Question

If $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\infty =\frac{{\pi }^2}{6}$, then $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...\infty$ 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)

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

Lets call the first series as S and the series whose sum has to be evaluated as T.
Then
$S=\left(1/1^2\right)+\left(1/2^2\right)+\left(1/3^2\right)+\left(1/4^2\right)+\left(1/5^2\right)......$
$S=\left(1/1^2\right)+\left(1/3^2\right)+\left(1/5^2\right)+\left(1/7^2\right)+\left(1/2^2\right)\left[\right(1/1^2)+(1/2^2)+(1/3^2)......]$
$S=T+S/4$
$3S/4=T$
$T=3/4\times \left[{\pi }^2/6\right]$
$={\pi }^2/8$