Prove $\sum_{k=m}^n{k\choose k-m}{2n\choose 2k}=4^{n-m}\frac{n(2n-m-1)!}{(2n-2m)!m!}.$ For $n\in\mathbb{Z}_{\ge 1}$ and $m\in\{0,1,\ldots,n\}$ I'd like to prove
$$\sum_{k=m}^n{k\choose k-m}{2n\choose 2k}=4^{n-m}\frac{n(2n-m-1)!}{(2n-2m)!m!}.$$
Though I've confirmed the identity for all $n\le 20$, sadly I don't have a proof.
 A: Nowadays it is a matter of machinery.
For a pencil-and-paper approach... the sum is equal to the coefficient of $x^{2n}$ in
$$\left(\sum_{k=m}^{\infty}\binom{k}{k-m}x^{2k}\right)\left(\sum_{k=0}^{2n}\binom{2n}{k}x^k\right)=\frac{x^{2m}}{(1-x^2)^{m+1}}(1+x)^{2n}=\frac{x^{2m}(1+x)^{2n-m-1}}{(1-x)^{m+1}}$$
which, by Cauchy integral formula, is equal to
$$\frac{1}{2\pi i}\oint_{|z|=r}\frac{(1+z)^{2n-m-1}\,dz}{z^{2n-2m+1}(1-z)^{m+1}}=\frac{1}{2\pi i}\oint_{|w|=1/r}\frac{w(w+1)^{2n-m-1}}{(w-1)^{m+1}}dw$$
(where $0<r<1$; the latter is obtained by substitution $z=1/w$).
Thus it is the coefficient of $z^m$ in $(z+1)(z+2)^{2n-m-1}$, equal to
$$2^{2n-2m-1}\binom{2n-m-1}{m}+2^{2n-2m}\binom{2n-m-1}{m-1},$$
which simplifies exactly to your formula.
A: I upvoted  the accepted answer  for its instructive management  of the
poles (pole at  zero disappears due to substitution but  a pole at one
appears inside the substituted contour and  the power series at one is
particularly easy to find).  What follows uses formal power series and
a different substitution and documents all the steps. Let's start with
$$\sum_{k=m}^n {k\choose k-m} {2n\choose 2k}
= \sum_{k=m}^n {k\choose k-m} {2n\choose 2n-2k}
\\ = \sum_{k=m}^n {k\choose k-m} [z^{2n-2k}] (1+z)^{2n}
= [z^{2n}] (1+z)^{2n} \sum_{k=m}^n {k\choose k-m} z^{2k}.$$
Now we may certainly extend $k$ beyond $n$ because there is no contribution
to $[z^{2n}]$ in that case. We get
$$[z^{2n}] (1+z)^{2n} \sum_{k\ge m} {k\choose k-m} z^{2k}
= [z^{2n}] (1+z)^{2n} z^{2m} \sum_{k\ge 0} {k+m\choose k} z^{2k}
\\ = [z^{2n-2m}] (1+z)^{2n}  \frac{1}{(1-z^2)^{m+1}}
= [z^{2n-2m}] (1+z)^{2n-m-1}  \frac{1}{(1-z)^{m+1}}.$$
This is
$$\mathrm{Res}_{z=0} \frac{1}{z^{2n-2m+1}}
(1+z)^{2n-m-1}  \frac{1}{(1-z)^{m+1}}.$$
The subsitution $z/(1+z) = w$ or $z=w/(1-w)$ now yields
$$\mathrm{Res}_{w=0} \frac{1}{w^{2n-2m+1}}
\frac{1}{(1-w)^{m-2}}
\frac{(1-w)^{m+1}}{(1-2w)^{m+1}}
\frac{1}{(1-w)^2}
\\ = \mathrm{Res}_{w=0} \frac{1}{w^{2n-2m+1}}
\frac{1-w}{(1-2w)^{m+1}}.$$
This is
$$[w^{2n-2m}] \frac{1-w}{(1-2w)^{m+1}}
= [w^{2n-2m}] \frac{1}{(1-2w)^{m+1}}
- [w^{2n-2m-1}] \frac{1}{(1-2w)^{m+1}}
\\ = 2^{2n-2m} {2n-m\choose m}
- 2^{2n-2m-1} {2n-m-1\choose m}
\\ = 4^{n-m} \frac{(2n-m-1)!}{(2n-2m)! m!}
\left(2n-m - \frac{1}{2} (2n-2m)\right).$$
which is indeed
$$\bbox[5px,border:2px solid #00A000]{
n 4^{n-m} \frac{(2n-m-1)!}{(2n-2m)! m!}.}$$
A: As mentioned by metamorphy, this can be done completely with machines now. So basically you do not have to prove it, a few lines of machine code can do it for you.
In the following, I will explain how CAS can prove this using creative telescoping algorithms.
The package that I used is called Sigma.
First the left-hand-side can be written as 
$$
S(n_0)=\underset{k=0}{\overset{n_0}{\sum }}
f(n_0,k)
$$
where $n_0=n-m$ and
$$
f(n_0,k)=
\left(
\begin{array}{c}
 k+m \\
 k \\
\end{array}
\right) \left(
\begin{array}{c}
 2 \left(m+n_0\right) \\
 2 (k+m) \\
\end{array}
\right)
.
$$
Next, let
$$
c_0(n)=2 (1 + m + n_0) (m + 2 n_0) (1 + m + 2 n_0),
$$
and
$$
c_1(n)=-(1 + n_0) (m + n_0) (1 + 2 n_0)
$$
and
$$
g(n_0,k)=-\frac{k (2 k+2 m-1) \left(m+n_0+1\right) \binom{k+m}{k} \left(2 k m+4 k n_0+2 k-4 m n_0-3 m-6 n_0^2-7 n_0-2\right) \binom{2 m+2 n_0}{2 k+2 m}}{\left(2 k-2 n_0-1\right) \left(k-n_0-1\right)}.
$$
Then you can verify in your favorite CAS system that
$$
g\left(n_0,k+1\right)-g\left(n_0,k\right)=c_0\left(n_0\right) f\left(n_0,k\right)+c_1\left(n_0\right) f\left(n_0+1,k\right)
.
$$
This recursion is found by using the Mathematica package Sigma. This is where the computer becomes most useful--they find these recursions for you.
Now we sum over $k=1,..n_0-1$, we get.
$$
c_0(n_0) S\left(n_0\right)+c_1(n_0)  S\left(n_0+1\right)=0
.
$$
This is a simple recursion, plugin $S(0)=f(0,0)=1$, we get
$$
S(n_0)=
\frac{4^{n_0} \left(m+n_0\right) \left(m+2 n_0-1\right)!}{m! \left(2 n_0\right)!}
=
\frac{n 4^{n-m} (-m+2 n-1)!}{m! (2 n-2 m)!}
.
$$
where we substitute by $n_0=n-m$. This is exactly the right-hand-side.
