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 will follow these steps:

### Step 1: Simplify the Expression
We start by rewriting the expression:
\[ \frac{e^{7x} + e^{x}}{e^{3x}} = e^{7x - 3x} + e^{x - 3x} = e^{4x} + e^{-2x} \]

### Step 2: Expand \( e^{4x} \) and \( e^{-2x} \)
Using the Taylor series expansion for \( e^x \):
\[ e^{x} = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]
we can expand \( e^{4x} \) and \( e^{-2x} \).

For \( e^{4x} \):
\[ e^{4x} = \sum_{k=0}^{\infty} \frac{(4x)^k}{k!} = \sum_{k=0}^{\infty} \frac{4^k x^k}{k!} \]

For \( e^{-2x} \):
\[ e^{-2x} = \sum_{k=0}^{\infty} \frac{(-2x)^k}{k!} = \sum_{k=0}^{\infty} \frac{(-2)^k x^k}{k!} \]

### Step 3: Combine the Expansions
Now we combine the two expansions:
\[ e^{4x} + e^{-2x} = \sum_{k=0}^{\infty} \frac{4^k x^k}{k!} + \sum_{k=0}^{\infty} \frac{(-2)^k x^k}{k!} \]
This can be rewritten as:
\[ = \sum_{k=0}^{\infty} \left( \frac{4^k + (-2)^k}{k!} \right) x^k \]

### Step 4: Identify the Coefficient of \( x^n \)
The coefficient of \( x^n \) in the expansion is given by:
\[ \frac{4^n + (-2)^n}{n!} \]

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

---