Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series. 

Here is a slightly different proof  that is simpler than the other one
I posted earlier.
Suppose we seek to verify that
$$\sum_{q\ge 0} {p+q\choose q} {m+2p\choose m-2q}
= 2^{m-1} \frac{2p+m}{m} {m+p-1\choose p}.$$
This is
$$\sum_{q\ge 0} {p+q\choose q} {m+2p\choose 2p+2q}.$$
We introduce
$${m+2p\choose 2p+2q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m-2q+1}} \frac{1}{(1-z)^{2p+2q+1}} \; dz.$$
This integral  controls the range, being  zero when $2q\gt  m$ and we
may extend the range of $q$ to infinity. We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+1}} \frac{1}{(1-z)^{2p+1}} 
\sum_{q\ge 0} {p+q\choose q} \frac{z^{2q}}{(1-z)^{2q}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+1}} \frac{1}{(1-z)^{2p+1}} 
\frac{1}{(1-z^2/(1-z)^2)^{p+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1-z}{z^{m+1}} 
\frac{1}{((1-z)^2-z^2)^{p+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1-z}{z^{m+1}} 
\frac{1}{(1-2z)^{p+1}}
\; dz.$$
Extracting coefficients we get
$$2^m {m+p\choose p} - 2^{m-1} {m-1+p\choose p}
\\ = 2^{m-1} {m-1+p\choose p} 
\left(2\frac{m+p}{m} - 1 \right)
\\ = 2^{m-1} {m-1+p\choose p} 
\frac{m+2p}{m}.$$
This is the claim.
