Sequence And Series Question 547
Question: In the expansion of $ \frac{a+bx+cx^{2}}{e^{x}} $ , the coefficient of $ x^{n} $ will be
Options:
A) $ \frac{a,{{(-1)}^{n}}}{n,!}+\frac{b{{(-1)}^{n-1}}}{(n-1),!}+\frac{c{{(-1)}^{n-2}}}{(n-2),!} $
B) $ \frac{a}{n,!}+\frac{b}{(n-1),!}+\frac{c}{(n-2),!} $
C) $ \frac{,{{(-1)}^{n}}}{n,!}+\frac{{{(-1)}^{n-1}}}{(n-1),!}+\frac{{{(-1)}^{n-2}}}{(n-2),!} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ (a+bx+cx^{2}){e^{-x}} $ $ =(a+bx+cx^{2}){ 1-\frac{x}{1!}+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+…. } $ Thus coefficient of xn = $ a,.,\frac{{{(-1)}^{n}}}{n,!}+b\frac{{{(-1)}^{n-1}}}{(n-1),!}+c\frac{{{(-1)}^{n-2}}}{(n-2),!} $ .
BETA
