Q. $\left( 1 + \frac{C_{1}}{C_{0}} \right)\left( 1 + \frac{C_{2}}{C_{1}} \right)\left( 1 + \frac{C_{3}}{C_{2}} \right)\ldots\left( 1 + \frac{C_{n}}{C_{n - 1}} \right)\ is\ equal\ to$
A) $\frac{n + 1}{n!}$
B) $\frac{{(n + 1)}^{n}}{(n - 1)!}$
C) $\frac{{(n - 1)}^{n}}{n!}$
D) $\frac{{(n + 1)}^{n}}{n!}$
Explanation
(d)
$\left( 1 + \frac{C_{1}}{C_{0}} \right)\left( 1 + \frac{C_{2}}{C_{1}} \right)\left( 1 + \frac{C_{3}}{C_{2}} \right)\ldots\left( 1 + \frac{C_{n}}{C_{n - 1}} \right)$
$= \left( 1 + \frac{n}{1} \right)\left( 1 + \frac{n(n - 1)}{2n} \right)\ldots\left( 1 + \frac{1}{n} \right)$
$= \left( \frac{1 + n}{1} \right)\left( \frac{1 + n}{2} \right)\ldots\left( \frac{1 + n}{n} \right) = \frac{{(1 + n)}^{n}}{n!}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.