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

Options

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

### Step 1: Simplify the Expression
We start with the expression:
\[ \frac{e^{7x} + e^x}{e^{3x}} \]
We can rewrite this by separating the terms in the numerator:
\[ = \frac{e^{7x}}{e^{3x}} + \frac{e^x}{e^{3x}} = e^{7x - 3x} + e^{x - 3x} = e^{4x} + e^{-2x} \]

### Step 2: Expand Each Exponential Function
Next, we will expand both \( e^{4x} \) and \( e^{-2x} \) using their Taylor series expansions.

The Taylor series expansion for \( e^{kx} \) is given by:
\[ e^{kx} = \sum_{m=0}^{\infty} \frac{(kx)^m}{m!} \]

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

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

### Step 3: Combine the Expansions
Now we will combine the two expansions:
\[ e^{4x} + e^{-2x} = \sum_{m=0}^{\infty} \frac{4^m x^m}{m!} + \sum_{m=0}^{\infty} \frac{(-2)^m x^m}{m!} \]

### Step 4: Find the Coefficient of \( x^n \)
To find the coefficient of \( x^n \), we can look at the combined series:
\[ \text{Coefficient of } x^n = \frac{4^n}{n!} + \frac{(-2)^n}{n!} \]

### Step 5: Write the Final Answer
Thus, the coefficient of \( x^n \) in the expansion is:
\[ \frac{4^n + (-2)^n}{n!} \]