In the expansion of frac{a + bx}{e^x},the coe | vedclass

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(C) The expression is given by $(a + bx)e^{-x}$.Using the expansion $e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}$,we have:$(a + bx) \left( 1 -

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$(-1)^r \frac{a - br}{r!}$

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(C) The expression is given by $(a + bx)e^{-x}$.Using the expansion $e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}$,we have:$(a + bx) \left( 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + \frac{(-1)^r x^r}{r!} + \dots \right)$.The coefficient of $x^r$ is obtained by multiplying $a$ with the coefficient of $x^r$ in $e^{-x}$ and $bx$ with the coefficient of $x^{r-1}$ in $e^{-x}$.Coefficient of $x^r = a \cdot \frac{(-1)^r}{r!} + b \cdot \frac{(-1)^{r-1}}{(r-1)!}$.$= \frac{(-1)^r a}{r!} + \frac{(-1)^{r-1} br}{r!}$.$= \frac{(-1)^r}{r!} (a - br)$.

In the expansion of $\frac{a + bx}{e^x}$,the coefficient of $x^r$ is

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