<s>[INST] Want to use Herbert Wilf's snake oil method to show $\sum_k \binom{2n+1}{2k}\binom{m+k}{2n} = \binom{2m+1}{2n}$ I was trying to use the snake oil method to show that $$\sum_k \binom{2n+1}{2k}\binom{m+k}{2n} = \binom{2m+1}{2n}$$
tl;dr I wasn't able to and desperately need help
My approach was to try to establish a more general identity regarding $$\sum_k \binom{2n+1}{2k}\binom{m+k}{s}$$
So I let $$F(x) = \sum_s \sum_k \binom{2n+1}{2k}\binom{m+k}{s} x^s$$
By changing the order of summation, I obtained
$$F(x) = \sum_k \sum_s \binom{2n+1}{2k}\binom{m+k}{s} x^s = \sum_k \binom{2n+1}{2k}(1+x)^{m+k}$$
After this, I noted that $$\sum_{k \geq 0} \binom{n}{2k}x^{2k} = \frac{(1+x)^n + (1-x)^n}{2}$$
Now, $$F(x) = \frac{(1+x)^m}{2}((1+\sqrt{1+x})^{2n+1} + (1-\sqrt{1+x})^{2n+1})$$
I'm having trouble finding the coefficient of $x^{2n}$ in this generating function. 
Alternate approach: I could have created a generating function where the summation was on $m$ but that proved to be fruitless.  
 [/INST] Following request by OP for snake oil method we  evaluate
$$\sum_{0\le 2k\le q+1} {q+1\choose 2k} {m+k\choose q}$$
and introduce
$$F(z, u) = \sum_{q\ge 0} \sum_{m\ge 0}
z^q u^m \sum_{0\le k, 2k\le q+1}
{q+1\choose 2k} {m+k\choose q}
\\ = \sum_{q\ge 0} \sum_{m\ge 0}
{m\choose q} z^q u^m 
+ \sum_{q\ge 0} \sum_{m\ge 0}
z^q u^m \sum_{1\le k, 2k\le q+1}
{q+1\choose 2k} {m+k\choose q}
\\ = \sum_{m\ge 0} u^m (1+z)^m
+ \sum_{m\ge 0} u^m \sum_{k\ge 1}
\sum_{q\ge 2k-1} z^q
{q+1\choose 2k} {m+k\choose q}
\\ = \frac{1}{1-u(1+z)}
+ \sum_{m\ge 0} u^m \sum_{k\ge 1} z^{2k-1}
\sum_{q\ge 0} z^q
{q+2k\choose 2k} {m+k\choose q+2k-1}
\\ = \frac{1}{1-u(1+z)}
+ \sum_{k\ge 1} z^{2k-1}
\sum_{q\ge 0} z^q
{q+2k\choose 2k} \sum_{m\ge 0} u^m {m+k\choose q+2k-1}$$
Continuing with the sum we find
$$\sum_{k\ge 1} z^{2k-1}
\sum_{q\ge 0} z^q
{q+2k\choose 2k} \sum_{m\ge q+k-1} u^m {m+k\choose q+2k-1}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1}
\sum_{q\ge 0} z^q u^q 
{q+2k\choose 2k} \sum_{m\ge 0} u^m {m+q+2k-1\choose q+2k-1}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1}
\sum_{q\ge 0} {q+2k\choose 2k} z^q u^q \frac{1}{(1-u)^{q+2k}}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1} \frac{1}{(1-u)^{2k}}
\sum_{q\ge 0}  {q+2k\choose 2k} z^q u^q \frac{1}{(1-u)^{q}}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1} \frac{1}{(1-u)^{2k}}
\frac{1}{(1-uz/(1-u))^{2k+1}}
\\ = \sum_{k\ge 1} z^{2k-1} u^{k-1} 
\frac{1-u}{(1-u(1+z))^{2k+1}}
\\ = (1-u) \sum_{k\ge 0} z^{2k+1} u^{k} 
\frac{1}{(1-u(1+z))^{2k+3}}
\\ = \frac{(1-u)z}{(1-u(1+z))^3} \sum_{k\ge 0} z^{2k} u^{k} 
\frac{1}{(1-u(1+z))^{2k}}
\\ = \frac{(1-u)z}{(1-u(1+z))^3}
\frac{1}{1-uz^2/(1-u(1+z))^2}
\\ = \frac{(1-u)z}{1-u(1+z)}
\frac{1}{(1-u(1+z))^2-uz^2}$$
Restoring the term in front now yields
$$\frac{1}{1-u(1+z)}
+ \frac{z}{1-u(1+z)}
\frac{1}{1-u(1+z)^2}
\\ = \frac{1}{1-u(1+z)}
\left(1+ \frac{z}{1-u(1+z)^2}\right)
\\ = \frac{1+z}{1-u(1+z)^2}.$$
Extracting coefficients we get
$$[z^q] [u^m] \frac{1+z}{1-u(1+z)^2}
= [z^q] (1+z) (1+z)^{2m} = [z^q] (1+z)^{2m+1} = {2m+1\choose q}.$$
Now put $q=2n$ to obtain
$$\bbox[5px,border:2px solid #00A000]{
{2m+1\choose 2n}.}$$
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