The coefficient of `x^(n)` in the expansion of
` log_(e)(1+3x+2x^2)` is

Answer

To find the coefficient of \( x^n \) in the expansion of \( \log_e(1 + 3x + 2x^2) \), we can follow these steps:

### Step 1: Rewrite the logarithmic expression
We start with the expression \( \log_e(1 + 3x + 2x^2) \). We can factor the quadratic expression inside the logarithm:

\[ 1 + 3x + 2x^2 = (1 + x)(1 + 2x) \]

### Step 2: Use logarithmic properties
Using the property of logarithms that states \( \log(mn) = \log(m) + \log(n) \), we can rewrite the expression as:

\[ \log_e(1 + 3x + 2x^2) = \log_e(1 + x) + \log_e(1 + 2x) \]

### Step 3: Expand each logarithmic term
Next, we can use the Taylor series expansion for \( \log(1 + u) \), which is given by:

\[ \log(1 + u) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} u^k}{k} \]

Applying this to both terms:

1. For \( \log_e(1 + x) \):

\[ \log_e(1 + x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} x^k}{k} \]

2. For \( \log_e(1 + 2x) \):

\[ \log_e(1 + 2x) = \sum_{m=1}^{\infty} \frac{(-1)^{m-1} (2x)^m}{m} = \sum_{m=1}^{\infty} \frac{(-1)^{m-1} 2^m x^m}{m} \]

### Step 4: Combine the expansions
Now we can combine the two series:

\[ \log_e(1 + 3x + 2x^2) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} x^k}{k} + \sum_{m=1}^{\infty} \frac{(-1)^{m-1} 2^m x^m}{m} \]

### Step 5: Find the coefficient of \( x^n \)
To find the coefficient of \( x^n \), we need to look at the contributions from both series:

- From \( \log_e(1 + x) \), the coefficient of \( x^n \) is \( \frac{(-1)^{n-1}}{n} \).
- From \( \log_e(1 + 2x) \), the coefficient of \( x^n \) is \( \frac{(-1)^{n-1} 2^n}{n} \).

Thus, the total coefficient of \( x^n \) in the expansion is:

\[ \text{Coefficient of } x^n = \frac{(-1)^{n-1}}{n} + \frac{(-1)^{n-1} 2^n}{n} = \frac{(-1)^{n-1}(1 + 2^n)}{n} \]

### Final Answer
The coefficient of \( x^n \) in the expansion of \( \log_e(1 + 3x + 2x^2) \) is:

\[ \frac{(-1)^{n-1}(1 + 2^n)}{n} \]