How to prove this (seemingly elusive) double binomial identity? When manipulating polynomials, I came across a double binomial identity as follows,
$$ \sum_{i=0}^{n} \binom{2n+1}{2i+1} \binom{i}{k}  = \binom{2n - k }{k} 2^{2 n-2 k}  , $$
where $n$ and $k$ are given nonnegative integers such that $2n \geq k$.
I tried to prove it but failed. I worked on this problem for two days and tried mathematical induction, k-degree derivative at zero, algebra methods but they all didn't work for me.

I haven’t made all the calculations so I may be wrong. But here’s how a proof could work: let $c_{n,k}$ be the LHS. It’s easy to see that $c_{n,k}$ is the coefficient in front of $x^n$ of the series $\sum_l{\binom{2n+1}{2l}x^l}\sum_l{\binom{l}{k}x^l}$.
So $c_{n,k}$ is the coefficient in front of $x^{2n}$ of $\sum_l{\binom{2n+1}{2l}x^{2l}}\sum_l{\binom{l}{k}x^{2l}}$, which can be rewritten as $\frac{(1+x)^{2n+1}+(1-x)^{2n+1}}{2}\frac{x^{2k}}{(1-x)^{k+1}}$. Because of parity considerations, it means that $c_{n,k}$ is the coefficient in front of $x^{2n-2k}$ of $\frac{(1+x)^{2n-k}}{(1-x)^{k+1}}$.
When differentiating twice, this yields the equality $(2n-2k)(2n-2k-1)c_{n,k}=(2n-k)(2n-k-1)c_{n-1,k}+2(2n-k)(k+1)c_{n,k+1}+(k+1)(k+2)c_{n+1,k+2}$.
And then you can use induction on $n-k$, since $(n-1)-k <n-k,n-(k+1)<n-k,(n+1)-(k+2) < n-k$.
