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Mathematics Question on Sum of First n Terms of an AP

The sum of the infinite series 12(13+14)−14(132+142)+16(133+143)−...\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{4}\left(\frac{1}{3^{2}}+\frac{1}{4^{2}}\right)+\frac{1}{6}\left(\frac{1}{3^{3}}+\frac{1}{4^{3}}\right)-... is equal to

A

12log 2\frac{1}{2} log\,2

B

log35log \frac{3}{5}

C

log53log \frac{5}{3}

D

12log53\frac{1}{2} log \frac{5}{3}

Answer

12log53\frac{1}{2} log \frac{5}{3}

Explanation

Solution

Consider 12(13+14)−14(132+142)+16(133+143)−...\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{4}\left(\frac{1}{3^{2}}+\frac{1}{4^{2}}\right)+\frac{1}{6}\left(\frac{1}{3^{3}}+\frac{1}{4^{3}}\right)-... =(12.13−14.132+16.133...)+(12.14−14.142+16.143−)...=\left(\frac{1}{2}. \frac{1}{3}-\frac{1}{4}.\frac{1}{3^{2}}+\frac{1}{6}. \frac{1}{3^{3}}...\right)+\left(\frac{1}{2}. \frac{1}{4}-\frac{1}{4}. \frac{1}{4^{2}}+\frac{1}{6}. \frac{1}{4^{3}}-\right)... =12(13−12(132)+13(133)...)+12(14−12.142+13.143..)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2}\left(\frac{1}{3^{2}}\right)+\frac{1}{3}\left(\frac{1}{3^{3}}\right)...\right)+\frac{1}{2}\left(\frac{1}{4}-\frac{1}{2}. \frac{1}{4^{2}}+\frac{1}{3}. \frac{1}{4^{3}}..\right) =12(x−x22+x33...)+12(y−y22+Y33)=\frac{1}{2}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}...\right)+\frac{1}{2}\left(y-\frac{y^{2}}{2}+\frac{Y^{3}}{3}\right) where x=13,y=14x=\frac{1}{3}, y=\frac{1}{4} =12log(1+x)+12log(1+y)=\frac{1}{2}log\left(1+x\right)+\frac{1}{2}log\left(1+y\right) =12log(1+13)+12log(1+14)=12log53=\frac{1}{2}log\left(1+\frac{1}{3}\right)+\frac{1}{2} log\left(1+\frac{1}{4}\right)=\frac{1}{2}log \frac{5}{3}