The coefficient of `x^(3)` in the expansion of `(1-x+x^(2))^(5)` is

Options

A. `-20`
B. `-30`
C. `-50`
D. 10

Answer

To find the coefficient of \( x^3 \) in the expansion of \( (1 - x + x^2)^5 \), we can use the multinomial expansion theorem.

### Step-by-Step Solution:

1. **Identify the terms in the expansion**:
The expression can be viewed as \( (a + b + c)^n \) where \( a = 1 \), \( b = -x \), and \( c = x^2 \) with \( n = 5 \).

2. **Use the multinomial expansion**:
The multinomial expansion states that:
\[ (a + b + c)^n = \sum_{i+j+k=n} \frac{n!}{i!j!k!} a^i b^j c^k \]
Here, we are looking for the terms where \( i + j + k = 5 \) and the total power of \( x \) is 3, which means:
\[ j + 2k = 3 \]

3. **Set up the equations**:
From \( i + j + k = 5 \) and \( j + 2k = 3 \), we can express \( i \) in terms of \( j \) and \( k \):
\[ i = 5 - j - k \]

4. **Substituting \( j \) from the second equation**:
Rearranging \( j + 2k = 3 \) gives us:
\[ j = 3 - 2k \]
Substitute this into the equation for \( i \):
\[ i = 5 - (3 - 2k) - k = 5 - 3 + 2k - k = 2 + k \]

5. **Finding possible values for \( k \)**:
Since \( i, j, k \) must be non-negative integers, we need \( 2 + k \geq 0 \) and \( 3 - 2k \geq 0 \). Thus:
- From \( 3 - 2k \geq 0 \), we get \( k \leq 1.5 \), so \( k \) can be 0 or 1.

6. **Evaluate for \( k = 0 \)**:
- If \( k = 0 \):
\[ j = 3, \quad i = 2 \]
The term is:
\[ \frac{5!}{2!3!0!} (1)^2 (-x)^3 (x^2)^0 = \frac{120}{2 \cdot 6} \cdot 1 \cdot (-x)^3 = -10x^3 \]

7. **Evaluate for \( k = 1 \)**:
- If \( k = 1 \):
\[ j = 1, \quad i = 2 \]
The term is:
\[ \frac{5!}{2!1!1!} (1)^2 (-x)^1 (x^2)^1 = \frac{120}{2 \cdot 1 \cdot 1} \cdot 1 \cdot (-x) \cdot (x^2) = -60x^3 \]

8. **Combine the coefficients**:
The total coefficient of \( x^3 \) is:
\[ -10 - 60 = -70 \]

Thus, the coefficient of \( x^3 \) in the expansion of \( (1 - x + x^2)^5 \) is **-70**.