The coefficient of `x^(6)` in the expansion of `(1+x+x^(2))^(6)` is

Options

A. 131
B. 141
C. 151
D. 167

Answer

To find the coefficient of \( x^6 \) in the expansion of \( (1 + x + x^2)^6 \), we can use the multinomial expansion.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression \( (1 + x + x^2)^6 \) can be expanded using the multinomial theorem, which states that:
\[ (a_1 + a_2 + a_3)^n = \sum \frac{n!}{k_1! k_2! k_3!} a_1^{k_1} a_2^{k_2} a_3^{k_3} \]
where \( k_1 + k_2 + k_3 = n \).

2. **Identifying Terms**:
In our case, \( a_1 = 1 \), \( a_2 = x \), \( a_3 = x^2 \), and \( n = 6 \). We need to find combinations of \( k_1, k_2, k_3 \) such that:
\[ k_2 + 2k_3 = 6 \]
and
\[ k_1 + k_2 + k_3 = 6. \]

3. **Setting Up Equations**:
From the second equation, we can express \( k_1 \) as:
\[ k_1 = 6 - k_2 - k_3. \]
Substituting \( k_1 \) into the first equation gives:
\[ k_2 + 2k_3 = 6. \]

4. **Finding Valid Combinations**:
We can solve for \( k_2 \) and \( k_3 \):
- If \( k_3 = 0 \): \( k_2 = 6 \) → \( (k_1, k_2, k_3) = (0, 6, 0) \)
- If \( k_3 = 1 \): \( k_2 = 4 \) → \( (k_1, k_2, k_3) = (1, 4, 1) \)
- If \( k_3 = 2 \): \( k_2 = 2 \) → \( (k_1, k_2, k_3) = (2, 2, 2) \)
- If \( k_3 = 3 \): \( k_2 = 0 \) → \( (k_1, k_2, k_3) = (3, 0, 3) \)

5. **Calculating Coefficients**:
For each valid combination, we calculate the coefficient:
- For \( (0, 6, 0) \):
\[ \frac{6!}{0!6!0!} = 1 \]
- For \( (1, 4, 1) \):
\[ \frac{6!}{1!4!1!} = \frac{720}{1 \cdot 24 \cdot 1} = 30 \]
- For \( (2, 2, 2) \):
\[ \frac{6!}{2!2!2!} = \frac{720}{2 \cdot 2 \cdot 2} = 90 \]
- For \( (3, 0, 3) \):
\[ \frac{6!}{3!0!3!} = \frac{720}{6 \cdot 1 \cdot 6} = 20 \]

6. **Summing the Coefficients**:
Now, we sum all the coefficients:
\[ 1 + 30 + 90 + 20 = 141. \]

### Final Answer:
The coefficient of \( x^6 \) in the expansion of \( (1 + x + x^2)^6 \) is **141**.