The sum of the series ${log}_{4} 2 - {log}_{8} 2 + {log}_{16} 2 - . . .$ is

e^{2}

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{log}_{e} 2

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{log}_{e} 3 - 2

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1 - {log}_{e} 2

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Solution

The correct option is B $1 - {log}_{e} 2$
${log}_{4} 2 - {log}_{8} 2 + {log}_{16} 2 - . . .$

$= (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - . . .) - 1 + 1$

$= - (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + . . .) + 1$

$= 1 - log (1 + 1) = 1 - {log}_{e} 2$

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