Q. If ${(1 + x)}^{15} = a_{0} + a_{1}x + .\ldots + a_{15}x^{15}$, then $\sum_{r = 1}^{15}r\frac{a_{r}}{a_{r - 1}}$ is equal to

A) 110
B) 115
C) 120
D) 135
Explanation

(c)
$(1 + x)^{15} = a_{0} + a_{1}x + a_{2}x^{2} +$….$+ a_{15}x^{15}$
$\Rightarrow \ \ ^{15}C_{0} + \ ^{15}C_{1}x + \ ^{15}C_{2}x^{2} +$….$+ \ ^{15}C_{15}x^{15}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a_{0} + a_{1}x + a_{2}x^{2} +$….$+ a_{15}x^{15}$
Equating the coefficient of various powers of $x$, we get
$a_{0} = \ ^{15}C_{0},\ a_{1} = \ ^{15}C_{1},\ a_{2} = \ ^{15}C_{2}$, ….$a_{15} = \ ^{15}C_{15}$
$\therefore\sum_{r = 1}^{15}{r\frac{a_{r}}{a_{r - 1}} = \sum_{r = 1}^{15}{r\frac{\ ^{15}C_{r}}{\ ^{15}C_{r - 1}}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sum_{r = 1}^{15}r\frac{\frac{15!}{r!(15 - r)!}}{\frac{15}{(r - 1)(15 - r + 1)!}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \sum_{r = 1}^{15}\frac{r(r - 1)!(15 - r + 1)!}{r!(15 - r)!}$
$\ \ \ \ \ \ \ \ \ \ \ \ = \sum_{r = 1}^{15}{15 - r + 1}$
$\ \ \ \ \ \ \ \ \ \ \ = 15 + 14 + 13 +$…..$+ 2 + 1$
$\ \ \ \ \ \ \ \ \ \ = \frac{15(15 + 1)}{2} = 120$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.