If n is not a multiple of 3 then the coefficient of `x^(n)` in the expansio of `log_(e)(1+x+x^(2))` is

Options

A. `1/n`
B. `2/n`
C. `-1/n`
D. `-2/n`

Answer

To find the coefficient of \( x^n \) in the expansion of \( \log_e(1 + x + x^2) \) when \( n \) is not a multiple of 3, we can follow these steps:

### Step 1: Rewrite the logarithm
We start with the expression:
\[ \log_e(1 + x + x^2) \]
We can factor the expression inside the logarithm. Notice that:
\[ 1 + x + x^2 = \frac{1 - x^3}{1 - x} \]
Thus, we can rewrite the logarithm as:
\[ \log_e(1 + x + x^2) = \log_e\left(\frac{1 - x^3}{1 - x}\right) \]
Using the properties of logarithms, this can be separated into:
\[ \log_e(1 - x^3) - \log_e(1 - x) \]

### Step 2: Expand the logarithms
Next, we can expand both logarithmic terms using their Taylor series expansions:
1. For \( \log_e(1 - x) \):
\[ \log_e(1 - x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right) = -\sum_{k=1}^{\infty} \frac{x^k}{k} \]
2. For \( \log_e(1 - x^3) \):
\[ \log_e(1 - x^3) = -\left(x^3 + \frac{x^6}{2} + \frac{x^9}{3} + \ldots\right) = -\sum_{k=1}^{\infty} \frac{x^{3k}}{k} \]

### Step 3: Combine the expansions
Now, substituting these expansions back into our expression:
\[ \log_e(1 + x + x^2) = -\left(-\sum_{k=1}^{\infty} \frac{x^{3k}}{k}\right) + \sum_{k=1}^{\infty} \frac{x^k}{k} \]
This simplifies to:
\[ \log_e(1 + x + x^2) = \sum_{k=1}^{\infty} \frac{x^k}{k} - \sum_{k=1}^{\infty} \frac{x^{3k}}{k} \]

### Step 4: Identify the coefficient of \( x^n \)
To find the coefficient of \( x^n \) in this series, we need to consider the contributions from both series:
- The first series contributes \( \frac{1}{n} \) for \( x^n \).
- The second series contributes \( -\frac{1}{n} \) only when \( n \) is a multiple of 3.

Since we are given that \( n \) is not a multiple of 3, the second series does not contribute anything to the coefficient of \( x^n \).

### Conclusion
Thus, the coefficient of \( x^n \) in the expansion of \( \log_e(1 + x + x^2) \) when \( n \) is not a multiple of 3 is:
\[ \frac{1}{n} \]