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 $\binom{p}{q}=\binom{p-1}{q-1}\frac{p}{q}$.