Question

If $\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+......\infty =\frac{{\pi }^2}{24}$, then$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+......\infty$ is equal to

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

Answer 1

The correct option is B $\frac{{\pi }^2}{8}$

Given, $\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........=\frac{{\pi }^2}{24}$
$\frac{1}{2^2}\left[\begin{array}{c}\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+......\end{array}\right]=\frac{{\pi }^2}{24}$
$\Rightarrow \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.......=\frac{{\pi }^2}{6}$
$\Rightarrow \left(\begin{array}{c}\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots \end{array}\right)+\left(\begin{array}{c}\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\dots \end{array}\right)=\frac{{\pi }^2}{6}$
$\Rightarrow \left(\begin{array}{c}\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots \end{array}\right)+\frac{{\pi }^2}{24}=\frac{{\pi }^2}{6}$
$\Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots =\frac{{\pi }^2}{8}$