Binomial Theorem And Its Simple Applications Question 199
Question: If $ x\ne 0 $ , then the sum of the series $ 1+\frac{x}{2!}+\frac{2x^{2}}{3!}+\frac{3x^{3}}{4!}+…….\infty $ is
Options:
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} $
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Answer:
Correct Answer: D
Solution:
- [d] The general term of the series $ \frac{x}{2!}+\frac{2x^{2}}{3!}+\frac{3x^{3}}{4!}+…..\infty $ is $ T_{n}=\frac{nx^{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{2x^{2}}{3!}+\frac{3x^{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{xe^{x}-e^{x}+1+x}{x}=\frac{(x-1)e^{x}+(1+x)}{x} $
BETA
