Jee Corner (Main + Advanced) 3

Solving jee question in jee style(No pen paper required),
$\dfrac{1}{1^4} + \dfrac{1}{2^4} + \dfrac{1}{3^4} + \ldots + = \dfrac{\pi^4}{90}$ $\dfrac{1}{1^4} + \dfrac{1}{3^4} + \ldots = \dfrac{\pi^4}{90} - \left(\dfrac{1}{2^4} + \dfrac{1}{4^4} + \ldots \right)$ .
Therefore, answer must be less than $\dfrac{\pi^4}{90}$ and $\dfrac{\pi^4}{96}$ is the only option satisfying this condition.
If you want actual solution, you can see @Devarsh Wali solution.

You can take 1/16 common from the even expansions of the terms that will become pi^4 /90 So finally we get pi^2 /90 (1-1/16) = (pi^2/90 * (15/16)= pi^2 /96