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Sum of Reciprocals of Odd Squares = pi^2/8 | JEE

JEE Maths question with a full step-by-step solution.

Question
If 112+122+132+⋯=π26\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{6}, then 112+132+152+⋯\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots is equal to
Aπ28\frac{\pi^2}{8}correct
Bπ212\frac{\pi^2}{12}
Cπ23\frac{\pi^2}{3}
Dπ22\frac{\pi^2}{2}
Solution
Step 1: The even part factors out 14\frac14:
122+142+⋯=14(112+122+⋯ )=14⋅π26=π224.\frac{1}{2^2}+\frac{1}{4^2}+\cdots=\frac14\left(\frac{1}{1^2}+\frac{1}{2^2}+\cdots\right)=\frac14\cdot\frac{\pi^2}{6}=\frac{\pi^2}{24}.
 Step 2: Odd part == total −- even:
π26−π224=4π2−π224=π28.\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{4\pi^2-\pi^2}{24}=\frac{\pi^2}{8}.
 Correct answer: (1)
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