If `|x|lt1` then the coefficient of `x^(3)` in the expansion of `log(1+x+x^(2))` is ascending power of x is

Options

A. `2/3`
B. `4/3`
C. `(-2)/(3)`
D. `-(4)/(3)`

Answer

To find the coefficient of \( x^3 \) in the expansion of \( \log(1 + x + x^2) \) for \( |x| < 1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic expression
We start with the expression:
\[ \log(1 + x + x^2) \]
We can factor out \( x \) from the expression inside the logarithm:
\[ \log(1 + x + x^2) = \log\left(1 + x(1 + x)\right) \]

### Step 2: Use the logarithmic expansion
We know that the Taylor series expansion for \( \log(1 + u) \) is:
\[ \log(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \ldots \]
Let \( u = x(1 + x) \). Then we substitute \( u \) in the expansion:
\[ \log(1 + x(1 + x)) = x(1 + x) - \frac{(x(1 + x))^2}{2} + \frac{(x(1 + x))^3}{3} - \ldots \]

### Step 3: Expand \( u \)
Now, we need to calculate \( (x(1 + x))^n \) for \( n = 1, 2, 3 \):
1. For \( n = 1 \):
\[ x(1 + x) = x + x^2 \]
2. For \( n = 2 \):
\[ (x(1 + x))^2 = (x + x^2)^2 = x^2 + 2x^3 + x^4 \]
3. For \( n = 3 \):
\[ (x(1 + x))^3 = (x + x^2)^3 = x^3 + 3x^4 + 3x^5 + x^6 \]

### Step 4: Substitute back into the logarithmic expansion
Now substituting these back into the logarithmic expansion:
\[ \log(1 + x + x^2) = (x + x^2) - \frac{(x^2 + 2x^3 + x^4)}{2} + \frac{(x^3 + 3x^4 + 3x^5 + x^6)}{3} - \ldots \]

### Step 5: Collect the \( x^3 \) terms
Now we need to collect the \( x^3 \) terms from the expansion:
1. From \( (x + x^2) \), there are no \( x^3 \) terms.
2. From \( -\frac{(x^2 + 2x^3 + x^4)}{2} \), the \( x^3 \) term is:
\[ -\frac{2}{2} = -1 \]
3. From \( \frac{(x^3 + 3x^4 + 3x^5 + x^6)}{3} \), the \( x^3 \) term is:
\[ \frac{1}{3} \]

### Step 6: Combine the coefficients
Now, we combine the coefficients of \( x^3 \):
\[ \text{Coefficient of } x^3 = -1 + \frac{1}{3} = -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3} \]

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