Find the value of $A+B+C$ in the following question?

More generally, the following power series expansion holds (see for example HERE): for $|x|\leq 1$ $$2(\arcsin(x))^2=\sum_{n=1}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}.$$ Hence, for $x=1/2$, we find that $$\sum_{n=1}^{\infty} \frac{1}{n^2\binom{2n}{n}}=2(\arcsin(1/2))^2=2\left(\frac{\pi}{6}\right)^2=\frac{1}{3}\zeta(2).$$ Finally it easy to obtain $A+B+C=1+3+2=6$.

P.S. See also How to prove by arithmetical means that $\sum\limits_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}}$

The famous identity $$\zeta(2)=\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}$$ can be proved in many ways: creative telescoping, complex analysis, Lagrage's inversion theorem or Legendre polynomials, just to mention a few of them. Have a look at the first section of my notes.

A self-contained proof: $$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}=\sum_{n\geq 1}\frac{(n-1)!^2}{(2n)!}&=&\sum_{n\geq 1}\frac{\Gamma(n)^2}{2n\,\Gamma(2n)}\\&=&\sum_{n\geq 1}\frac{B(n,n)}{2n}\\&=&\sum_{n\geq 1}\frac{1}{2n}\int_{0}^{1}x^{n-1}(1-x)^{n-1}\,dx\\&=&-\frac{1}{2}\int_{0}^{1}\frac{\log(1-x+x^2)}{x(1-x)}\,dx\end{eqnarray*}$$ and now it is enough to notice that $\frac{1}{x(1-x)}=\frac{1}{x}+\frac{1}{1-x}$ and that $1-x+x^2$ is a cyclotomic polynomial, in order to exploit this lemma: $$ \int_{0}^{1}\frac{\log\Phi_n(x)}{x}\,dx = \frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)\,\text{rad}(n)}{n^2}.$$