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 look at the terms where the power of $x$ is $k$:$1$. From $\sum \frac{x^n}{n!}$,the coefficient is $\frac{1}{k!}$.$2$. From $-2 \sum \frac{x^{n+1}}{n!}$,we set $n+1=k \implies n=k-1$,so the coefficient is $-\frac{2}{(k-1)!}$.$3$. From $-\sum \frac{x^{n+2}}{n!}$,we set $n+2=k \implies n=k-2$,so the coefficient is $-\frac{1}{(k-2)!}$.Summing these,the coefficient of $x^k$ is $\frac{1}{k!} - \frac{2}{(k-1)!} - \frac{1}{(k-2)!}$.Multiplying the terms to have a common denominator $k!$:$\frac{1}{k!} - \frac{2k}{k!} - \frac{k(k-1)}{k!} = \frac{1-2k-(k^2-k)}{k!} = \frac{1-2k-k^2+k}{k!} = \frac{1-k-k^2}{k!}$.

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

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