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 these
Correct Answer: A
Solution :
Key Idea: \[{{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+....\] \[{{e}^{-x}}=1-x+\frac{{{x}^{2}}}{2!}-....\] Now, \[\frac{(a-bx)}{{{e}^{x}}}=(a-bx){{e}^{-x}}\] \[=(a-bx)\left[ 1-x+\frac{{{x}^{2}}}{2!}-.... \right]\] \[\therefore \]Coefficient of\[{{x}^{n}}\]is \[{{(-1)}^{n}}\frac{a}{n!}+\frac{(-b){{(-1)}^{n-1}}}{(n-1)!}\] \[=\frac{{{(-1)}^{n}}}{(n-1)!}\left( \frac{a}{n}+\frac{b}{1} \right)\] \[=\frac{{{(-1)}^{n}}}{(n-1)1}\frac{(a+bn)}{n}=\frac{{{(-1)}^{n}}}{n!}(a+bn)\]
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