The coefficient of x^{10} in the expansion of | vedclass

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(B) We know the expansion of $e^{-x}$ is given by $e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + (-1)^n \frac{x^n}{n!} + \dots$ Now,consider th

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$\frac{-28}{10!}$

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(B) We know the expansion of $e^{-x}$ is given by $e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + (-1)^n \frac{x^n}{n!} + \dots$ Now,consider the expression $(2+3x)e^{-x}$: $(2+3x)e^{-x} = 2e^{-x} + 3xe^{-x}$ To find the coefficient of $x^{10}$: $1$. In $2e^{-x}$,the term containing $x^{10}$ is $2 	imes \frac{(-1)^{10} x^{10}}{10!} = \frac{2}{10!} x^{10}$. $2$. In $3xe^{-x}$,the term containing $x^{10}$ is $3x 	imes \frac{(-1)^9 x^9}{9!} = 3x 	imes \frac{-x^9}{9!} = \frac{-3}{9!} x^{10}$. To combine these,express $\frac{-3}{9!}$ with a denominator of $10!$: $\frac{-3}{9!} = \frac{-3 	imes 10}{10!} = \frac{-30}{10!}$. Adding the coefficients: $\frac{2}{10!} - \frac{30}{10!} = \frac{2-30}{10!} = \frac{-28}{10!}$.

The coefficient of $x^{10}$ in the expansion of $(2+3x)e^{-x}$ is

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