Q. The expression $\left( 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + ...\infty \right)^{2}$ will be represented in ascending power of $x$ as
A) $1 + \frac{2^{2}x^{2}}{2!} + \frac{2^{4}x^{4}}{4!} + ...\infty$
B) $1 + \frac{(2x)^{2}}{2!} + \frac{2^{2}x^{4}}{4!} + ...\infty$
C) $1 + \frac{(2x)^{2}}{2.2!} + \frac{2x^{4}}{4!} + ...\infty$
D) $1 + \frac{(2x)^{2}}{2.2!} + \frac{(2x)^{4}}{2.4!} + ...\infty$
Explanation
(d)
Given, $\left( 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + ... \right)^{2} = \left( \frac{e^{x} + e^{- x}}{2} \right)^{2}$
$= \frac{1}{4}(e^{2x} + e^{- 2x} + 2)$
$= \frac{1}{4}\left\lbrack 4 + 2\left\{ \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} + ... \right\} \right\rbrack$
$= 1 + \frac{(2x)^{2}}{2.2!} + \frac{(2x)^{4}}{2.4!} + ...$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.