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Question: The coefficient of \(x^{n}\)in the expansion of \(\frac{e^{7x} + e^{x}}{e^{3x}}\)is...

The coefficient of xnx^{n}in the expansion of e7x+exe3x\frac{e^{7x} + e^{x}}{e^{3x}}is

A

4n−1+(−2)nn!\frac{4^{n - 1} + ( - 2)^{n}}{n!}

B

4n−1+2nn!\frac{4^{n - 1} + 2^{n}}{n!}

C

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

D

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

Answer

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

Explanation

Solution

We have e7x+exe3x=e4x+e−2x=∑n=0∞(4x)nn!+∑n=0∞(−2x)nn!\frac{e^{7x} + e^{x}}{e^{3x}} = e^{4x} + e^{- 2x} = \sum_{n = 0}^{\infty}{\frac{(4x)^{n}}{n!} +}\sum_{n = 0}^{\infty}\frac{( - 2x)^{n}}{n!}

∴\thereforecoefficient of xnx^{n} in e7x+exe3x=4n+(−2)nn!\frac{e^{7x} + e^{x}}{e^{3x}} = \frac{4^{n} + ( - 2)^{n}}{n!}