Q. $1 + \frac{\left( \log_{e}n \right)^{2}}{2!} + \frac{\left( \log_{e}n \right)^{4}}{4!} + ...$ is equal to
A) $n$
B) $\frac{1}{n}$
C) $\frac{1}{2}(n + n^{- 1})$
D) $\frac{1}{2}(e^{n} + e^{- n})$
Explanation
(c)
$1 + \frac{\left( \log_{e}n \right)^{2}}{2!} + \frac{\left( \log_{e}n \right)^{4}}{4!} + ...$
$= \frac{e^{\log n} + e^{- \log n}}{2}$
$= \frac{e^{\log n} + e^{\log n^{- 1}}}{2} = \frac{n + n^{- 1}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.