In the expansion of (e^x - 1)^2,the coefficie | vedclass

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

Vedclass · Accepted Answer

$7/12$

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(B) The expression is $(e^x - 1)^2 = e^{2x} - 2e^x + 1$. Using the expansion $e^t = 1 + \frac{t}{1!} + \frac{t^2}{2!} + \frac{t^3}{3!} + \frac{t^4}{4!} + \dots$,we have: $e^{2x} = 1 + \frac{2x}{1!} + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + \dots$ $-2e^x = -2 \left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \right)$ Adding these to the constant term $1$: $(e^{2x} - 2e^x + 1) = (1 - 2 + 1) + x(2 - 2) + x^2 \left( \frac{4}{2} - \frac{2}{2} \right) + x^3 \left( \frac{8}{6} - \frac{2}{6} \right) + x^4 \left( \frac{16}{24} - \frac{2}{24} \right) + \dots$ The coefficient of $x^4$ is $\frac{16 - 2}{24} = \frac{14}{24} = \frac{7}{12}$.

In the expansion of $(e^x - 1)^2$,the coefficient of $x^4$ will be

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