Q52 / 160 · AP EAMCET 2006 · Mathematics

The coefficient of $x^n$ in $\frac{1-2 x}{e^x}$ is :

  • A
    $\frac{(1+2 n)}{n !}$
  • B
    $(-1)^n \cdot \frac{(1+2 n)}{n !}$
  • Correct answer
    C
    $(-1)^n \cdot \frac{(1-2 n)}{n !}$
  • D
    $(-1)^n \cdot \frac{(1+4 n)}{n !}$

Solution

$\frac{1-2 x}{e^x}=(1-2 x) e^{-x}$ $=(1-2 x)\left(1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\ldots\right.$ $\left.+(-1)^n \frac{x^n}{n !}+\ldots\right)$ $\therefore$ Coefficient of $x^n$ in $\frac{1-2 x}{e^x}=\frac{(-1)^n}{n !}-\frac{2(-1)^n}{(n-1) !}$ $=(-1)^n \frac{(1-2 n)}{n !}$

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

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