The coefficient of x^n in the expansion of fr | vedclass

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Question

(D) Given expression: $\frac{e^{7x}+e^x}{e^{3x}} = e^{4x} + e^{-2x}$.Using the Taylor series expansion for $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$,

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$\frac{4^{n}+(-2)^{n}}{n!}$

Vedclass · Answer

(D) Given expression: $\frac{e^{7x}+e^x}{e^{3x}} = e^{4x} + e^{-2x}$.Using the Taylor series expansion for $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$, we have:$e^{4x} = \sum_{n=0}^{\infty} \frac{(4x)^n}{n!} = \sum_{n=0}^{\infty} \frac{4^n x^n}{n!}$.$e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^n x^n}{n!}$.Adding these two series, the coefficient of $x^n$ is $\frac{4^n + (-2)^n}{n!}$.

The coefficient of $x^n$ in the expansion of $\frac{e^{7x}+e^x}{e^{3x}}$ is

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