Q. The coefficient of $x^{n}$ in the expansion of $\frac{(a - bx)}{e^{x}}$ is
A) $\frac{( - 1)^{n}}{n!}(a + bn)$
B) $\frac{( - 1)^{n}}{n!}(b + an)$
C) $\frac{( - 1)^{n + 1}}{n!}(a + bn)$
D) None of the above
Explanation
(a)
$(a - bx)e^{- x} =$($a - bx$)$\left\lbrack 1 - x + \frac{x^{2}}{2!} - ... \right\rbrack$
$\therefore$ Coefficient of$\ x^{n}$is$( - 1)^{n}\frac{a}{n!} + \frac{( - b)( - 1)^{n - 1}\ }{(n - 1)!}$
$= \frac{( - 1)^{n}}{(n - 1)!}\frac{(a + bn)}{n} = \frac{( - 1)^{n}}{n!}(a + bn)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.