The sum of the series 2x/(1 + x²) + (1/3)(2x/(1 + x²))³ + (1/5)(2x/(1 + x²))⁵ + ... to ∞ isश्रेणी 2x/(1 + x²) + (1/3)(2x/(1 + x²))³ + (1/5)(2x/(1 + x²))⁵ + ... अनन्त तक का योगफल है
श्रेणी 2x/(1 + x²) + (1/3)(2x/(1 + x²))³ + (1/5)(2x/(1 + x²))⁵ + ... अनन्त तक का योगफल है
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Correct answer: (B) logₑ((1 + x)/(1 − x))logₑ((1 + x)/(1 − x))
Explanation
With t = 2x/(1 + x²), the series t + t³/3 + t⁵/5 + ... = (1/2) log((1 + t)/(1 − t)). Here (1 + t)/(1 − t) = ((1 + x)/(1 − x))², so the sum is logₑ((1 + x)/(1 − x)) (for |x| < 1).t = 2x/(1 + x²) रखने पर श्रेणी t + t³/3 + t⁵/5 + ... = (1/2) log((1 + t)/(1 − t)) है। यहाँ (1 + t)/(1 − t) = ((1 + x)/(1 − x))², इसलिए योग logₑ((1 + x)/(1 − x)) है (|x| < 1 के लिए)।