Prove $\sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}\frac{1}{(2k+1)(2n-2k+1)}=\frac{2^{\left(4n+1\right)}}{\left(n+1\right)^{2}\binom{2n+2}{n+1}}$ How it can be shown that:

$$\sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}\frac{1}{(2k+1)(2n-2k+1)}=\frac{2^{\left(4n+1\right)}}{\left(n+1\right)^{2}\binom{2n+2}{n+1}}$$

Which is indeed Bruckman’s Formula Version 2.
It seems that there exist a relation between this expression and Rothe–Hagen identity, but I'm not sure.(it's appreciated if someone proves the expression in the most elementary ways, without using hypergeometric functions and integrals)
 A: An elementary approach is WZ-pair. Let
$$F(n,k)=\binom{2k}{k}\binom{2(n-k)}{n-k}\binom{2(n+1)}{n+1}\frac{(n+1)^2}{16^n(2k+1)(2(n-k)+1)}$$
Then we have to show that for any $n\geq 0$,
$$\sum_{k=0}^{n}F(n,k)=1.$$
Verify that 
$$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k) \tag{1}$$
where
$$G(n,k)=-F(n,k)\cdot \frac{k(2k+1)(3n-2k+4)(2(n-k)+1)^2}{4(2(n-k)+3)(n+1)^3(n-k+1)}.$$
The identity (1) seems terrible but it is not. No summation is involved. Moreover
$$\binom{2(k+1)}{k+1}=2\binom{2k+1}{k}=\binom{2k+1}{k+1}=\frac{2k+1}{k+1}\binom{2k}{k}$$
implies that all the binomial coefficients in (1) can be easily simplified.
It follows that
$$\begin{align}\sum_{k=0}^{n+1}F(n+1,k)-\sum_{k=0}^{n}F(n,k)&=\sum_{k=0}^{n+1}
(G(n,k+1)-G(n,k))\\
&=G(n,n+1)-G(n,0)=0
\end{align}$$
because the sum on the right is telescopic.
Hence we may conclude that
$$\sum_{k=0}^{n}F(n,k)=\sum_{k=0}^{0}F(0,k)=F(0,0)=1.$$
Another approach (less elementary). Note that the Taylor series of arcsin at $0$ is
$$\arcsin (x)= \sum_{n=0}^\infty \binom{2k}{k}\frac{x^{2k+1}}{2^{2k}(2k+1)}.$$ 
The given identity is related to the generating function of $(\arcsin (x))^2$ (see Cauchy product).
