The coefficient of `x^(n)` in the expansion of `(e^(7x)+e^(x))/(e^(3x))` is

Options

A. `(4^(n-1) + (-2)^(n))/(n!)`
B. `(4^(n-1) + 2^(n))/(n!)`
C. `(4^(n-1) + (-2)^(n-1))/(n!)`
D. `(4^(n) + (-2)^(n))/(n!)`

Answer

To find the coefficient of \( x^n \) in the expansion of \( \frac{e^{7x} + e^x}{e^{3x}} \), we can follow these steps:

1. **Rewrite the Expression**:
We start by rewriting the given expression:
\[ \frac{e^{7x} + e^x}{e^{3x}} = e^{7x - 3x} + e^{x - 3x} = e^{4x} + e^{-2x} \]

2. **Expand Each Exponential**:
We can expand \( e^{4x} \) and \( e^{-2x} \) using their Taylor series:
\[ e^{4x} = \sum_{k=0}^{\infty} \frac{(4x)^k}{k!} = \sum_{k=0}^{\infty} \frac{4^k x^k}{k!} \]
\[ e^{-2x} = \sum_{m=0}^{\infty} \frac{(-2x)^m}{m!} = \sum_{m=0}^{\infty} \frac{(-2)^m x^m}{m!} \]

3. **Combine the Series**:
Now we combine the two expansions:
\[ e^{4x} + e^{-2x} = \sum_{k=0}^{\infty} \frac{4^k x^k}{k!} + \sum_{m=0}^{\infty} \frac{(-2)^m x^m}{m!} \]

4. **Find the Coefficient of \( x^n \)**:
The coefficient of \( x^n \) in the combined series is the sum of the coefficients from each series:
\[ \text{Coefficient of } x^n = \frac{4^n}{n!} + \frac{(-2)^n}{n!} \]

5. **Simplify the Result**:
We can factor out \( \frac{1}{n!} \):
\[ \text{Coefficient of } x^n = \frac{1}{n!} (4^n + (-2)^n) \]

Thus, the coefficient of \( x^n \) in the expansion of \( \frac{e^{7x} + e^x}{e^{3x}} \) is:
\[ \frac{4^n + (-2)^n}{n!} \]