Prove an equation about binomial coefficients Could we prove:
$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$

Here is a similar approach using basic complex variables.
Suppose we seek to evaluate
$$\sum_{k\ge 0} {2k\choose k} {n+k\choose m+2k} 
\frac{(-1)^k}{k+1}$$
where $n\ge m.$

Introduce the integral representation
$${n+k\choose m+2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz.$$
This yields the following integral for the sum:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\sum_{k\ge 0} {2k\choose k} \frac{(-1)^k}{k+1}
\frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}}
\sum_{k\ge 0} {2k\choose k} \frac{(-1)^k}{k+1}
\frac{(1+z)^k}{z^{2k}} \; dz.$$
Now the OGF of the Catalan numbers is
$$\sum_{k\ge 0} {2k\choose k} \frac{1}{k+1} w^k
= \frac{1-\sqrt{1-4w}}{2w}.$$
Substituting this closed form that we computed
into the integral yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}}
\frac{1-\sqrt{1+4(1+z)/z^2}}{(-1)\times 2(1+z)/z^2} \; dz
\\= - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m-1}}
\frac{1-\sqrt{1+4(1+z)/z^2}}{2(1+z)} \; dz
\\= - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^m}
\frac{z-\sqrt{z^2+4(1+z)}}{2(1+z)} \; dz.$$
Observe that $z^2+4(1+z) = (z+2)^2$ so the integral becomes
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^m}
\frac{z-(z+2)}{2(1+z)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^m} \; dz.$$
This last integral can be evaluated by inspection and yields
$$[z^{m-1}] (1+z)^{n-1} =
{n-1\choose m-1}.$$
Note that for the convergence of the Catalan GF series we require $|(1+z)/z^2|\lt 1/4$ (distance to the nearest singularity). Now $|(1+z)/z^2|\le (1+\epsilon)/\epsilon^2$ so we may take $\epsilon = N \ge 5.$
Remark: the same functions appear here as in @robjohn's answer,
who was first.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
