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Given that 1+122+132+142+...=π261 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}, the value of 1+132+152+172+...1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ... is

Solution

✅ Correct Option: 4
Solution figure for IPMAT Indore 2025 MCQ question 1 (Algebra)


Note: This IPMAT Indore 2025 question is repeated from IPMAT Indore 2020.

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