Question

If $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....\infty =\frac{{\pi }^2}{6}$ then, $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty$ is equal to

• A. $\frac{{2\pi }^2}{15}$
• B. $\frac{{\pi }^2}{8}$
• C. $\frac{{\pi }^2}{18}$
• D. $\frac{{\pi }^2}{6}$

Answer 1

The correct option is A $\frac{{2\pi }^2}{15}$

Given that

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

This can be written as

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

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

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

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

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

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

$\Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty =\frac{2{\pi }^2}{15}$