In the expansion of frac{1 - 2x + 3x^2}{e^x}, | vedclass

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Question

(B) The given expression is $(1 - 2x + 3x^2)e^{-x}$.Using the expansion $e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}

Vedclass · Accepted Answer

$-\frac{71}{120}$

Vedclass · Answer

(B) The given expression is $(1 - 2x + 3x^2)e^{-x}$.Using the expansion $e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots$,We have $(1 - 2x + 3x^2) \left( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120} + \dots ight)$.The coefficient of $x^5$ is obtained by multiplying terms from the first polynomial with corresponding terms from the expansion:$= 1 	imes \left( -\frac{1}{120} ight) + (-2) 	imes \left( \frac{1}{24} ight) + 3 	imes \left( -\frac{1}{6} ight)$,$= -\frac{1}{120} - \frac{2}{24} - \frac{3}{6}$,$= -\frac{1}{120} - \frac{10}{120} - \frac{60}{120} = -\frac{71}{120}$.

In the expansion of $\frac{1 - 2x + 3x^2}{e^x}$,the coefficient of $x^5$ will be

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