Q. If $x = 1 + 2 + \frac{4}{2!} + \frac{8}{3!} + \frac{16}{4!} + \ldots$, then $x^{- 1}$ is equal to
A) $e^{- 2}$
B) $e^{2}$
C) $e^{\frac{1}{2}}$
D) None of these
Explanation
(a)
Given, $x = 1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...$
$\Rightarrow x = e^{2}$
$\Rightarrow x^{- 1} = e^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.