Question

It is given that ${\sum }_{r=1}^{\infty }\frac{1}{(2r-1{)}^2}=\frac{{\pi }^2}{8}$, then ${\sum }_{r=1}^{\infty }\frac{1}{r^2}$ is equal to

• A. $\frac{{\pi }^2}{24}$
• B. $\frac{{\pi }^2}{3}$
• C. $\frac{{\pi }^2}{6}$
• D. None of these

Answer 1

Answer: (C)

$\frac{{\pi }^2}{6}$

${}_{r=1}^{\infty }\Sigma \frac{1}{{(2r-1)}^2}=\frac{{\pi }^2}{8}$

$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+........=\frac{{\pi }^2}{8}$

$I=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5}+\frac{1}{6^2}+\frac{1}{7^2}$

$I=(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}.......)+(\frac{1}{1^2}+\frac{1}{3^2}+........)$

$=\frac{1}{2^2}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}..........)+\frac{{\pi }^2}{8}$

$\frac{4I-I}{4}=\frac{{\pi }^2}{8}$

$\frac{4I-I}{4}=\frac{{\pi }^2}{8}$

$\frac{3I}{4}=\frac{{\pi }^2}{8}$

$I=\frac{{\pi }^2}{6}$