<s>[INST] Binomial Identity $\sum\binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$ I'm looking for a reference with the proof of the following binomial identity:
$$\sum_{k=0}^n \binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$$
I've looked in a number of textbooks that have a lot of binomial identities but I can't seem to find this specific one. Any help would be greatly appreciated!
 [/INST] This can also be done using a basic complex variables technique, which
gives a close variation on what was posted by @GrigoryM.
Suppose we seek to evaluate
$$\sum_{k=0}^n {2n+1\choose 2k+1} {m+k\choose 2n}.$$
Introduce the integral representation
$${m+k\choose 2n}
= \frac{1}{2\pi i} 
\int_{|z|=\varepsilon} \frac{1}{z^{2n+1}} (1+z)^{m+k} \; dz.$$
This gives the following integral
$$\frac{1}{2\pi i} 
\int_{|z|=\varepsilon} 
\sum_{k=0}^n {2n+1\choose 2k+1}
\frac{1}{z^{2n+1}} (1+z)^{m+k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\varepsilon} \frac{(1+z)^m}{z^{2n+1}}
\sum_{k=0}^n {2n+1\choose 2k+1} (1+z)^k \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\varepsilon} \frac{(1+z)^{m-1/2}}{z^{2n+1}}
\sum_{k=0}^n {2n+1\choose 2k+1} \sqrt{1+z}^{2k+1} \; dz.$$
The sum is
$$\sum_{k=0}^{2n+1} {2n+1\choose k} \sqrt{1+z}^k
\frac{1}{2} (1-(-1)^k)
\\ = \frac{1}{2} ((1+\sqrt{1+z})^{2n+1}
- (1-\sqrt{1+z})^{2n+1})$$
and we get for the integral
$$\frac{1}{2\pi i} 
\int_{|z|=\varepsilon} \frac{(1+z)^{m-1/2}}{2z^{2n+1}}
\left((1+\sqrt{1+z})^{2n+1} - (1-\sqrt{1+z})^{2n+1})\right) \; dz.$$
By  way  of  ensuring  analyticity   we  observe  that  we  must  have
$\varepsilon \lt  1$ owing to  the branch  cut $(-\infty, -1]$  of the
square  root. Now  put $1+z  = w^2$  so that  $dz =  2w\; dw$  and the
integral becomes
$$\frac{1}{2\pi i} 
\int_{|w-1|=\gamma} \frac{w^{2m-1}}{(w^2-1)^{2n+1}}
\left((1+w)^{2n+1} - (1-w)^{2n+1})\right) \; w \; dw.$$
This is
$$\frac{1}{2\pi i} 
\int_{|w-1|=\gamma} w^{2m} 
\left(\frac{1}{(w-1)^{2n+1}} + \frac{1}{(w+1)^{2n+1}}\right) \;  dw.$$
Treat the two terms in the parentheses in turn. The first contributes
$$[(w-1)^{2n}] w^{2m}
= [(w-1)^{2n}] \sum_{q=0}^{2m} {2m\choose q} (w-1)^q
= {2m\choose 2n}.$$
The second term  is analytic on and inside the  circle that $w$ traces
round the value $1$  with no poles (pole is at  $w=-1$) and hence does
not contribute anything.  This concludes the argument.

Remark.  We  must   document  the  choice  of   $\gamma$  so  that
$|w-1|=\gamma$    is   entirely    contained   in    the   image    of
$|z|=\varepsilon$, which since $w=1+\frac{1}{2}  z + \cdots$ makes one
turn around $w=1$ and may then  be continuously deformed to the circle
$|w-1|=\gamma.$ We need  a bound on where this image  comes closest to
one.   We have  $w =  1  + \frac{1}{2}  z +  \sum_{q\ge 2}  (-1)^{q+1}
\frac{1}{4^q} \frac{1}{2q-1}  {2q\choose q}  z^q.$ The modulus  of the
series term is bounded  by $\sum_{q\ge 2} \frac{1}{4^q} \frac{1}{2q-1}
{2q\choose q} |z|^q  = 1 - \frac{1}{2} |z|  - \sqrt{1-|z|}.$ Therefore
choosing   $\gamma  =   \frac{1}{2}\varepsilon  -   1  +   \frac{1}{2}
\varepsilon + \sqrt{1-\epsilon} = \sqrt{1-\varepsilon} + \varepsilon -
1$ will  fit the  bill. For  example with $\varepsilon  = 1/2$  we get
$\gamma = (\sqrt{2}-1)/2.$ It is a matter of arithmetic to verify that
with the formula we have $\gamma \lt 1$.
</s>