A) \[\frac{(1+2n)}{n!}\]
B) \[{{(-1)}^{n}}.\frac{\left( 1+2n \right)}{n!}\]
C) \[{{(-1)}^{n}}.\frac{(1-2n)}{n!}\]
D) \[{{(-1)}^{n}}.\frac{(1+4n)}{n!}\]
Correct Answer: B
Solution :
Now, \[\frac{1-2x}{{{e}^{x}}}=(1-2x){{e}^{-x}}\] \[=(1-2x)\left( 1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+...+{{(-1)}^{n}}\frac{{{x}^{n}}}{n!}+... \right)\] \[\therefore \]Coefficient of \[{{x}^{n}}\]in \[\frac{1-2x}{{{e}^{x}}}=\frac{{{(-1)}^{n}}}{n!}+\frac{2{{(-1)}^{n}}}{(n-1)!}\] \[={{(-1)}^{n}}\frac{(1+2n)}{n!}\]
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