The coefficient of x^k in the expansion of fr | vedclass

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(A) We have $\frac{1-2x-x^2}{e^{-x}} = (1-2x-x^2)e^x$.Expanding $e^x$ as $\sum_{n=0}^{\infty} \frac{x^n}{n!}$,we get:$(1-2x-x^2) \sum_{n=0}^{\infty} \

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$\frac{1-k-k^2}{k!}$

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(A) We have $\frac{1-2x-x^2}{e^{-x}} = (1-2x-x^2)e^x$.Expanding $e^x$ as $\sum_{n=0}^{\infty} \frac{x^n}{n!}$,we get:$(1-2x-x^2) \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} - 2 \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} - \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$.To find the coefficient of $x^k$,we extract terms where the power of $x$ is $k$:From the first sum,the coefficient is $\frac{1}{k!}$.From the second sum,we need $n+1=k$,so $n=k-1$,giving $-2 	imes \frac{1}{(k-1)!}$.From the third sum,we need $n+2=k$,so $n=k-2$,giving $-1 	imes \frac{1}{(k-2)!}$.Summing these coefficients:$\frac{1}{k!} - \frac{2}{(k-1)!} - \frac{1}{(k-2)!} = \frac{1 - 2k - k(k-1)}{k!} = \frac{1 - 2k - k^2 + k}{k!} = \frac{1 - k - k^2}{k!}$.

The coefficient of $x^k$ in the expansion of $\frac{1-2x-x^2}{e^{-x}}$ is

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