log _4 2 - log _8 2 + log _{16} 2 - ldots is | vedclass

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(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$ Using the property $\log _b a = \frac{1}{\log _a b}$,we get: $= \frac{1}{\log

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$1 - \log _e 2$

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(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$ Using the property $\log _b a = \frac{1}{\log _a b}$,we get: $= \frac{1}{\log _2 4} - \frac{1}{\log _2 8} + \frac{1}{\log _2 16} - \ldots$ $= \frac{1}{\log _2 2^2} - \frac{1}{\log _2 2^3} + \frac{1}{\log _2 2^4} - \ldots$ $= \frac{1}{2 \log _2 2} - \frac{1}{3 \log _2 2} + \frac{1}{4 \log _2 2} - \ldots$ Since $\log _2 2 = 1$,the series becomes: $= \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots$ We know the Taylor series expansion $\log _e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$ For $x = 1$,$\log _e(1 + 1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots$ $\Rightarrow \log _e 2 = 1 - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots)$ $\Rightarrow \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots = 1 - \log _e 2$

$\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$ is equal to

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