The coefficient of $x^{n}$ in the expansion of $(1 - x) e^{x}$ is

\frac{1}{n!}

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\frac{1}{(n + 1)!}

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\frac{1 - n}{n!}

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\frac{n - 1}{n!}

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Solution

The correct option is C $\frac{1 - n}{n!}$
$(1 - x) e^{x} = e^{x} - x e^{x} = [1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots] - x [1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots]$
Coefficient of $x^{n}$ will be
$= \frac{1}{n!} - \frac{1}{(n - 1)!} = \frac{1}{n \times (n - 1)!} - \frac{1}{(n - 1)!} = \frac{1 - n}{n!}$

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The coefficient of xn in the expansion of (1−x)ex is

The correct option is C 1−nn!(1−x)ex=ex−xex=[1+x1!+x22!+⋯]−x[1+x1!+x22!+⋯] Coefficient of xn will be =1n!−1(n−1)!=1n×(n−1)!−1(n−1)!=1−nn!

The coefficient of $x^{n}$ in the expansion of $(1 - x) e^{x}$ is