2n-2k}$.
