Q. If $x,y,z$ are three consecutive positive integers, then $\log_{e}\sqrt{x} + \log_{e}\sqrt{z} + \left( \frac{1}{2xz + 1} \right) + \frac{1}{3}\left( \frac{1}{2xz + 1} \right)^{3} + \frac{1}{5}\left( \frac{1}{2xz + 1} \right)^{5} + \ ...$is

A) $\log_{e}\sqrt{y}$
B) $\log_{e}y$
C) $\log_{e}y^{2}$
D) None of these
Explanation

(b)
Let $x = n,\ y = n + 1$ and $z = n + 2$, where $n$ is a positive integer.
$\therefore\ \log_{e}\sqrt{x} + \log_{e}\sqrt{z} + \left( \frac{1}{2xz + 1} \right) + \frac{1}{3}\left( \frac{1}{2xz + 1} \right)^{3} + \frac{1}{5}\left( \frac{1}{2xz + 1} \right)^{5} + ...$
$= \log_{e}\sqrt{xz} + \frac{1}{2}\log_{e}\left\lbrack \frac{1 + \frac{1}{2xz + 1}}{1 - \frac{1}{2xz + 1}} \right\rbrack$
$= \log_{e}\sqrt{xz} + \frac{1}{2}\log_{e}\left( \frac{2xz + 2}{2xz} \right)$
$= \log_{e}\sqrt{n(n + 2)} + \log_{e}\sqrt{\frac{n(n + 2) + 1}{n(n + 2)}}$
$= \log_{e}\sqrt{(n + 1)^{2}} = \log_{e}{(n + 1)}$
$= \log_{e}y$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.