Q. $\ ^{n}C_{0} - \frac{1}{2}\ ^{n}C_{1} + \frac{1}{3}\ ^{n}C_{2} - ... + {( - 1)}^{n}\frac{\ ^{n}C_{n}}{n + 1}$ is equal to
A) $n$
B) $\frac{1}{n}$
C) $\frac{1}{n + 1}$
D) $\frac{1}{n - 1}$
Explanation
(c)
Given, $\ ^{n}C_{0} - \frac{1}{2}\ ^{n}C_{1} + \frac{1}{3}\ ^{n}C_{2} - ... + {( - 1)}^{n}\frac{\ ^{n}C_{n}}{n + 1}$
At $n = 1,\ \ ^{1}C_{0} - \frac{1}{2}\ ^{1}C_{1} = 1 - \frac{1}{2} = \frac{1}{2}$
At $n = 2,\ \ ^{2}C_{0} - \frac{1}{2}\ ^{2}C_{1} + \frac{1}{3}\ ^{2}C_{2} = 1 - 1 + \frac{1}{3} = \frac{1}{3}$
Which is satisfied only in option (c)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.