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

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(B) We know that the expansion of $e^{-x}$ is given by $e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that the expansion of $e^{-x}$ is given by $e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + \frac{(-1)^n x^n}{n!} + \dots$ Now,consider the expression $(2+3x)e^{-x}$: $(2+3x)e^{-x} = 2e^{-x} + 3xe^{-x}$ The term containing $x^{10}$ in $2e^{-x}$ is $2 	imes \frac{(-x)^{10}}{10!} = \frac{2}{10!} x^{10}$. The term containing $x^{10}$ in $3xe^{-x}$ is $3x 	imes \frac{(-x)^9}{9!} = 3x 	imes \frac{-x^9}{9!} = -\frac{3}{9!} x^{10}$. To combine these,we write the second term with a denominator of $10!$: $-\frac{3}{9!} = -\frac{3 	imes 10}{10!} = -\frac{30}{10!}$. Thus,the coefficient of $x^{10}$ is $\frac{2}{10!} - \frac{30}{10!} = \frac{2-30}{10!} = \frac{-28}{10!}$.

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

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