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Question: \[\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \frac{1}{4.5} + ........\infty =\]...

11.2−12.3+13.4−14.5+........∞=\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \frac{1}{4.5} + ........\infty =

A

log⁡e(4e)\log_{e}\left( \frac{4}{e} \right)

B

log⁡ee4\log_{e}\frac{e}{4}

C

log⁡e4\log_{e}4

D

log⁡e2\log_{e}2

Answer

log⁡e(4e)\log_{e}\left( \frac{4}{e} \right)

Explanation

Solution

We know that, log⁡e2=11.2+13.4+15.6+.......∞\log_{e}2 = \frac{1}{1.2} + \frac{1}{3.4} + \frac{1}{5.6} + .......\infty ..........(i)

Also log⁡e2=1−(12.3)−(14.5)−(16.7)−.......∞\log_{e}2 = 1 - \left( \frac{1}{2.3} \right) - \left( \frac{1}{4.5} \right) - \left( \frac{1}{6.7} \right) - .......\infty ………(ii)

By adding (i) and (ii), we get,

2log⁡e2=1+(11.2−12.3)+(13.4−14.5)+........⇒2log⁡e2−1=11.2−12.3+13.4−14.5+........⇒11.2−12.3+13.4−14.5+........∞=log⁡e4−log⁡ee=log⁡e(4e)2\log_{e}2 = 1 + \left( \frac{1}{1.2} - \frac{1}{2.3} \right) + \left( \frac{1}{3.4} - \frac{1}{4.5} \right) + ........ \Rightarrow 2\log_{e}2 - 1 = \frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \frac{1}{4.5} + ........ \Rightarrow \frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \frac{1}{4.5} + ........\infty = \log_{e}4 - \log_{e}e = \log_{e}\left( \frac{4}{e} \right)