TITLE: Proof of a combinatorial identity (possibly using trigonometric identities)
QUESTION [10 upvotes]: For integers $n \geq k \geq 0$, can anyone provide a proof for the following identity?
$$\sum_{j=0}^k\left(\begin{array}{c}2n+1\\\ 2j\end{array}\right)\left(\begin{array}{c}n-j\\\ k-j\end{array}\right) = 2^{2k} \left(\begin{array}{c} n+k\\\ 2k \end{array}\right)$$
I've verified this identity numerically for many values of $n$ and $k$, and suspect it to be true.
I found similar identities in http://www.math.wvu.edu/~gould/Vol.6.PDF, most notably:
$$\sum_{j=0}^k\left(\begin{array}{c}2n\\\ 2j\end{array}\right)\left(\begin{array}{c}n-j\\\ k-j\end{array}\right) = 2^{2k} \frac{n}{n+k}\left(\begin{array}{c} n+k\\\ 2k \end{array}\right)$$
which is Eq. (3.20) in the above link, and
$$\sum_{j=0}^k\left(\begin{array}{c}2n+1\\\ 2j+1\end{array}\right)\left(\begin{array}{c}n-j\\\ k-j\end{array}\right) = 2^{2k} \frac{2n+1}{n-k}\left(\begin{array}{c} n+k\\\ 2k+1 \end{array}\right)$$
which is Eq. (3.34) in the above link.  The derivations of these two identities seem to rely on trigonometric identities, which I've been having trouble reconstructing.

REPLY [3 votes]: In terms of hypergeometric series, the sum is 
$\binom nk {}_2F_1(-n-1/2, -k; 1/2;1)$, so the identity is a case of the Chu-Vandermonde theorem. It may be obtained from the more "usual-looking" case of Vandermonde's theorem 
$$\sum_j \binom{k-1/2}{k-j}\binom {n+1/2}{j} = \binom{n+k}{k}
$$
by multiplying both sides by $\binom nk/\binom{k-1/2}{k}$ and simplifying.
Similarly, the other two identities are also cases of Vandermonde's theorem. The sums are
$\binom nk {}_2F_1(-n+1/2, -k; 1/2;1)$ and $(2n+1)\binom nk {}_2F_1(-n+1/2, -k; 3/2;1)$
Incidentally, there is a fourth identity to go along with these three:
$$\sum_{j=0}^k \binom{2n+2}{2j+1}\binom{n-j}{k-j} = 2^{2k+1}\binom{n+k+1}{2k+1}.
$$
All four are equivalent to cases of Vandermonde's theorem.