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 $$\sum\limits_{j=k}^n\,(-1)^{j-k}\,\binom{j}{k}\,\binom{2n-j}{j}\,2^{2(n-j)}=\binom{2n+1}{2k+1}\,.$$


Suppose we seek to show that
$$\sum_{q=k}^n (-1)^{q-k} {q\choose k} {2n-q\choose q}
2^{2(n-q)} = {2n+1\choose 2k+1}.$$
Re-index to get
$$\sum_{q=0}^{n-k} (-1)^{q} {q+k\choose k} {2n-k-q\choose k+q}
2^{2(n-k-q)}.$$
Introduce
$${2n-k-q\choose k+q}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k-2q+1}} \frac{1}{(1-z)^{k+q+1}}
\; dz.$$
Observe  that this  vanishes  when  $q\gt n-k$  so  it provides  range
control and we may extend $q$ to infinity, getting for the sum
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} \frac{1}{(1-z)^{k+1}}
\sum_{q\ge 0} (-1)^q {q+k\choose q} 
\frac{z^{2q}}{(1-z)^q} 2^{2(n-k-q)}
\; dz
\\ = \frac{2^{2(n-k)}}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} \frac{1}{(1-z)^{k+1}}
\frac{1}{(1+z^2/(1-z)/2^2)^{k+1}}
\; dz
\\ = \frac{2^{2(n-k)}}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} 
\frac{1}{(1-z+z^2/2^2)^{k+1}}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} 
\frac{1}{(4-4z+z^2)^{k+1}}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} 
\frac{1}{(z-2)^{2k+2}}
\; dz
\\ = \frac{2^{2n-2k}}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} 
\frac{1}{(z/2-1)^{2k+2}}
\; dz
\\ = \frac{(-1)^{2k+2} 2^{2n-2k}}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} 
\frac{1}{(1-z/2)^{2k+2}}
\; dz.$$
This is
$$2^{2n-2k} [z^{2n-2k}] \frac{1}{(1-z/2)^{2k+2}}
= 2^{2n-2k} {2n-2k+2k+1\choose 2k+1} \frac{1}{2^{2n-2k}}
\\ = {2n+1\choose 2k+1}$$
which is the claim.
Remark. We  discover having reached the end  of the computation
that we never needed to substitute the variable in the integral, which
means  it   could  be   done  using  formal   power  series   and  the
coefficient-of method.
Remark II. For the geometric series to converge we need with $\epsilon\ll 1$ (pole at $z=1$) that $|z^2/(1-z)/4| \lt 1$ which requires $\epsilon^2/(1-\epsilon)/4\lt 1.$ With $\epsilon = 1/Q$ this is $1/Q/(Q-1)/4\lt 1$ which is seen to hold for $Q\ge 2.$
