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. $\frac{{\pi }^2}{12}$
• B. $\frac{{\pi }^2}{24}$
• C. $\frac{{\pi }^2}{8}$
• D. $\frac{{\pi }^2}{4}$

Answer 1

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

Let $S={\sum }^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}\Rightarrow \frac{S}{2^2}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\infty$

$\therefore S^{{}^{\prime }}=S-\frac{S}{4}=\frac{3S}{4}=\frac{1}{1^2}+(\frac{1}{2^2}-\frac{1}{2^2})+\frac{1}{3^2}+(\frac{1}{4^2}-\frac{1}{4^2})+...\infty$

ie, $\therefore S^{{}^{\prime }}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...\infty =\frac{3S}{4}=\frac{3{\pi }^2}{4\times 6}=\frac{{\pi }^2}{8}$

So, Option C is the correct one.