Given that $1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}$, the value of $1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...$ is

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D - $\frac{\pi^2}{8}$. Sum of all 1/n^2 = even terms + odd terms; even terms = (1/4)(pi^2/6). So odd terms = (3/4)(pi^2/6) = pi^2/8.

Answer

A - $\frac{\pi^2}{6} - 1$

Answer

B - $\frac{\pi}{6}$

Answer

C - $\frac{\pi^2}{12}$

Given that \(1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}\), the value of \(1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...\) is

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