Infinity and beyond : 2

$\begin{aligned} S & = 1^2 - \frac {2^2}{5} + \frac {3^2}{5^2} - \frac {4^2}{5^3} + \frac {5^2}{5^4} - \frac {6^2}{5^5} + \cdots \\ \frac S5 & = \frac {1^2}5 - \frac {2^2}{5^2} + \frac {3^2}{5^3} - \frac {4^2}{5^4} + \frac {5^2}{5^5} - \frac {6^2}{5^6} + \cdots \\ S + \frac S5 & = 1^2 - \frac {2^2-1^2}{5} + \frac {3^2-2^2}{5^2} - \frac {4^2-3^3}{5^3} + \frac {5^2-4^2}{5^4} - \frac {6^2-5^2}{5^5} + \cdots \\ \frac {6S}5 & = 1^2 - \frac {(2-1)(2+1)}{5} + \frac {(3-2)(3+2)}{5^2} - \frac {(4-3)(4+3)}{5^3} + \frac {(5-4)(5+4)}{5^4} - \frac {(6-5)(6+5)}{5^5} + \cdots \\ & = 1 - \frac 3{5} + \frac 5{5^2} - \frac 7{5^3} + \frac 9{5^4} - \frac {11}{5^5} + \cdots \\ \frac {6S}{25} & = \frac 15 - \frac 3{5^2} + \frac 5{5^3} - \frac 7{5^4} + \frac 9{5^5} - \frac {11}{5^6} + \cdots \\ \frac {6S}5 + \frac {6S}{25} & = 1 - \frac 25 + \frac 2{5^2} - \frac 2{5^3} + \frac 2{5^4} - \frac 2{5^5} + \cdots \\ \frac {36}{25}S & = 1 - 2\left(\frac 15 + \frac 1{5^3} + \frac 1{5^5} + \cdots \right) + 2\left(\frac 1{5^2} + \frac 1{5^4} + \frac 1{5^6} + \cdots \right) \\ & = 1 - 2\left(\frac 15 - \frac 1{5^2} \right) \left(1 + \frac 1{5^2} + \frac 1{5^4} + \frac 1{5^6} + \cdots \right) \\ & = 1 -\frac 8{25} \left(\frac 1{1-\frac 1{25}} \right) = \frac 23 \\ \implies S & = \frac {25}{36} \cdot \frac 23 = \frac {25}{54} \end{aligned}$

$\implies A+B = 25+54 = \boxed{79}$

Differentiate $(1+x)^{-1} = 1-x+x^2-x^3+x^4- . . . .\rightarrow \dfrac{-1}{(1+x)^2} = -1 + 2x - 3x^2+4x^3- . . .$ . Multiply both sides by -x and differentiate again $\dfrac{1}{(1+x)^2} - \dfrac{2x}{(1+x)^3} = 1-2^2x+3^2x^2-4^2x^3+. . . .$ . Put $x=\frac{1}{5}$ to get the required summation in RHS equal to the LHS $\frac {25}{54} .$ hence answer 25+54=79.