Looking for a short proof of a harmless looking binomial identity I managed to prove for   this MSE post the rather harmless looking binomial identity for natural  $1\leq k\leq  n$:
\begin{align*}
\color{blue}{\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}=\binom{n+k}{n-k}\frac{4^kn}{n+k}}\tag{1}
\end{align*}
using the coefficient of operator method. Admittedly, there are a lot of intermediate steps used to show the validity of (1).
Question: I'm wondering if there is a more direct, less lengthy derivation than the one I've provided below.

We obtain for $1\leq k\leq n$:
  \begin{align*}
\color{blue}{\sum_{j=0}^k}&\color{blue}{\binom{2n}{2j}\binom{n-j}{k-j}}\\
&=\sum_{j=0}^n\binom{2n}{2j}\binom{n-j}{n-k}\tag{2}\\
&=\sum_{j=0}^n\binom{2n}{2j}[z^{n-k}](1+z)^{n-j}\tag{3}\\
&=[z^{n-k}](1+z)^n\sum_{j=0}^n\binom{2n}{2j}\frac{1}{(1+z)^j}\\
&=\frac{1}{2}[z^{n-k}](1+z)^n\left(\left(1+\frac{1}{\sqrt{1+z}}\right)^{2n}+\left(1-\frac{1}{\sqrt{1+z}}\right)^{2n}\right)\\
&=\frac{1}{2}[z^{n-k}]\left(\left(1+\sqrt{1+z}\right)^{2n}+\left(1-\sqrt{1+z}\right)^{2n}\right)\\
&=\frac{1}{2}[z^{n-k}]\left(1+\sqrt{1+z}\right)^{2n}\tag{4}\\
&=\frac{1}{2}[z^{-1}]z^{-n+k-1}\left(1+\sqrt{1+z}\right)^{2n}\tag{5}\\
&=\frac{1}{2}[w^{-1}]\left(w^2-1\right)^{n-k-1}(1+w)^{2n}2w\tag{6}\\
&=[w^{-1}]w(w-1)^{-n+k-1}(w+1)^{n+k-1}\\
&=[u^{-1}](u+1)u^{-n+k-1}(u+2)^{n+k-1}\tag{7}\\
&=\left([u^{n-k}]+[u^{n-k-1}]\right)\sum_{j=0}^{n+k-1}\binom{n+k-1}{j}u^j2^{n+k+1-j}\\
&=\binom{n+k-1}{n-k}2^{2k-1}+\binom{n+k-1}{n-k-1}2^{2k}\tag{8}\\
&=\binom{n+k}{n-k}\frac{2k}{n+k}2^{2k-1}+\binom{n+k}{n-k}\frac{n-k}{n+k}2^{2k}\tag{9}\\
&\,\,\color{blue}{=\binom{n+k}{n-k}\frac{4^kn}{n+k}}
\end{align*}
  and the claim follows.

Comment:


*

*In (2) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$. We also set the upper index to $n$ without changing anything, since we are adding zeros only.

*In (3) we use the coefficient of operator  method.

*In (4) we skip $\left(1-\sqrt{1+z}\right)^{2n}=cz^{2n}+\cdots$ since it has only powers of $z$ greater than $n$ and does not contribute to $[z^{n-k}]$.

*In (5) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (6) we use the transformation of variable formula $[z^{-1}]f(z)=[w^{-1}]f(g(w))g^\prime(w)$ with $1+z=w^2, \frac{dz}{dw}=2w$.

*In (7) we use the  transformation of variable formula again, with $w-1=u, \frac{dw}{du}=1$.

*In (8) we select the coefficients accordingly.

*In (9) we use the binomial identities $\binom{p-1}{q}=\binom{p}{q}\frac{p-q}{p}$ and $\binom{p}{q}=\binom{p-1}{q-1}\frac{p}{q}$.

Here is an alternate solution, where the number of steps is about
the same as what OP provided. Could use additional streamlining by
removing some of the simpler proceedings. Start as follows:
$$\sum_{j=0}^k {2n\choose 2j} {n-j\choose k-j}
= \sum_{j=0}^k {2n\choose 2k-2j} {n-k+j\choose j}
\\ = [z^{2k}] (1+z)^{2n} \sum_{j=0}^k z^{2j}  {n-k+j\choose j}.$$
Here the coefficient extractor enforces the range:
$$[z^{2k}] (1+z)^{2n} \sum_{j\ge 0} z^{2j}  {n-k+j\choose j}
\\ = [z^{2k}] (1+z)^{2n} \frac{1}{(1-z^2)^{n-k+1}}
= [z^{2k}] (1+z)^{n+k-1} \frac{1}{(1-z)^{n-k+1}}.$$
This is
$$\mathrm{Res}_{z=0} \frac{1}{z^{2k+1}}
(1+z)^{n+k-1} \frac{1}{(1-z)^{n-k+1}}
\\ = (-1)^{n-k+1} \mathrm{Res}_{z=0} \frac{1}{z^{2k+1}}
(1+z)^{n+k-1} \frac{1}{(z-1)^{n-k+1}}.$$
Now the residue at infinity is zero so this is minus the residue
at one:
$$(-1)^{n-k} \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{2k+1}}
(2+(z-1))^{n+k-1} \frac{1}{(z-1)^{n-k+1}}
\\ = (-1)^{n-k} \sum_{j=0}^{n-k} {n+k-1\choose j} 2^{n+k-1-j}
(-1)^{n-k-j} {n-k-j+2k\choose 2k}
\\ = 2^{n+k-1} \sum_{j=0}^{n-k} {n+k-1\choose j} 2^{-j}
(-1)^{j} {n+k-j\choose n-k-j}.$$
Coefficient extractor enforces range:
$$2^{n+k-1} [z^{n-k}] (1+z)^{n+k}
\sum_{j\ge 0} {n+k-1\choose j} 2^{-j}
(-1)^{j} \frac{z^j}{(1+z)^j}
\\ = 2^{n+k-1} [z^{n-k}] (1+z)^{n+k}
\left(1-\frac{z}{2(1+z)}\right)^{n+k-1}
\\ = [z^{n-k}] (1+z) (2+z)^{n+k-1}
\\ = [z^{n-k}] (2+z)^{n+k-1} + [z^{n-k-1}] (2+z)^{n+k-1}
\\ = {n+k-1\choose n-k} 2^{n+k-1-(n-k)}
+ {n+k-1\choose n-k-1} 2^{n+k-1-(n-k-1)}
\\ = \frac{1}{2} 4^k \frac{2k}{n+k} {n+k\choose n-k}
+ \frac{n-k}{n+k} 4^k {n+k\choose n-k}
\\ = \frac{4^k n}{n+k} {n+k\choose n-k}.$$
