If log`(1-x+x^(2))=a_(1)x+a_(2)x^(2)+a_(3)x^(3)+…`
then `a_(3)+a_(6)+a_(9)+..` is equal to

Options

A. log 2
B. `2/3 log 2`
C. `1/3 log 2`
D. `2 log 2

Answer

To solve the problem, we need to find the sum \( a_3 + a_6 + a_9 + \ldots \) where \( \log(1 - x + x^2) = a_1 x + a_2 x^2 + a_3 x^3 + \ldots \).

### Step-by-Step Solution:

1. **Rewrite the Logarithm:**
We start with the expression:
\[ \log(1 - x + x^2) \]
We can manipulate this expression by rewriting it as:
\[ \log\left(\frac{1 + (1 - x + x^2)}{1 + x}\right) \]

2. **Simplifying the Expression:**
We multiply and divide by \( 1 + x \):
\[ \log\left(\frac{(1 + x)(1 - x + x^2)}{1 + x}\right) = \log\left(1 + x^2 - x + x^3\right) - \log(1 + x) \]

3. **Expand Both Logarithms:**
We can expand both logarithmic expressions using their Taylor series:
- For \( \log(1 + x) \):
\[ \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
- For \( \log(1 + x^2 - x + x^3) \):
We can expand this using the series expansion as well, but we need to focus on the coefficients of \( x^3 \) and higher.

4. **Finding Coefficients:**
To find \( a_3, a_6, a_9, \ldots \), we need to find the coefficients of \( x^3, x^6, x^9, \ldots \) in the expansion.

- The coefficient \( a_3 \) comes from the expansion of \( \log(1 + x^2 - x + x^3) \):
\[ a_3 = 1 - \frac{1}{3} \]
- The coefficient \( a_6 \) is:
\[ a_6 = -\frac{1}{2} + \frac{1}{6} \]
- The coefficient \( a_9 \) is:
\[ a_9 = \frac{1}{3} - \frac{1}{9} \]

5. **Summing the Coefficients:**
We sum these coefficients:
\[ a_3 + a_6 + a_9 + \ldots = \left(1 - \frac{1}{3}\right) + \left(-\frac{1}{2} + \frac{1}{6}\right) + \left(\frac{1}{3} - \frac{1}{9}\right) + \ldots \]
This can be rearranged and simplified.

6. **Recognizing the Series:**
The series can be recognized as a combination of logarithmic series. We can express it as:
\[ \sum_{n=1}^{\infty} \left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots\right) = \log(2) \]

7. **Final Result:**
After simplifying, we find:
\[ a_3 + a_6 + a_9 + \ldots = \frac{2}{3} \log(2) \]

### Final Answer:
Thus, the value of \( a_3 + a_6 + a_9 + \ldots \) is:
\[ \frac{2}{3} \log(2) \]