Q103 / 200 · Manipal (MET) 2011 · Mathematics

$1+\frac{1+x}{2!}+\frac{1+x+x^2}{3!}+\frac{1+x+x^2+x^3}{4!}+\ldots$ is equal to

  • A
    $\frac{e^x+1}{x-1}$
  • B
    $\frac{e^x+1}{x+1}$
  • C
    $\frac{e^x-e}{x+1}$
  • Correct answer
    D
    $\frac{e^x-e}{x-1}$

Solution

$T_n= \frac{1+x+x^2+\ldots+x^{n-1}}{n!}$ $=\frac{x^n-1}{x-1} \cdot \frac{1}{n!}$ $=\frac{1}{1-x}\left(\frac{x^n}{n!}-\frac{1}{n!}\right)$ $\Rightarrow \sum_{n=1}^{\infty} \frac{1}{1-x}\left(\frac{x^n}{n!}-\frac{1}{n!}\right)$ $=\frac{1}{(x-1)}\left[\sum_{n=1}^{\infty} \frac{x^n}{n!}-\sum_{n=1}^{\infty} \frac{1}{n!}\right]$ $=\frac{1}{(x-1)}\left(e^x-e\right)$

Hence, the correct answer is option: 'D'.

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