The coefficient of `x^(18)` in the product `(1+x)(1-x)^(10)(1+x+x^(2))^(9)` is

Options

A. 84
B. `-126`
C. `-84`
D. `126`

Answer

To find the coefficient of \( x^{18} \) in the expression \( (1+x)(1-x)^{10}(1+x+x^2)^{9} \), we can break down the problem step by step.

### Step 1: Expand \( (1-x)^{10} \)

Using the Binomial Theorem, we can expand \( (1-x)^{10} \):
\[ (1-x)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-x)^k = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k \]

### Step 2: Expand \( (1+x+x^2)^{9} \)

Next, we need to expand \( (1+x+x^2)^{9} \). This can be done using the multinomial expansion:
\[ (1+x+x^2)^{9} = \sum_{a+b+c=9} \frac{9!}{a!b!c!} (1)^a (x)^b (x^2)^c = \sum_{a+b+c=9} \frac{9!}{a!b!c!} x^{b+2c} \]

### Step 3: Combine the expansions

Now, we combine the expansions of \( (1+x) \), \( (1-x)^{10} \), and \( (1+x+x^2)^{9} \):
\[ (1+x)(1-x)^{10}(1+x+x^2)^{9} = (1+x) \left( \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^k \right) \left( \sum_{a+b+c=9} \frac{9!}{a!b!c!} x^{b+2c} \right) \]

### Step 4: Find the coefficient of \( x^{18} \)

We need to find the coefficient of \( x^{18} \) in the combined expression. This can be done by considering the contributions from \( (1+x) \):
1. From \( 1 \): We need the coefficient of \( x^{18} \) from \( (1-x)^{10}(1+x+x^2)^{9} \).
2. From \( x \): We need the coefficient of \( x^{17} \) from \( (1-x)^{10}(1+x+x^2)^{9} \).

#### Case 1: Coefficient of \( x^{18} \)

To find the coefficient of \( x^{18} \) from \( (1-x)^{10}(1+x+x^2)^{9} \):
- From \( (1-x)^{10} \), we have \( \binom{10}{k} (-1)^k \).
- From \( (1+x+x^2)^{9} \), we need \( b + 2c = 18 \) with \( a + b + c = 9 \).

This gives us the equations:
- \( b + 2c = 18 \)
- \( a + b + c = 9 \)

From \( a + b + c = 9 \), we can express \( a = 9 - b - c \).

Substituting \( c = \frac{18 - b}{2} \) into \( a + b + c = 9 \):
\[ a + b + \frac{18 - b}{2} = 9 \]
Multiply through by 2 to eliminate the fraction:
\[ 2a + 2b + 18 - b = 18 \implies 2a + b = 0 \implies b = -2a \]

Since \( b \) must be non-negative, \( a \) must be 0, which gives \( b = 0 \) and \( c = 9 \).

Thus, the only contribution to \( x^{18} \) comes from \( c = 9 \) and \( a = 0 \), which gives:
\[ \text{Coefficient} = \binom{10}{0}(-1)^0 \cdot \frac{9!}{0!0!9!} = 1 \]

#### Case 2: Coefficient of \( x^{17} \)

To find the coefficient of \( x^{17} \):
- From \( (1-x)^{10} \), we have \( \binom{10}{k} (-1)^k \).
- From \( (1+x+x^2)^{9} \), we need \( b + 2c = 17 \).

This gives us:
- \( b + 2c = 17 \)
- \( a + b + c = 9 \)

Substituting \( c = \frac{17 - b}{2} \) into \( a + b + c = 9 \):
\[ a + b + \frac{17 - b}{2} = 9 \]
Multiply through by 2:
\[ 2a + 2b + 17 - b = 18 \implies 2a + b = 1 \implies b = 1 - 2a \]

The values \( a \) and \( b \) must be non-negative integers. The only solution is \( a = 0, b = 1, c = 8 \).

Thus, the contribution to \( x^{17} \) is:
\[ \text{Coefficient} = \binom{10}{1}(-1)^1 \cdot \frac{9!}{0!1!8!} = -10 \cdot 9 = -90 \]

### Final Calculation

Combining both contributions:
\[ \text{Total Coefficient of } x^{18} = 1 - 90 = -89 \]

### Conclusion

The coefficient of \( x^{18} \) in the expression \( (1+x)(1-x)^{10}(1+x+x^2)^{9} \) is \( -89 \).