Q. The sum of series $1 + \frac{1}{4.2!} + \frac{1}{16.4!} + \frac{1}{64.6!} + \ldots\infty$ is

A) $\frac{e + 1}{2\sqrt{e}}$
B) $\frac{e - 1}{2\sqrt{e}}$
C) $\frac{e + 1}{\sqrt{e}}$
D) $\frac{e - 1}{\sqrt{e}}$
Explanation

(a)
We know that, $\frac{e^{x} + e^{- x}}{2} = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + ...$
Put $x = \frac{1}{2},\$we get
$\frac{e^{1/2} + e^{- 1/2}}{2} = 1 + \left( \frac{1}{2} \right)^{2}\frac{1}{2!} + \left( \frac{1}{2} \right)^{4}\frac{1}{4!} + ...$
$\Rightarrow \ \frac{e + 1}{2\sqrt{e}} = 1 + \frac{1}{4.2!} + \frac{1}{16.4!} + ...\infty$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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