<s>[INST] Proving the combinatorial identity: $\sum_{k=0}^m (-1)^k 2^{2k-1}\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right]=(-1)^m$ Let $m$ be a positive integer. I have trouble proving that
$$\sum_{k=0}^m (-1)^k 2^{2k-1}\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right]=(-1)^m$$
Anyone?
 [/INST] Suppose we seek to evaluate
$$\sum_{k=0}^m (-1)^k 2^{2k-1} 
\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right].$$
There are two pieces here, the first is
$$\frac{1}{2}\sum_{k=0}^m (-1)^k 2^{2k} 
{m+k-1\choose 2k}$$
and the second is
$$\frac{1}{2}\sum_{k=0}^m (-1)^k 2^{2k} 
{m+k\choose 2k}.$$
We treat these in turn.
Introduce
$${m+k-1\choose 2k}
= {m+k-1\choose m-1-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m-k}} (1+z)^{m+k-1} \; dz.$$
This is zero when $k\ge m$ so we may let $k$ go to infinity.

We thus get for the first piece
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m-1}}{z^m} 
\sum_{k\ge 0} (-1)^k 2^{2k} z^k (1+z)^k
\; dz
\\ = \frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m-1}}{z^m} 
\frac{1}{1+4z(1+z)}
\; dz
\\ = \frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m-1}}{z^m} 
\frac{1}{(2z+1)^2}
\; dz.$$
Similarly the second piece is
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m}}{z^{m+1}} 
\frac{1}{(2z+1)^2}
\; dz.$$
Adding these we obtain
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{m-1}}{z^m} 
\left(1+\frac{1+z}{z}\right)
\frac{1}{(2z+1)^2}
\; dz
\\ = \frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{m-1}}{z^{m+1}} 
\frac{1}{2z+1}
\; dz.$$
Extracting the residue we get
$$\frac{1}{2}
\sum_{q=0}^{m-1} {m-1\choose q} (-1)^{m-q} 2^{m-q}
= -\sum_{q=0}^{m-1} {m-1\choose q} (-1)^{m-1-q} 2^{m-1-q}
\\ = - (1-2)^{m-1} = (-1)^m.$$
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