Question

If $\displaystyle\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$ and $S_i=\displaystyle\sum_{k=1}^{\infty} \frac{i}{\left(36 k^2-1\right)^i}$, then $S_1+S_2$ is equal to

• A. $\frac{\pi^2}{9}-\frac{1}{2}$
• B. $\frac{\pi^2}{12}+\frac{1}{2}$
• C. $\frac{\pi^2}{15}+\frac{1}{2}$
• D. $\frac{\pi^2}{18}-\frac{1}{2}$

Answer 1

Answer: (D)

Correct answer is (d) $\frac{\pi^2}{18}-\frac{1}{2}$