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