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Question 78 of 95

Q.The value of 1−12(23)+13(23)2−14(23)3+…1 - \dfrac{1}{2}\left(\dfrac{2}{3}\right) + \dfrac{1}{3}\left(\dfrac{2}{3}\right)^2 - \dfrac{1}{4}\left(\dfrac{2}{3}\right)^3 + \ldots is:

(a) log⁡(53)\log\left(\dfrac{5}{3}\right)
(b) 32log⁡(53)\dfrac{3}{2}\log\left(\dfrac{5}{3}\right)
(c) 53log⁡(53)\dfrac{5}{3}\log\left(\dfrac{5}{3}\right)
(d) 23log⁡(23)\dfrac{2}{3}\log\left(\dfrac{2}{3}\right)
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2020MCQ· 1mImportance★★★★★
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Recognising the series as a scaled logarithm series gives the sum 32log⁡(53)\dfrac32\log\left(\dfrac53\right).

Recall the standard expansion ln⁡(1+x)=x−x22+x33−x44+…\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\ldots for ∣x∣<1|x|<1.

The given series is 1−12(23)+13(23)2−14(23)3+…=∑n=1∞(−1)n−1n(23)n−11-\dfrac12\left(\dfrac23\right)+\dfrac13\left(\dfrac23\right)^2-\dfrac14\left(\dfrac23\right)^3+\ldots=\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\left(\dfrac23\right)^{n-1}.

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