Question

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

• A. ${\pi }^2/8$
• B. ${\pi }^2/12$
• C. ${\pi }^2/3$
• D. ${\pi }^2/9$

Answer 1

The correct option is A ${\pi }^2/8$

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}..........\infty =\frac{{\pi }^2}{6}$
Now $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}...........$
$=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}..........\infty -\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}..........\infty \right)$
$=\frac{{\pi }^2}{6}-\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}..........\infty \right)$
$=\frac{{\pi }^2}{6}-\frac{1}{4}.\frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}$