A power series equality involving binomial coefficients I believe the following identity holds for any non-negative integers $m,k$:
$$\binom{m+k}{k}^2 = \sum_{n=0}^m \binom{k}{m-n}^2\binom{2k+n}{n}.$$
It seems like there should be a slick double counting way of proving it, but I am not able to see it. I got this identity because I believe the following power series equality holds for $|x| < 1$,
$$(1-x)^{2k-1} \sum_{m=0}^\infty \binom{m+k-1}{k-1}^2 x^m = \sum_{j=0}^{k-1} \binom{k-1}{j}^2 x^j. $$
The binomial identity above comes from dividing by $(1-x)^{2k-1}$, applying the binomial theorem replacing $k$ with $k+1$ and comparing the coefficients of both sides.
But this power series equality doesn't seem any easier to prove than the binomial coefficient identity, since I don't really have a handle on the sums on either side.
Does anyone know a reference for an identity of this nature or have an idea for a double counting argument?
 A: Permit me to contribute an algebraic proof. We seek
$${m+k\choose k}^2 =
\sum_{q=0}^m {k\choose m-q}^2 {2k+q\choose q}$$
Starting with the RHS we find
$$\sum_{q=0}^m {k\choose q}^2 {2k+m-q\choose m-q}
\\ = \sum_{q=0}^m {k\choose q}
[z^k] z^q (1+z)^k [w^{m}] w^q (1+w)^{2k+m-q}.$$
Now we may extend $q$ to infinity because the coefficient extractor
$[w^m]$ enforces the upper limit. We get
$$[z^k] (1+z)^k [w^m] (1+w)^{2k+m}
\sum_{q\ge 0} {k\choose q} z^q
w^q (1+w)^{-q}
\\ = [z^k] (1+z)^k [w^m] (1+w)^{2k+m}
(1 + zw/(1+w))^k
\\ = [z^k] (1+z)^k [w^m] (1+w)^{k+m}
(1 + w + zw)^k$$
Re-expanding we find
$$[z^k] (1+z)^k [w^m] (1+w)^{k+m}
\sum_{q=0}^k {k\choose q} w^q (1+z)^q.$$
We may set the upper limit of the sum to $m.$ (If $k\lt m$ the values
$k\lt q\le m$ produce zero from the binomial coefficient and we may
raise $q$ to $m.$ If $k\gt m$  the values $m\lt q\le k$ produce zero by
the coefficient extractor  $[w^m]$ and we may lower $q$ to $m.$) We get
$$[z^k] (1+z)^k [w^m] (1+w)^{k+m}
\sum_{q=0}^m {k\choose q} w^q (1+z)^q
\\ = \sum_{q=0}^m {k\choose q} {k+q\choose k} {k+m\choose m-q}.$$
Now observe that
$${k+q\choose k} {k+m\choose m-q}
= \frac{(k+m)!}{k! \times q! \times (m-q)!}
= {m+k\choose k} {m\choose m-q}.$$
This yields for our sum
$${m+k\choose k} \sum_{q=0}^m {k\choose q} {m\choose m-q}.$$
Using Vandermonde we obtain at last
$$\bbox[5px,border:2px solid #00A000]{
{m+k\choose k}^2.}$$
