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

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(B) We have $\frac{1-2x}{e^x} = (1-2x)e^{-x}$.Using the expansion $e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = \sum_{k=0}^{\infty} (-1)^k \frac{x

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$(-1)^n \cdot \frac{(1+2n)}{n!}$

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(B) We have $\frac{1-2x}{e^x} = (1-2x)e^{-x}$.Using the expansion $e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}$,we get:$(1-2x) \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!} = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!} - 2x \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}$.The coefficient of $x^n$ is obtained from the first term when $k=n$ and from the second term when $k=n-1$:Coefficient of $x^n = \frac{(-1)^n}{n!} - 2 \cdot \frac{(-1)^{n-1}}{(n-1)!}$.Since $(-1)^{n-1} = -(-1)^n$,we have:Coefficient of $x^n = \frac{(-1)^n}{n!} + 2 \cdot \frac{(-1)^n}{(n-1)!} = \frac{(-1)^n}{n!} [1 + 2n] = (-1)^n \cdot \frac{(1+2n)}{n!}$.

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

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