Question

$If{\sum }_{r=1}^{\infty }\frac{1}{(2r-1{)}^2}=\frac{{\pi }^2}{8}$ then the value of x = ${\sum }_{r=1}^{\infty }\frac{1}{r^2}$ is

• A. $\frac{{\pi }^2}{8}+\frac{1}{4}x$
• B. $\frac{{\pi }^2}{4}+\frac{1}{3}x$
• C. $\frac{{\pi }^2}{8}-\frac{1}{4}x$
• D. $\frac{{\pi }^2}{4}+\frac{2}{3}x$

Answer 1

The correct option is A $\frac{{\pi }^2}{8}+\frac{1}{4}x$

$Here\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}....\infty =\frac{{\pi }^2}{8}Let\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....\infty =xThenx=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....\infty =(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty )+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\infty )=\frac{{\pi }^2}{8}+\frac{1}{4}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\infty )=\frac{{\pi }^2}{8}+\frac{1}{4}x$