Proving binomial summation identity using generating functions An exercise for class requires me to prove the following identity using generating functions:
$$\sum_{k=0}^{m/2} (-1)^k {n \choose k} {n+m-2k-1 \choose n-1} = {n \choose m}$$ for all $m \leq n$ and $m$ is even.
I've tried a bunch of things but can't wrap my head around this one. 

I assume one would start with
$${\sum_{k=0}^{n} {n \choose m}} x^m = (1+x)^n
= \frac{(1-x^2)^n}{(1-x)^n}$$
and then interpreting the expression on the right as the product of generating functions to somehow arrive at 
$$\sum_{m=0}^{n} \left( \sum_{k=0}^{m/2} (-1)^k {n \choose k} {n+m-2k-1 \choose n-1} \right) x^m$$
but now idea how to actually do it.
Some help would be greatly appreciated! :-)
 A: Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor m/2\rfloor} (-1)^k
{n\choose k} {n+m-2k-1\choose n-1}$$
where $n\ge m.$
Re-write this as
$$\sum_{k=0}^{\lfloor m/2\rfloor} (-1)^k
{n\choose k} {n+m-2k-1\choose m-2k}.$$
Now introduce
$${n+m-2k-1\choose m-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-2k+1}} (1+z)^{n+m-2k-1} \; dz.$$
Observe that this is zero when $2k\gt m$ so we may extend the range of
$k$ to infinity.

We thus get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1+z)^{n+m-1} 
\sum_{k\ge 0} {n\choose k} (-1)^k \frac{z^{2k}}{(1+z)^{2k}}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1+z)^{n+m-1} 
\left(1-\frac{z^2}{(1+z)^2}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1+z)^{n-m+1}}
\left(1+2z\right)^n \; dz.$$
Extracting the residue we get
$$\sum_{q=0}^m {n\choose q} 2^q {m-q+n-m\choose m-q}
(-1)^{m-q}
\\ = \sum_{q=0}^m {n\choose q} 2^q {n-q\choose m-q}
(-1)^{m-q}.$$
To conclude introduce the integral
$${n-q\choose m-q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-q+1}} (1+z)^{n-q} \; dz.$$
Observe that this is zero when $q\gt m$ so we may extend the range of
$q$ to infinity to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1+z)^{n} 
\sum_{q\ge 0} {n\choose q} 2^q (-1)^{m-q} \frac{z^q}{(1+z)^q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(-1)^m}{z^{m+1}} (1+z)^{n} 
\left(1-2\frac{z}{1+z}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(-1)^m}{z^{m+1}}
\left(1-z\right)^n \; dz
= (-1)^m {n\choose m} (-1)^m = {n\choose m}.$$
Alternative solution.
Introduce
$${n+m-2k-1\choose n-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-2k+1}} \frac{1}{(1-z)^n} \; dz.$$
This is again zero when $2k\gt m.$ We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^n} 
\sum_{k\ge 0} {n\choose k} (-1)^k z^{2k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^n} 
(1-z^2)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} 
(1+z)^n \; dz
\\ = {n\choose m}.$$
This is probably what the problem had in mind.
