Coefficient of `x^(n)` in `(e^(5x)+e^(x))/(e^(2x))` is .....

Options

A. 1) `1/(n!) (3^(n) +1)`
B. 2) `(3^n+5^n)/(n!)`
C. 3) `(3^n+(-1)^n)/(n!)`
D. 4) `1/(n!)`

Answer

To find the coefficient of \( x^n \) in the expression \( \frac{e^{5x} + e^x}{e^{2x}} \), we can follow these steps:

### Step 1: Simplify the Expression
We start by simplifying the given expression:

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

### Step 2: Expand the Exponential Functions
Next, we will expand \( e^{3x} \) and \( e^{-x} \) using their Taylor series expansions.

The Taylor series expansion for \( e^x \) is:

\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]

Thus, for \( e^{3x} \):

\[ e^{3x} = \sum_{k=0}^{\infty} \frac{(3x)^k}{k!} = \sum_{k=0}^{\infty} \frac{3^k x^k}{k!} \]

And for \( e^{-x} \):

\[ e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} \]

### Step 3: Combine the Series
Now, we combine the two series:

\[ e^{3x} + e^{-x} = \sum_{k=0}^{\infty} \frac{3^k x^k}{k!} + \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} \]

### Step 4: Find the Coefficient of \( x^n \)
To find the coefficient of \( x^n \), we can look at the combined series:

\[ \sum_{k=0}^{\infty} \left( \frac{3^k + (-1)^k}{k!} \right) x^k \]

The coefficient of \( x^n \) in this series is:

\[ \frac{3^n + (-1)^n}{n!} \]

### Final Answer
Thus, the coefficient of \( x^n \) in the expression \( \frac{e^{5x} + e^x}{e^{2x}} \) is:

\[ \frac{3^n + (-1)^n}{n!} \]