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 $m,n \in N$
 A: When $n=1$ it's easy to evaluate for all $m\ge1$. Then it's easy to prove the $n+1$ by induction, given the $n$ case (for all $m$): just write
$$
\binom{n+k}{m+2k} = \binom{n-1+k}{m+2k} + \binom{n-1+k}{m-1+2k}.
$$
A: The generating function for the Central Binomial Coefficients is
$$
\sum_{k=0}^\infty\binom{2k}{k}x^k=(1-4x)^{-1/2}\tag{1}
$$
Integrating $(1)$ and dividing by $x$ yields
$$
\sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{k+1}=\frac{1-\sqrt{1-4x}}{2x}\tag{2}
$$
Sum the formula against $x^m$:
$$
\begin{align}
&\sum_{m=-\infty}^n\left(\sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}\right)x^m\tag{3}\\
&=\sum_{k=0}^\infty\sum_{m=-2k}^{n-k}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}x^m\tag{4}\\
&=\sum_{k=0}^\infty x^{-2k}\sum_{m=0}^{n+k}\binom{2k}{k}\binom{n+k}{m}\frac{(-1)^k}{k+1}x^m\tag{5}\\
&=\sum_{k=0}^\infty x^{-2k}\binom{2k}{k}\frac{(-1)^k}{k+1}(x+1)^{n+k}\tag{6}\\
&=(x+1)^n\sum_{k=0}^\infty\binom{2k}{k}\frac{(-1)^k}{k+1}\left(\frac{x+1}{x^2}\right)^{k}\tag{7}\\
&=(x+1)^n\frac{-1+\sqrt{1+4\frac{x+1}{x^2}}}{2\frac{x+1}{x^2}}\tag{8}\\
&=x(x+1)^{n-1}\frac{-x+(x+2)}{2}\tag{9}\\[9pt]
&=x(x+1)^{n-1}\tag{10}\\[6pt]
&=\sum_{m=1}^n\binom{n-1}{m-1}x^m\tag{11}
\end{align}
$$
Justification:
$\:\ (3)$: sum against $x^m$
$\:\ (4)$: change order of summation
$\:\ (5)$: substitute $m\mapsto m-2k$
$\:\ (6)$: apply the Binomial Theorem
$\:\ (7)$: reorganize
$\:\ (8)$: apply $(2)$
$\:\ (9)$: simplify
$(10)$: simplify
$(11)$: apply the Binomial Theorem
Equating coefficients of $x^m$ yields
$$
\sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}=\binom{n-1}{m-1}\tag{12}
$$

Note that the sum has non-zero terms for $m\lt1$, but these cancel; e.g. $n=3,m=-1$:
$$
\begin{align}
&\binom{0}{0}\binom{3}{-1}\frac11-\binom{2}{1}\binom{4}{1}\frac12+\binom{4}{2}\binom{5}{3}\frac13-\binom{6}{3}\binom{6}{5}\frac14+\binom{8}{4}\binom{7}{7}\frac15\\[6pt]
&=0-4+20-30+14\\[9pt]
&=0
\end{align}
$$
A: 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.
