The correct option is B 1 log_e2 Let S =1/2.3+1/4.5+1/6.7+……∞ S=3 2/3.2+ #160;5 4/5.4+ #160;7 6/7.6+........ S= #160;1/2 #160;1/3+ ...

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1/2.3+1/4.5+1...

Question

$\frac{1}{2.3} + \frac{1}{4.5} + \frac{1}{6.7} + \dots \infty =$

1 + {log}_{e} 2

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1 - {log}_{e} 2

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2 + {log}_{e} 2

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2 - {log}_{e} 2

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Solution

The correct option is B $1 - {log}_{e} 2$
Let S $= \frac{1}{2.3} + \frac{1}{4.5} + \frac{1}{6.7} + \dots \dots \infty$
$S = \frac{3 - 2}{3.2} + \frac{5 - 4}{5.4} + \frac{7 - 6}{7.6} + . . . . . . . .$
$S = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} - \frac{1}{7} + . . . . .$
We know $log (1 + X) = \sum (- 1)^{n + 1} \frac{x^{n}}{n}$
$S = 1 - [1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + . . . . .]$
$S = 1 - {log}_{e} (1 + 1)$
$\Rightarrow S = 1 - {log}_{e} 2$

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