The sum of the series
${log}_{4} 2 - {log}_{8} 2 + {log}_{16} 2 - {log}_{32} 2 + . . . .$ is

e^{2}

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{log}_{e} 2 + 1

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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 D $1 - {log}_{e} 2$
We have
${log}_{4} 2 - {log}_{8} 2 + {log}_{16} 2 - {log}_{32} 2 + . . . . = \frac{1}{{log}_{2} 4} - \frac{1}{{log}_{2} 8} + \frac{1}{{log}_{2} 16} - \frac{1}{{log}_{2} 32} + . . . . .$
$= \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = 1 - {log}_{e} 2$

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