If n is odd natural number , then coefficient of `x^n` in `(e^(5x)+e^x)/(e^(3x))` is

Options

A. `2^n/(n!)`
B. `2/n`
C. `n/2`
D. 0

Answer

To find the coefficient of \( x^n \) in the expression

\[ \frac{e^{5x} + e^x}{e^{3x}}, \]

we can start by simplifying the expression.

### Step 1: Simplify the Expression
We can rewrite the expression as:

\[ e^{5x} + e^x = e^{3x} \cdot \left( e^{2x} + 1 \right). \]

Thus, we have:

\[ \frac{e^{5x} + e^x}{e^{3x}} = e^{2x} + e^{-3x}. \]

### Step 2: Expand Each Exponential Function
Now we will expand \( e^{2x} \) 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 \( e^{2x} \):

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

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 Series
Now, we need to find the coefficient of \( x^n \) in the combined series:

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

This can be combined as:

\[ \sum_{m=0}^{\infty} \left( \frac{2^m + (-2)^m}{m!} \right) x^m. \]

### Step 4: Find the Coefficient of \( x^n \)
The coefficient of \( x^n \) is:

\[ \frac{2^n + (-2)^n}{n!}. \]

### Step 5: Analyze for Odd \( n \)
Since \( n \) is given to be an odd natural number, we can analyze \( (-2)^n \):

\[ (-2)^n = -2^n. \]

Thus, we have:

\[ 2^n + (-2)^n = 2^n - 2^n = 0. \]

### Conclusion
Therefore, the coefficient of \( x^n \) in the given expression is:

\[ \frac{0}{n!} = 0. \]

So, the final answer is:

\[ \text{The coefficient of } x^n \text{ is } 0. \]