Q. The inequality $n! > 2^{n - 1}$ is true for
A) $n > 2$
B) $n \in N$
C) $n > 3$
D) None of these
Explanation
(a)
Let $P(n) \equiv n! > 2^{n - 1}$
$P(3) \equiv 6 > 4$
Let $P(k) \equiv k! > 2^{k - 1}$ is true.
$\therefore P(k + 1) = (k + 1)! = \left( k^{'} + 1 \right)k!$
$> (k + 1)2^{k - 1}$
$> 2^{k}$ $(as\ k + 1 > 2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.