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.