Question

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

• A. $\frac{{\pi }^2}{12}$
• B. $\frac{{\pi }^2}{24}$
• C. $\frac{{\pi }^2}{8}$
• D. none of these

Answer 1

Answer: (C)

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

Given:

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....\infty =\frac{{\pi }^2}{6}$

$\Rightarrow [\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty ]+[\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+....\infty ]=\frac{{\pi }^2}{6}$

$\Rightarrow [\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty ]+\frac{1}{2^2}[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....\infty ]=\frac{{\pi }^2}{6}$

$\Rightarrow [\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty ]+\frac{1}{2^2}\left(\frac{{\pi }^2}{6}\right)=\frac{{\pi }^2}{6}$

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