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.

Compute the Generating Function
$$
\begin{align}
\hspace{-1cm}\sum_{i,m}\binom{m}{2i+1}\binom{i}{k}x^m
&=\sum_{i,m}\binom{i}{k}\binom{-2i-2}{m-2i-1}(-1)^{m-2i-1}x^{m-2i-1}x^{2i+1}\tag{1a}\\
&=\sum_i\binom{i}{k}\frac{x^{2i+1}}{(1-x)^{2i+2}}\tag{1b}\\
&=\frac1x\left(\frac{x}{1-x}\right)^{2k+2}\sum_i\binom{-k-1}{i-k}(-1)^{i-k}\left(\frac{x}{1-x}\right)^{2(i-k)}\tag{1c}\\
&=\frac1x\left(\frac{x}{1-x}\right)^{2k+2}\left(1-\left(\frac{x}{1-x}\right)^2\right)^{-k-1}\tag{1d}\\
&=\frac{x^{2k+1}}{(1-2x)^{k+1}}\tag{1e}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\binom{m}{2i+1}=(-1)^{m-2i-1}\binom{-2i-2}{m-2i-1}$ using negative binomial coefficients
$\text{(1b)}$: sum in $m$ (Binomial Theorem)
$\text{(1c)}$: $\binom{i}{k}=(-1)^{i-k}\binom{-k-1}{i-k}$
$\text{(1d)}$: sum in $i$ (Binomial Theorem)
$\text{(1e)}$: simplify
Extract the Coefficients
$$
\begin{align}
\sum_i\binom{m}{2i+1}\binom{i}{k}
&=\left[x^m\right]\frac{x^{2k+1}}{(1-2x)^{k+1}}\tag{2a}\\
&=(-2)^{m-2k-1}\binom{-k-1}{m-2k-1}\tag{2b}\\
&=2^{m-2k-1}\binom{m-k-1}{k}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: the sum is the coefficient of $x^m$
$\text{(2b)}$: compute the coefficient: $\left[x^{m-2k-1}\right](1-2x)^{-k-1}$
$\text{(2c)}$: $(-1)^{m-2k-1}\binom{-k-1}{m-2k-1}=\binom{m-k-1}{m-2k-1}=\binom{m-k-1}{k}$
Apply to the Question
Thus, plugging $m=2n+1$ into $(2)$, we get
$$
\sum_i\binom{2n+1}{2i+1}\binom{i}{k}=2^{2n-2k}\binom{2n-k}{k}\tag3
$$
