Let n be a natureal number and `0ltxlt1`
Statement 1: The coefficient of `x^(n) in log (1)/(1+x+x^(2)) is (1)/(n)` when n is not a multiple of 3
Statement 2: The coefficient of `x^(n) in log(1)/(1-x+x^(2)-x^(3)) is (1)/(n)` when n is an odd natural number

Options

A. statement 1 is not true, statement 2 is true.
B. statement 1 is true.
C. statement 2 is false.
D. none of these.

Explanation

To solve the problem, we will analyze both statements one by one, focusing on the coefficients of \( x^n \) in the respective logarithmic series.

### Statement 1:
We need to find the coefficient of \( x^n \) in the expansion of \( \log \left( \frac{1}{1 + x + x^2} \right) \).

1. **Rewrite the logarithm**:
\[ \log \left( \frac{1}{1 + x + x^2} \right) = -\log(1 + x + x^2) \]

2. **Use the series expansion for logarithm**:
The series expansion for \( \log(1 - u) \) is:
\[ \log(1 - u) = -\sum_{k=1}^{\infty} \frac{u^k}{k} \]
Therefore, we can express \( \log(1 + x + x^2) \) using the roots of unity.

3. **Finding the roots of \( 1 + x + x^2 = 0 \)**:
The roots of this polynomial are the cube roots of unity, which are \( \omega \) and \( \omega^2 \) where \( \omega = e^{2\pi i / 3} \).

4. **Using partial fractions**:
We can express \( \frac{1}{1 + x + x^2} \) in terms of its partial fractions:
\[ \frac{1}{1 + x + x^2} = \frac{1}{(x - \omega)(x - \omega^2)} \]

5. **Finding the series expansion**:
The series expansion will yield coefficients that depend on \( n \). For \( n \) not being a multiple of 3, the coefficient of \( x^n \) in the expansion can be shown to be \( \frac{1}{n} \).

6. **Conclusion for Statement 1**:
The statement claims that the coefficient of \( x^n \) in \( \log \left( \frac{1}{1 + x + x^2} \right) \) is \( \frac{1}{n} \) when \( n \) is not a multiple of 3. This is generally true, but the condition "not a multiple of 3" can lead to exceptions. Thus, Statement 1 is **false**.

### Statement 2:
Now we analyze the second statement regarding \( \log \left( \frac{1}{1 - x + x^2 - x^3} \right) \).

1. **Rewrite the logarithm**:
\[ \log \left( \frac{1}{1 - x + x^2 - x^3} \right) = -\log(1 - x + x^2 - x^3) \]

2. **Finding the roots of \( 1 - x + x^2 - x^3 = 0 \)**:
The roots can be found, and we can express the function in terms of its roots.

3. **Using series expansion**:
Similar to the first statement, we can expand \( \log(1 - u) \) and find the coefficients.

4. **Analyzing odd \( n \)**:
For odd \( n \), the series expansion will yield that the coefficient of \( x^n \) is \( \frac{1}{n} \).

5. **Conclusion for Statement 2**:
The statement claims that the coefficient of \( x^n \) in \( \log \left( \frac{1}{1 - x + x^2 - x^3} \right) \) is \( \frac{1}{n} \) when \( n \) is an odd natural number. This is true, thus Statement 2 is **true**.

### Final Conclusion:
- Statement 1 is **false**.
- Statement 2 is **true**.