How can we prove the identity $\sum_{r=0, r~ even}^{k}\binom{k}{r}\binom{r}{r/2}2^{k-r}=\binom{2k}{k}$? We know that $$\sum_{r=0}^{k}{\binom{k}{r}}^2=\binom{2k}{k}$$ Can we prove such identity for the above case. I can check the identity for some numerical values. For example $k=4$ case satisfies it.
 A: We're going to using $2$ ways to calculate the coefficient of $x^k$ in $(1+x)^{2k}$.
First, the binomial formula suggests it's $\binom{2k}{k}$
And notice $$(1+x)^{2k}=(1+2x+x^2)^k=\sum_{m+n+l=k} { k\choose m,n,l}(2x)^n(x^2)^l $$
This indicates the coefficients are $$\sum_{m+n+l=k\\n+2l=k} { k\choose m,n,l }2^n=\sum_{\;r=0\\ r\;\text{even}}\binom{k}{r}\binom{r}{r/2}2^{k-r} $$
It completes our proof.
A: We seek to prove the identity
$$\sum_{q=0}^{\lfloor n/2\rfloor}
{n\choose 2q} {2q\choose q} 2^{n-2q} = {2n\choose n}.$$
We have
$${n\choose 2q} {2q\choose q}
= \frac{n!}{(n-2q)! \times q! \times q!}
= {n\choose q} {n-q\choose n-2q}.$$
The LHS is
$$\sum_{q=0}^{\lfloor n/2\rfloor}
{n\choose q} {n-q\choose n-2q} 2^{n-2q}
= 2^n [z^n] (1+z)^n
\sum_{q=0}^{\lfloor n/2\rfloor}
{n\choose q} z^{2q} (1+z)^{-q} 2^{-2q}.$$
Here the coefficient extractor enforces the upper limit of the sum and
we obtain
$$2^n [z^n] (1+z)^n
\sum_{q\ge 0}
{n\choose q} z^{2q} (1+z)^{-q} 2^{-2q}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{z^2}{4(1+z)}\right)^n
= 2^n [z^n] \left(1+z+\frac{z^2}{4}\right)^n
\\ = 2^n [z^n] \left(1+\frac{z}{2}\right)^{2n}
= 2^n {2n\choose n} \frac{1}{2^n}
= {2n\choose n}.$$
This sum also appeared at the following MSE link.
Addendum. In hindsight we may simplify this starting from
$${2n\choose n} = [z^n] (1+z)^{2n}
= [z^n] (1+2z+z^2)^n
= [z^n] \sum_{q=0}^n {n\choose q} z^q (2+z)^q
\\ = \sum_{q=0}^n {n\choose q}
[z^{n-q}] (2+z)^q
= \sum_{q=0}^n {n\choose q} {q\choose n-q} 2^{2q-n}
\\ = \sum_{q=0}^n {n\choose q} {n-q\choose q} 2^{n-2q}
= \sum_{q=0}^n {n\choose q} {n-q\choose n-2q} 2^{n-2q}.$$
This is what the first answer to appear used.
