Question

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

• A. $\frac{{\pi }^2}{12}$
• B. $\frac{{\pi }^2}{24}$
• C. $\frac{{\pi }^2}{8}$
• D. none of these

Answer 1

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

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

$⟹(\frac{1}{1^2}+\frac{1}{3^2}+\cdots \infty )+(\frac{1}{2^2}+\frac{1}{4^2}+\cdots \infty )=\frac{{\pi }^2}{6}$

$⟹(\frac{1}{1^2}+\frac{1}{3^2}+\cdots \infty )+\frac{1}{2^2}(\frac{1}{1^2}+\frac{1}{2^2}+\cdots \infty )=\frac{{\pi }^2}{6}$

$⟹(\frac{1}{1^2}+\frac{1}{3^2}+\cdots \infty )+\frac{1}{4}\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{6}$

$⟹(\frac{1}{1^2}+\frac{1}{3^2}+\cdots \infty )=\frac{{\pi }^2}{6}-\frac{{\pi }^2}{24}=\frac{3{\pi }^2}{24}=\frac{{\pi }^2}{8}$