Question

If $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+$ upto $\infty =\frac{{\pi }^2}{6}$, then value of $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+......$ upto $\infty$ is

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{{\pi }^2}{6}$
• C. $\frac{{\pi }^2}{8}$
• D. $\frac{{\pi }^2}{12}$

Answer 1

Answer: (C)

$\frac{{\pi }^2}{8}$

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{{\pi }^2}{6}\Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+...+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...=\frac{{\pi }^2}{6}\Rightarrow (\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...)+\frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)=\frac{{\pi }^2}{6}\Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...=\frac{{\pi }^2}{6}-\left(\frac{{\pi }^2}{6}\right)\frac{1}{4}=\frac{{\pi }^2}{8}$