Combinatorial Summation!

Therefore, we have $A=1, B=1, A+B =2, (A+B)^3 = \boxed{8}$ .

$\displaystyle S =\sum_{n=0}^{\infty } \frac{2^{n+1}}{(2n+1) \binom{2n}{n} } = \sum_{n=0}^{\infty } \frac{1}{2^{n-1}} . \frac{2^{2n} ( n!)^2 }{ (2n+1)!}$

Using the fact that $\displaystyle \int_0^{ \frac{ \pi}{2} } \sin ^m x \text{ d}x = \frac{2^{2n} ( n!)^2 }{ (2n+1)!}$ for $m = 2n +1$ (Refer Wallis' Integrals ), we can write $S$ as ,

$\displaystyle S = \sum_{n=0}^{\infty } \frac{1}{2^{n-1}} . \int_0^{ \frac{ \pi}{2} } \sin ^{2n+1} x \text{ d}x$

$\displaystyle \begin{array}{c}\\ S && = \int_0^{ \frac{ \pi}{2} } 2 \sin x \left( \sum_{n=0}^{\infty } \left( \frac{\sin ^{2} x}{2} \right)^n \right) \text{ d}x \\ && = \int_0^{ \frac{ \pi}{2} } 2 \sin x . \frac{1}{1 - \frac{\sin ^{2} x}{2} } \text{ d}x \\ && = \int_0^{ \frac{ \pi}{2} } \frac{ 4 \sin x }{ 2 - \sin ^2 x } \text{ d}x \\ && = \int_0^{ \frac{ \pi}{2} } \frac{ 4 \sin x }{ 1 + \cos ^2 x } \text{ d}x \\ \end{array}$

Substitute $u = \cos x \Rightarrow \text{ d}u = -\sin x \text{ d}x$ in the integral to obtain,
$\displaystyle S = - \int_1^0 \frac{4}{1 + u^2} \text{ d}u = (-4). \left( -\frac{\pi }{4} \right) = \pi$ .

Compare the coefficients to evaluate the required value to get the answer as $\displaystyle \boxed{8}$ .