A) \[\frac{{{e}^{x}}+1}{x}\]
B) \[\frac{{{e}^{x}}\,(x-1)}{x}\]
C) \[\frac{{{e}^{x}}\,(x-1)+1}{x}\]
D) \[\frac{{{e}^{x}}\,(x-1)+1+x}{x}\]
Correct Answer: D
Solution :
| [d] The general term of the series |
| \[\frac{x}{2!}+\frac{2{{x}^{2}}}{3!}+\frac{3{{x}^{3}}}{4!}+.....\infty \] is |
| \[{{T}_{n}}=\frac{n{{x}^{n}}}{(n+1)!},n=1,2..........,\infty \] |
| \[=\frac{n+1-1}{(n+1)!}{{x}^{n}}=\frac{{{x}^{n}}}{n!}-\frac{1}{x}\frac{{{x}^{n+1}}}{(n+1)!}\] |
| \[\therefore 1+\frac{x}{2!}+\frac{2{{x}^{2}}}{3!}+\frac{3{{x}^{3}}}{4!}+.......\infty \] |
| \[=1+\sum\limits_{n=1}^{\infty }{\frac{{{x}^{n}}}{n!}-\frac{1}{x}\sum\limits_{n=1}^{\infty }{\frac{{{x}^{n+1}}}{(n+1)!}}}\] |
| \[=1+({{e}^{x}}-1)-\frac{1}{x}({{e}^{x}}-1-x)\] |
| \[=\frac{x{{e}^{x}}-{{e}^{x}}+1+x}{x}=\frac{(x-1){{e}^{x}}+(1+x)}{x}\] |
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