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?

Generating functions! Multiply both sides of the desired identity by $x^m$, sum over all nonnegative integers $n$, and check that you get the same function on both sides.
First, note that your formula is valid for $m\ge1$; for $m=0$ the answer is $1/2$. In particular, making a generating function out of the right-hand side yields
$$
\frac12 + \sum_{m=1}^\infty (-1)^m x^m = \frac12 + \frac{-x}{1+x} = \frac{1-x}{2+2x}.
$$
Suppose we knew the formula
$$
\sum_{m=0^\infty} x^m \sum_{k=0}^m (-1)^k 2^{2k-1}\binom{m+k}{2k} = \frac{1-x}{2 (x+1)^2}.
$$
Then the left-hand side equals
$$
\sum_{m=0^\infty} x^m \sum_{k=0}^m (-1)^k 2^{2k-1}\binom{m+k}{2k} + x \sum_{m=0^\infty} x^{m-1} \sum_{k=0}^m (-1)^k 2^{2k-1}\binom{m-1+k}{2k} = \frac{1-x}{2 (x+1)^2} + x\frac{1-x}{2 (x+1)^2} = \frac{1-x}{2x+2}.
$$
To establish the necessary formula, switch the order of summation on the left-hand side to get
\begin{align*}
\sum_{k=0}^\infty (-1)^k 2^{2k-1} \sum_{m=k^\infty} x^m \binom{m+k}{2k}
&= \sum_{k=0}^\infty (-1)^k 2^{2k-1} \sum_{m=0^\infty} x^{m+k} \binom{m+2k}{2k}
\\\
&= \sum_{k=0}^\infty (-1)^k 2^{2k-1} x^k \sum_{m=0^\infty} x^m \binom{m+2k}{m}
\\\
&= \sum_{k=0}^\infty (-1)^k 2^{2k-1} x^k \sum_{m=0^\infty} x^m (-1)^m \binom{-2k-1}{m}
\\\
&= \sum_{k=0}^\infty (-1)^k 2^{2k-1} x^k (1-x)^{-2k-1}
\\\
&= \frac1{2(1-x)} \sum_{k=0}^\infty \bigg(\frac{-4x}{1-x}^2\bigg)^k
\\\
&= \frac1{2(1-x)} \frac1{1-(-4x/(1-x)^2)} = \frac{1-x}{2 (x+1)^2}
\end{align*}
as desired.
