The coeffiecient of `x^(n)` in the expansion of ` log_(n)(1+x)` is

Options

A. ` ((-1)^(n-1))/n`
B. ` ((-1)^(n-1))/n log_(e)e`
C. ` ((-1)^(n-1))/n log_(e)a`
D. ` ((-1)^(n))/n log_(a)e`

Explanation

To find the coefficient of \( x^n \) in the expansion of \( \log_n(1+x) \), we can follow these steps:

### Step-by-Step Solution:

1. **Change of Base Formula**:
We start by using the change of base formula for logarithms:
\[ \log_n(1+x) = \frac{\log_e(1+x)}{\log_e(n)} \]
Here, \( \log_e \) denotes the natural logarithm (base \( e \)).

2. **Expand \( \log_e(1+x) \)**:
The Taylor series expansion of \( \log_e(1+x) \) around \( x=0 \) is given by:
\[ \log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} x^k}{k} \]

3. **Substituting into the Formula**:
Substituting the expansion of \( \log_e(1+x) \) into our expression, we have:
\[ \log_n(1+x) = \frac{1}{\log_e(n)} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \right) \]

4. **Finding the Coefficient of \( x^n \)**:
To find the coefficient of \( x^n \) in this series, we look at the term corresponding to \( x^n \):
\[ \text{Coefficient of } x^n = \frac{(-1)^{n-1}}{n \log_e(n)} \]

5. **Final Result**:
Therefore, the coefficient of \( x^n \) in the expansion of \( \log_n(1+x) \) is:
\[ \frac{(-1)^{n-1}}{n \log_e(n)} \]

### Summary:
The coefficient of \( x^n \) in the expansion of \( \log_n(1+x) \) is given by:
\[ \frac{(-1)^{n-1}}{n \log_e(n)} \]