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Question: The sum of series \({\log _4}2 - {\log _8}2 + {\log _{16}}2\)…………..is A. \({e^2}\) B. \({\log _e...

The sum of series log⁡42−log⁡82+log⁡162{\log _4}2 - {\log _8}2 + {\log _{16}}2…………..is
A. e2{e^2}
B. log⁡e2{\log _e}2
C. log⁡e3−2{\log _e}3 - 2
D. 1−log⁡e21 - {\log _e}2

Explanation

Solution

Hint: First simplify the sum using various logarithmic properties & then use the Maclaurin’s series to find the solution.

Complete step-by-step answer:
Here we have to find the sum of series of log⁡42−log⁡82+log⁡162……..{\log _4}2 - {\log _8}2 + {\log _{16}}2……..
This can be written as log⁡222−log⁡232+log⁡242……….{\log _{{2^2}}}2 - {\log _{{2^3}}}2 + {\log _{{2^4}}}2……….
Now we will be using the property of logarithm, log⁡ba=ln⁡aln⁡b{\log _b}a = \dfrac{{\ln a}}{{\ln b}} and log⁡bpa=1plog⁡ba{\log _{{b^p}}}a = \dfrac{1}{p}{\log _b}a
So using the above property we can write our series as
12ln⁡2ln⁡2−13ln⁡2ln⁡2+14ln⁡2ln⁡2................∞\dfrac{1}{2}\dfrac{{\ln 2}}{{\ln 2}} - \dfrac{1}{3}\dfrac{{\ln 2}}{{\ln 2}} + \dfrac{1}{4}\dfrac{{\ln 2}}{{\ln 2}}................\infty

ln⁡2\ln 2 gets cancelled from the numerator and denominator, so we get
⇒12−13+14............∞\Rightarrow \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4}............\infty
Let’s take negative common from the above sum, we get
−[−12+13−14................∞]- \left[ {\dfrac{{ - 1}}{2} + \dfrac{1}{3} - \dfrac{1}{4}................\infty } \right]………………………….. (1)
Now the Maclaurin’s series expansion for ln⁡(1+x)=x−x22+x33−x44+..................\ln (1 + x) = x - {\dfrac{x}{2}^2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ..................
Let’s add and subtract 1 from equation (1)
−[1+−12+13−14−1................∞]- \left[ {1 + \dfrac{{ - 1}}{2} + \dfrac{1}{3} - \dfrac{1}{4} - 1................\infty } \right]
Clearly from 1−12+13..........∞1 - \dfrac{1}{2} + \dfrac{1}{3}..........\infty leaving −1-1 forms Maclaurin’s series expansion for ln⁡(1+x)\ln (1 + x) at x=1x = 1
So we can write above as
−[ln⁡(1+1)−1]- \left[ {\ln (1 + 1) - 1} \right]
This is nothing but
−[ln⁡(2)−1]- \left[ {\ln (2) - 1} \right]
Multiplying the negative sign inside, we get
1−ln⁡(2)1 - \ln (2)
Using ln⁡a=log⁡ea\ln a = {\log _e}a we can write the above value as
1−log⁡e21 - {\log _e}2
So option (D) is correct.

Note: Whenever we come across such problems, the key point is to figure out which series expansion we are dealing with and to retrieve back the function from this series, important logarithm properties are also advised to be remembered as it helps solving such problems.