The coefficient of `x^(n)`, where n = 3k in the expansion of `log (1 + x + x^(2))` is equal to

Options

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

Answer

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

### Step 1: Rewrite the logarithmic expression

We start with the expression:
\[ \log(1 + x + x^2) \]

### Step 2: Use the property of logarithms

Using the property of logarithms, we can express \( \log(1 + x + x^2) \) in terms of simpler logarithmic functions. We can factor the expression inside the logarithm:
\[ 1 + x + x^2 = \frac{1 - x^3}{1 - x} \]
Thus,
\[ \log(1 + x + x^2) = \log\left(\frac{1 - x^3}{1 - x}\right) = \log(1 - x^3) - \log(1 - x) \]

### Step 3: Expand using Taylor series

Now we can expand both logarithmic functions using their Taylor series:
\[ \log(1 - x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\right) \]
\[ \log(1 - x^3) = -\sum_{m=1}^{\infty} \frac{x^{3m}}{m} = -\left(x^3 + \frac{x^6}{2} + \frac{x^9}{3} + \cdots\right) \]

### Step 4: Combine the series

Now we combine these expansions:
\[ \log(1 + x + x^2) = -\left(-\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\right) + \left(x^3 + \frac{x^6}{2} + \frac{x^9}{3} + \cdots\right)\right) \]

### Step 5: Identify the coefficients

Now we need to find the coefficient of \( x^{3k} \). From the expansion:
- The term \( -\log(1 - x^3) \) contributes \( -\frac{1}{m} \) for \( x^{3m} \).
- The term \( -\log(1 - x) \) contributes \( -\frac{1}{n} \) for \( x^n \).

### Step 6: Coefficient of \( x^{3k} \)

The coefficient of \( x^{3k} \) in \( \log(1 + x + x^2) \) will be:
\[ -\frac{1}{3k} + \text{(contributions from other terms)} \]

### Conclusion

Thus, the coefficient of \( x^{3k} \) in the expansion of \( \log(1 + x + x^2) \) is:
\[ -\frac{1}{3k} \]