Asking for a proof for a sum of products of binomials: an "interesting" identity?

Question

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)}{n - k}$. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \label{1}\tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \label{2}\tag2$$ So, I like to ask:

QUESTION. Can you provide a variety of proofs (algebraic, combinatorial, etc) to the identity \eqref{2}?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of \eqref{1} in an answer to Looking for a $q$-analogue of a binomial identity.

REMARK 2. Here is an equivalent fomulation of \eqref{2}: $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$

Answer 1

The sum can be expressed in terms of hypergeometric series as $$(n+1)\binom{2n}{n}\binom{2n+1}{n}\,{}_3F_2\left({-n,\,\tfrac12,\,\tfrac12\atop -n+\tfrac12,\tfrac32}\biggm| 1\right).$$ This means that $$\binom{2k}{k}\binom{2n-2k}{n-k}\binom{2n}{n}\frac{2n+1}{2k+1}$$ is equal to $$(n+1)\binom{2n}{n}\binom{2n+1}{n}\frac{(-n)_k (\frac12)_k^2}{k!\,(-n+\frac12)_k(\frac32)_k} $$ where $(a)_k$ is the rising factorial $a(a+1)\cdots (a+k-1)$. The hypergeometric series can be evaluated by Saalschütz's theorem.

Answer 2

Just as an alternative approach, let's record this. We employ the Wilf-Zeilberger methodology and it runs as follows.

Start by defining the function $$F(n,k):=\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{(2k+1)16^n}.$$ Zeilberger's algorithm generates the companion function $$G(n,k):=-\binom{2k}k\binom{2n-2k+1}{n-k}\binom{2n+1}n\frac{k}{4(n+1)16^n}$$ as well as the recurrence relation $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k).$$ Now, sum both sides over all integers $k$. It turns out that $\sum_{k=0}^{n+1}F(n+1,k)=\sum_{k=0}^nF(n,k)$ because the sum on the right-hand side cancel out to vanish.

For $n=0$, this common sum equals $1$. The identity (2) follows, immediately.

Answer 3

A generating function proof.

As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}&= [z^{2n}] \frac{\arcsin(z)}{z} \frac{1}{\sqrt{1-z^2}}\\ &= [z^{2n+1}]\arcsin(z)\,\arcsin^\prime(z)\\ &=(2n+2) [z^{2n+2}] \frac{1}{2} \big(\arcsin(z)\big)^2 \end{align*} The series expansion of $\frac{1}{2} \big(\arcsin(z)\big)^2$ was already given by Euler and is well known \begin{align*} \frac{1}{2} \big(\arcsin(z)\big)^2=\sum_{n\geq 0} \frac{4^n (n!)^2}{(2n+2)!}z^{2n+2}\end{align*} (See e.g. formula 1.645.1 in Gradshteyn-Ryzhik). Thus \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}=\frac{4^n}{(2n+1){2n \choose n}},\,\mbox{ as claimed.}\end{align*} (Of course, this may also be seen as a special case of hypergeometric series summation.)

ADDED: the Taylor expansion of $y(x)=\frac{1}{2}\big(\arcsin(x)\big)^2$ can be derived independently from the conjectured equality, by noting that $y$ solves the differential equation \begin{align*} (1-x^2)y^{\prime\prime} - xy^\prime=1\end{align*} with $y(0)=y^\prime(0)=0$, and using undetermined coefficients. (This is in fact what Euler did).