Given that $$1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ...... = \dfrac{\pi^2}{6}$$, the value of $$1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \dfrac{1}{7^2} + ......$$ is

Given,

$$1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ...... = \dfrac{\pi^2}{6}$$ ------>(1)

Multiplying all the terms by $$\dfrac{1}{4}$$ we get,

$$\dfrac{1}{4}+\dfrac{1}{2^2\cdot4}+\dfrac{1}{3^2\cdot4}+\dfrac{1}{4^2\cdot4}+......=\dfrac{\pi^2}{6\cdot4}=\dfrac{\pi^2}{24}$$

or, $$\dfrac{1}{2^2}+\dfrac{1}{2^2\cdot2^2}+\dfrac{1}{3^2\cdot2^2}+\dfrac{1}{4^2\cdot2^2}+......=\dfrac{\pi^2}{6\cdot4}=\dfrac{\pi^2}{24}$$

or, $$\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+......=\dfrac{\pi^2}{6\cdot4}=\dfrac{\pi^2}{24}$$ ------->(2)

Subtracting (2) from (1),

$$1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+......=\dfrac{\pi^2}{6}-\dfrac{\pi^2}{24}=\dfrac{3\pi^2}{24}=\dfrac{\pi^2}{8}\ $$