TITLE: Proving $\sum_{k=0}^n\sum_{l=0}^k{n \choose k}^2{k \choose l}{n \choose l}{2n-l \choose n}=\sum_{k=0}^n {n \choose k}^2{n+k \choose k}^2$
QUESTION [5 upvotes]: I'm trying to prove this identity
$$\sum_{k=0}^n\sum_{l=0}^k{n \choose k}^2{k \choose l}{n \choose l}{2n-l \choose n}=\sum_{k=0}^n {n \choose k}^2{n+k \choose k}^2$$
but I don't even know from where to start.
Any ideas?

REPLY [3 votes]: We seek to verify that
$$\sum_{k=0}^n {n\choose k}^2
\sum_{l=0}^k {k\choose l} {n\choose l} {2n-l\choose n}
= \sum_{k=0}^n {n\choose k}^2 {n+k\choose k}^2.$$
Starting with the inner term on the LHS we find
$$\sum_{l=0}^k {k\choose k-l} {n\choose l} {2n-l\choose n}
\\ = [z^k] (1+z)^k \sum_{l=0}^k z^l {n\choose l} {2n-l\choose n}
\\ = [z^k] (1+z)^k [w^n] (1+w)^{2n}
\sum_{l=0}^k z^l {n\choose l} (1+w)^{-l}.$$
The coefficient extractor $[z^k]$ enforces the upper limit of the sum
and we find
$$[z^k] (1+z)^k [w^n] (1+w)^{2n}
\sum_{l\ge 0} z^l {n\choose l} (1+w)^{-l}
\\ = [z^k] (1+z)^k [w^n] (1+w)^{2n}
\left(1+\frac{z}{1+w}\right)^n
\\ = [z^k] (1+z)^k [w^n] (1+w)^{n} (1+w+z)^n.$$
We get from the outer sum
$$\sum_{k=0}^n {n\choose k}^2
[z^k] (1+z)^k [w^n] (1+w)^{n} (1+w+z)^n
\\ = \sum_{k=0}^n {n\choose k}^2
[z^n] z^k (1+z)^{n-k} [w^n] (1+w)^{n} (1+w+z)^n
\\ = [z^n] (1+z)^n [w^n] (1+w)^{n} (1+w+z)^n
\sum_{k=0}^n {n\choose k}^2
z^k (1+z)^{-k}
\\ = [z^n] (1+z)^n [w^n] (1+w)^{n} (1+w+z)^n [v^n] (1+v)^n
\sum_{k=0}^n {n\choose k}
v^k z^k (1+z)^{-k}
\\ = [z^n] (1+z)^n [w^n] (1+w)^{n} (1+w+z)^n [v^n] (1+v)^n
\left(1+\frac{vz}{1+z}\right)^n
\\ = [z^n] [w^n] (1+w)^{n} (1+w+z)^n [v^n] (1+v)^n
(1+z+vz)^n.$$
Extracting the coefficient on $[z^n]$ we find
$$\sum_{k=0}^n ( [z^{n-k}] [w^n] (1+w)^n (1+w+z)^n )
( [z^k] [v^n] (1+v)^n (1+z(1+v))^n )
\\ = \sum_{k=0}^n
\left({n\choose n-k} [w^n] (1+w)^{n+k}\right)
\left({n\choose k} [v^n] (1+v)^{n+k}\right)
\\ = \sum_{k=0}^n {n\choose k}^2 {n+k\choose n}^2.$$
This is the claim.