A) \[\frac{1}{n!}\]
B) \[\frac{1}{(n-1)!}\]
C) \[\frac{1}{n!}-\frac{1}{(n-1)!}\]
D) \[1\]
Correct Answer: A
Solution :
We have, \[\frac{{{e}^{x}}}{1-x}={{B}_{0}}+{{B}_{1}}x+{{B}_{2}}{{x}^{2}}+...+{{B}_{n}}{{x}^{n}}+...\] \[\Rightarrow \]\[{{e}^{x}}{{(1-x)}^{-1}}={{B}_{0}}+{{B}_{1}}x+{{B}_{2}}{{x}^{2}}+...+{{B}_{n}}{{x}^{n}}+...\] \[\Rightarrow \]\[\left( 1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+...+\frac{{{x}^{n-1}}}{(n-1)!}+\frac{{{x}^{n}}}{n!}+... \right)\times \] \[(1+x+{{x}^{2}}+...+{{x}^{n-1}}+{{x}^{n}}+...\infty )\] \[={{B}_{0}}+{{B}_{1}}x+{{B}_{2}}{{x}^{2}}+...+{{B}_{n}}{{x}^{n}}+...\] On comparing the coefficients of\[{{x}^{n}}\]and\[{{x}^{n-1}}\]on both sides, we get \[\frac{1}{n!}+\frac{1}{(n-1)!}+...+\frac{1}{2!}+\frac{1}{1!}+1={{B}_{n}}\] and\[\frac{1}{(n-1)!}+\frac{1}{(n-2)!}+...+\frac{1}{2!}+\frac{1}{1!}+1\] \[={{B}_{n-1}}\] \[{{B}_{n}}-{{B}_{n-1}}\,=\frac{1}{n!}\]
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