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

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$\frac{4^{n - 1} + 2^{n}}{n!}$

$\frac{4^{n - 1} + {(- 1)}^{n} . 2}{n!}$

$\frac{4^{n - 1} + {(-)}^{n - 1}}{n!}$

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

answer is A.

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Detailed Solution

$\begin{array}{rcl} \frac{e^{7 x} + e^{x}}{e^{3 x}} \\ = & e^{4 x} + e^{- 2 x} \\ x^{n} coof & = & \frac{4^{n}}{n} + \frac{{(- 2)}^{n}}{n} \end{array}$

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