In the expansion of `(e^(7x)+e^(3x))/(e^(5x))` the constant term is

Options

A. `(4^(n-1)+(1-2))^(n)/(n!)`
B. `(4^(n-1)+(2n))/(n!)`
C. `(4^(n-1)+(1-2))^(n-1)/(n!)`
D. none of these

Answer

To find the constant term in the expression \((e^{7x} + e^{3x}) / (e^{5x})\), we can follow these steps:

### Step 1: Simplify the Expression
We start with the expression:
\[ \frac{e^{7x} + e^{3x}}{e^{5x}} \]
We can separate the terms in the numerator:
\[ = \frac{e^{7x}}{e^{5x}} + \frac{e^{3x}}{e^{5x}} = e^{(7x - 5x)} + e^{(3x - 5x)} = e^{2x} + e^{-2x} \]

### Step 2: Recognize the Series Expansion
The expression \(e^{2x} + e^{-2x}\) can be expanded using the Taylor series for \(e^x\):
\[ e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \ldots \]
\[ e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = 1 - 2x + \frac{(-2x)^2}{2!} - \frac{(-2x)^3}{3!} + \ldots \]

### Step 3: Combine the Series
Now, we combine the two expansions:
\[ e^{2x} + e^{-2x} = \left(1 + 2x + \frac{(2x)^2}{2!} + \ldots\right) + \left(1 - 2x + \frac{(-2x)^2}{2!} - \ldots\right) \]
When we combine these, the linear terms \(2x\) and \(-2x\) cancel out:
\[ = 1 + 1 + \frac{(2x)^2 + (-2x)^2}{2!} + \ldots = 2 + \frac{8x^2}{2} + \ldots \]

### Step 4: Identify the Constant Term
From the combined series, we can see that the constant term is:
\[ 2 \]

### Conclusion
Thus, the constant term in the expansion of \((e^{7x} + e^{3x}) / (e^{5x})\) is:
\[ \boxed{2} \]