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If \(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots\) up to \(\infty = \frac{\pi^2}{6}\), then the value of \(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots\) up to \(\infty\) is

Question and options

If 112+122+132+…\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots up to ∞=π26\infty = \frac{\pi^2}{6}, then the value of 112+132+152+…\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots up to ∞\infty is
  1. π28\frac{\pi^2}{8}
  2. π216\frac{\pi^2}{16}
  3. π212\frac{\pi^2}{12}
  4. π236\frac{\pi^2}{36}

Answer and explanation

Correct answer: A - \(\frac{\pi^2}{8}\)
Even terms sum to (1/4)(pi^2/6)=pi^2/24. Odd terms = pi^2/6 - pi^2/24 = pi^2/8.

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  • IPMAT Indore

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