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)

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

We have $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+.....upto\infty$
$=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}.....upto\infty$
$-\frac{1}{2^2}[1+\frac{1}{2^2}+\frac{1}{3^2}+....]=\frac{{\pi }^2}{6}-\frac{1}{4}\left(\frac{{\pi }^2}{6}\right)=\frac{{\pi }^2}{8}$
Hence, option 'C' is correct.