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. 
 A: Let $d_s = \binom{p+s}{s} \binom{2p+m}{2p+2s}$.  Using the recurrence relations for binomial, the ratio of successive terms is:
$$
 \frac{d_{s+1}}{d_s} = \frac{\left(s - m/2\right)\left(s -(m-1)/2\right)}{ (s+1)(s+p+1/2)  } = \frac{(s+a)(s+b)}{(s+1)(s+c)}
$$
The hypergeometric certificate above means that
$$
  \sum_{s=0}^\infty d_s = d_0 \sum_{s=0}^\infty \frac{(a)_s (b)_s}{s! (c)_s} = \binom{2p+m}{2p} {}_2 F_1\left( -\frac{m}{2}, -\frac{m-1}{2} ; p+\frac{1}{2} ; 1\right) 
$$
where $a = -\frac{m}{2}$, $b=-\frac{m-1}{2}$ and $c=p+\frac{1}{2}$.
Using Gauss's theorem, valid for $c>a+b$:
$$
   {}_2 F_1\left( a, b; c; 1\right) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}
$$
we obtain the required identity:
$$
  \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = \binom{2p+m}{2p} \frac{\Gamma\left(p+\frac{1}{2}\right) \Gamma\left( p+m \right)}{ \Gamma\left( p+\frac{m+1}{2} \right) \Gamma\left( p+\frac{m}{2} \right) }
$$
Applying the duplication formula for $\Gamma(2p+m+1)$ and $\Gamma(2p+1)$ arising from $\binom{2p+m}{2p}$ we arrive at the result:
$$
  \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = 2^{m-1} (m+2p) \frac{\Gamma(m+p)}{\Gamma(m+1) \Gamma(p+1)} = 2^{m-1} \frac{m+2p}{m+p} \binom{m+p}{p}
$$
A: Ok, here is an approach with generating functions. Let
$$
   g_1(z) = \sum_{s=0}^\infty \binom{p+s}{s} z^s = \frac{1}{\left(1-z\right)^{p+1}}
$$
$$
   g_2(z) = \sum_{s=0}^\infty \binom{2p+m}{s} z^s = \left(1+z\right)^{m+2p}
$$
Now
$$ \begin{eqnarray}
  \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} &=& \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{m-2s} = [z]^m g_1(z^2) g_2(z) = [z]^m \frac{\left(1+z\right)^{m+2p}}{(1-z^2)^{p+1}} \\
   &=& [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}}
 \end{eqnarray} 
$$
Here is a verification:
In[27]:= With[{p = 5, 
  m = 7}, {SeriesCoefficient[(1 + z)^(m + 2 p)/(1 - z^2)^(
   p + 1), {z, 0, m}], 
  Sum[Binomial[p + s, s] Binomial[2 p + m, 2 p + 2 s], {s, 
    0, \[Infinity]}]}]

Out[27]= {71808, 71808}

Let's continue:
$$ \begin{eqnarray}
    [z]^m \frac{\left(1+z\right)^{m+p-1}}{\left(1-z\right)^{p+1}} &=& \sum_{s=0}^\infty \binom{p+m-1}{m-s} \binom{p+s}{s} =  \sum_{s=0}^\infty \binom{p+m-1}{p+s-1} \binom{p+s}{s}\\ &=& \sum_{s=0}^\infty \frac{(p+s) (m+p-1)!}{p! s! (m-s)!} = 
        \sum_{s=0}^\infty \frac{p (m+p-1)!}{p! s! (m-s)!} + \sum_{s=0}^\infty \frac{s   (m+p-1)!}{p! s! (m-s)!} \\
   &=& \binom{m+p-1}{m} \left( \sum_{s=0}^\infty  \binom{m}{s} + \sum_{s=0}^\infty \frac{s}{p} \binom{m}{s}  \right) \\ 
    &=& \binom{m+p-1}{m}  \left( 2^m + 2^{m-1} \frac{m}{p} \right) 
\end{eqnarray}
$$
A: I will try to give an answer using basic complex variables here.
This calculation is very simple in spite of some more complicated intermediate expressions that appear.

Suppose we are trying to show that
$$\sum_{q=0}^\infty
{p+q\choose q} {2p+m\choose m-2q}
= 2^{m-1} \frac{2p+m}{m} {m+p-1\choose p}.$$

Introduce the integral representation
$${2p+m\choose m-2q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2p+m}}{z^{m-2q+1}} \; dz.$$

This gives for  the sum the integral (the  second binomial coefficient
enforces the range)
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2p+m}}{z^{m+1}} 
\sum_{q=0}^\infty {p+q\choose q} z^{2q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2p+m}}{z^{m+1}} \frac{1}{(1-z^2)^{p+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{p+m-1}}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(2+z-1)^{p+m-1}}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \; dz
\\ = 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+(z-1)/2)^{p+m-1}}{z^{m+1}} \frac{1}{(1-z)^{p+1}} \; dz
\\ = 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1-z)^{p+1}}
\sum_{q=0}^{p+m-1} {p+m-1\choose q} \frac{(z-1)^q}{2^q} \; dz
\\ = 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1-z)^{p+1}}
\sum_{q=0}^{p+m-1} {p+m-1\choose q} (-1)^q \frac{(1-z)^q}{2^q} \; dz
\\ = 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}}
\sum_{q=0}^{p+m-1} {p+m-1\choose q} 
(-1)^q \frac{(1-z)^{q-p-1}}{2^q} \; dz.$$
The only non-zero contribution is with $q$ ranging from $0$ to $p.$
This gives
$$ 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}}
\sum_{q=0}^p {p+m-1\choose q} 
(-1)^q \frac{1}{2^q} \frac{1}{(1-z)^{p+1-q}} \; dz$$
which on extracting coefficients yields
$$2^{p+m-1} \sum_{q=0}^p {p+m-1\choose q} 
(-1)^q \frac{1}{2^q} {m+p-q\choose p-q}.$$
Introduce the integral representation
$${m+p-q\choose p-q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+p-q}}{z^{p-q+1}} \; dz.$$

This gives for  the sum the integral (the  second binomial coefficient
enforces the range)
$$2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+p}}{z^{p+1}} 
\sum_{q=0}^\infty 
{p+m-1\choose q}\frac{(-1)^q}{2^q}
\left(\frac{z}{1+z}\right)^q \; dz
\\ = 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+p}}{z^{p+1}} 
\left(1-\frac{1}{2}\frac{z}{1+z}\right)^{p+m-1} \; dz
\\ = 2^{p+m-1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z^{p+1}} 
\left(1+z-1/2\times z\right)^{p+m-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z^{p+1}} 
\left(2+z\right)^{p+m-1} \; dz.$$
Extracting coefficients now yields
$${p+m-1\choose p} \times 2^{m-1}
+ {p+m-1\choose p-1} \times 2^m.$$
This  symmetric  form may  be  re-written  in  an asymmetric  form  as
follows,
$${p+m-1\choose p} \times 2^{m-1}
+ \frac{p}{m} {p+m-1\choose p} \times 2^m
\\ = 2^{m-1} \times
\left(1 + \frac{2p}{m}\right) {p+m-1\choose p}$$
as claimed.
The bonus feature of this calculation is that we evaluated two binomial sums instead of one.

We have  not made use of  the properties of complex  integrals here so
this  computation  can  also   be  presented  using  just  algebra  of
generating functions.

Apparently  this method is  due to  Egorychev although  some of  it is
probably folklore.
A: 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.
