Binomial identity involving central binomial coefficient I came across this nice binomial identity
$${‎‎\sum}_{k=0}^{2n} \frac{(-1)^k {2n \choose k} {2k \choose k}}{n+k \choose k} = 1$$
This is equivalent to the hypergeometric function ${}_2 F_1(-2n, 1/2; n+1; 4)$. I am having a hard time trying to prove this. Can someone help me with a hint to prove this?
 A: Let $F(n,k)=(-1)^k\binom{2n}k\binom{2k}{k}\big/\binom{n+k}{k}$ be the summand, and let $S_n=\sum_{k=0}^{2n}F(n,k)$ be the sum to be computed. Let
$$
G(n,k)=(-1)^k\binom{2n}{k-2}\binom{2k}{k}\Big/\binom{n+k}{k}.
$$
You can show, through tedious algebraic manipulations alone, that for all $n,k\ge 0$,
$$
3(F(n,k)-F(n+1,k))=G(n,k+1)-G(n,k).\tag{*}
$$
Specifically, write both sides in terms of factorials, then cancel common factorial terms until all that remains is a rational equation in $n,k$, which can be proven by clearing denominators and polynomial manipulation.
Summing both sides of $(*)$ from $k=0$ to $2n+2$, you get that
$$
3S_{n}-3S_{n+1}=G(n,2n+3)-G(n,0)=0-0=0.
$$
This proves that $S_{n}$ is independent of $n$, so you need only verify that $S_0=1$ to prove $S_n=1$ for all $n$.

If you are wondering where $G(n,k)$ came from, it is of the form $G(n,k)=F(n,k)\cdot R(n,k),$ where $$R(n,k)=k(k-1)/(2n-k+1)(2n-k+2),$$ after canceling some common factors, necessary to prevent division by zero. This function $R(n,k)$ which causes $(*)$ to be satisfied can be found using Zeilberger's algorithm. For example, in Maxima, you can use the commands
load(zeilberger)$ 
Zeilberger((-1)^k * binomial(2*n, k) * binomial(2*k, k) / binomial(n + k, k), k, n);

to get both the coefficients $[3,-3]$ on the LHS of $(*)$, and $R(n,k)$. You can see for yourself by copy pasting those commands into this online Maxima compiler.
My purpose in including this explanation is to help spread the word that a proof of "any" summation identity  with binomial coefficients can be found automatically with the right computer program. For anyone who encounters complicated binomial summations like this a lot, this power is very desirable! In fact, if you only have the summation, you can determine if a closed form exists using the same algorithm. The exact requirements on the "any" part  are described in the book $A = B$, available for free online: A = B on Herbert Wilf's website. This is a must-read for anyone who wants this power.
A: In trying to verify the identity
$$\sum_{k=0}^{2n} (-1)^k
{n+k\choose k}^{-1} {2n\choose k} {2k\choose k} = 1$$
we see that
$${n+k\choose k}^{-1} {2n\choose k}
= \frac{(2n)! / (2n-k)!}{(n+k)! / n!}
= {3n\choose n}^{-1} {3n\choose 2n-k}$$
so that we seek to prove
$$\sum_{k=0}^{2n} (-1)^k {3n\choose 2n-k} {2k\choose k}
= {3n\choose n}.$$
The LHS is
$$\sum_{k=0}^{2n} (-1)^k {3n\choose k}
{4n-2k\choose 2n-k}
= [z^{2n}] (1+z)^{4n}
\sum_{k=0}^{2n} (-1)^k {3n\choose k}
\frac{z^k}{(1+z)^{2k}}$$
Here the coefficient extractor enforces the range of the sum and we
find
$$[z^{2n}] (1+z)^{4n}
\sum_{k\ge 0} (-1)^k {3n\choose k}
\frac{z^k}{(1+z)^{2k}}
= [z^{2n}] (1+z)^{4n}
\left(1-\frac{z}{(1+z)^2}\right)^{3n}
\\ = [z^{2n}] \frac{1}{(1+z)^{2n}}
(1+z+z^2)^{3n}.$$
Expanding the second powered term
$$[z^{2n}] \frac{1}{(1+z)^{2n}}
\sum_{q=0}^{3n} {3n\choose q} (1+z)^{3n-q} z^{2q}$$
The coefficient extractor sets the upper limit of the sum to $n$ and we
get (note that the powers of $1+z$ do not have a pole at zero hence the
expansion about zero starts with $z^{2q}$ and there is no  contribution
to $[z^{2n}]$ when $q\gt n$):
$$[z^{2n}] \sum_{q=0}^{n} {3n\choose q} (1+z)^{n-q} z^{2q}
= \sum_{q=0}^{n} {3n\choose q}
{n-q\choose 2n-2q} = {3n\choose n}.$$
Observe that the the power $n-q$ to which $1+z$ is raised is a
non-negative integer and hence we are justified  in writing $[z^{2n}]
z^{2q} (1+z)^{n-q} = [z^{2n-2q}] (1+z)^{n-q} = {n-q\choose 2n-2q}.$  The
only $q$ in the range $0\le q\le n$ where this binomial coefficient  is
not zero is $q=n$, producing a contribution of ${3n\choose n}$ and we
have the claim.
