A) \[\frac{{{e}^{a}}{{b}^{101}}}{101!}\]
B) \[\frac{-{{e}^{a}}{{b}^{100}}}{100!}\]
C) \[\frac{{{e}^{a}}b}{101!}\]
D) \[\frac{-{{e}^{a}}{{b}^{101}}}{101!}\]
E) \[\frac{{{e}^{a}}{{b}^{100}}}{101!}\]
Correct Answer: D
Solution :
Let \[S=1+\frac{(a-bx)}{1!}+\frac{{{(a-bx)}^{2}}}{2!}+....\] \[={{e}^{a-bx}}={{e}^{a}}{{e}^{-bx}}\] \[\therefore \]The coefficient of\[{{x}^{101}}\]in S = coefficient of\[{{x}^{101}}\]in \[{{e}^{a}}{{e}^{-bx}}\] \[=\frac{-{{e}^{a}}{{b}^{101}}}{101!}\]
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