Q. Let $C_{1},\ C_{2},\ C_{3}$ are the usual binomial coefficients. Let $S = C_{1} + 2C_{2} + 3C_{3} + ... + nC_{n}$, then $S$ equals
A) ${n2}^{n}$
B) $2^{n - 1}$
C) ${n2}^{n - 1}$
D) $2^{n + 1}$
Explanation
(c)
Let $S = C_{1} + 2C_{2} + 3C_{3} + \ldots + nC_{n} = \sum_{r = 1}^{n}\ r \bullet \ ^{n}C_{r}$
$= \sum_{r = 1}^{n}\ r \bullet \frac{n}{r}\ ^{n - 1}C_{r - 1}\ \ \ \left\lbrack \because\ ^{n}C_{r} = \frac{n}{r}\ ^{n - 1}C_{r - 1} \right\rbrack$
$= n\sum_{r = 1}^{n}\ \ ^{n - 1}C_{r - 1}$
$= n\left\lbrack \ ^{n - 1}C_{0} + \ ^{n - 1}C_{1} + \ ^{n - 1}C_{2} + \ldots + \ ^{n - 1}C_{n - 1} \right\rbrack$
$= n2^{n - 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.