In the expansion of frac{e^{5x} + e^x}{e^{3x} | vedclass

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(B) Given expression: $\frac{e^{5x} + e^x}{e^{3x}} = e^{2x} + e^{-2x}$.Using the exponential series expansion $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^

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(B) Given expression: $\frac{e^{5x} + e^x}{e^{3x}} = e^{2x} + e^{-2x}$.Using the exponential series expansion $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!} + \dots$,we have:$e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + \dots$$e^{-2x} = 1 - (2x) + \frac{(-2x)^2}{2!} - \frac{(-2x)^3}{3!} + \frac{(-2x)^4}{4!} - \dots$Adding these two series,the odd powers of $x$ cancel out:$e^{2x} + e^{-2x} = 2 \left[ 1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \dots \right]$$= 2 + 2 \cdot \frac{4x^2}{2} + 2 \cdot \frac{16x^4}{24} + \dots$$= 2 + 4x^2 + \frac{32x^4}{24} + \dots$$= 2 + 4x^2 + \frac{4}{3}x^4 + \dots$The coefficient of $x^4$ is $\frac{4}{3}$.

In the expansion of $\frac{e^{5x} + e^x}{e^{3x}}$,the coefficient of $x^4$ is

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