The coefficient of `x^(5)` in the expansion of `(2-x+3x^(2))^(6)` is

Options

A. `-4692`
B. 4692
C. 2346
D. -`5052`

Answer

To find the coefficient of \( x^5 \) in the expansion of \( (2 - x + 3x^2)^6 \), we can use the multinomial expansion. Here’s a step-by-step solution:

### Step 1: Identify the terms
The expression can be rewritten as \( (2 + (-x) + 3x^2)^6 \). We can treat \( 2 \), \( -x \), and \( 3x^2 \) as separate terms for the multinomial expansion.

### Step 2: Apply the multinomial theorem
According to the multinomial theorem, the expansion of \( (a + b + c)^n \) is given by:
\[ \sum \frac{n!}{k_1! k_2! k_3!} a^{k_1} b^{k_2} c^{k_3} \]
where \( k_1 + k_2 + k_3 = n \).

In our case:
- \( a = 2 \)
- \( b = -x \)
- \( c = 3x^2 \)
- \( n = 6 \)

### Step 3: Determine the combinations that yield \( x^5 \)
We need to find combinations of \( k_1 \), \( k_2 \), and \( k_3 \) such that:
\[ k_2 + 2k_3 = 5 \]
and
\[ k_1 + k_2 + k_3 = 6 \]

### Step 4: Solve the equations
1. From \( k_1 + k_2 + k_3 = 6 \), we can express \( k_1 \) as:
\[ k_1 = 6 - k_2 - k_3 \]

2. Substitute \( k_1 \) into the first equation:
\[ k_2 + 2k_3 = 5 \]

Now we can find valid combinations of \( k_2 \) and \( k_3 \):

- **Case 1:** \( k_3 = 0 \)
- \( k_2 + 2(0) = 5 \) → \( k_2 = 5 \)
- \( k_1 = 6 - 5 - 0 = 1 \) → Combination: \( (1, 5, 0) \)

- **Case 2:** \( k_3 = 1 \)
- \( k_2 + 2(1) = 5 \) → \( k_2 = 3 \)
- \( k_1 = 6 - 3 - 1 = 2 \) → Combination: \( (2, 3, 1) \)

- **Case 3:** \( k_3 = 2 \)
- \( k_2 + 2(2) = 5 \) → \( k_2 = 1 \)
- \( k_1 = 6 - 1 - 2 = 3 \) → Combination: \( (3, 1, 2) \)

### Step 5: Calculate the coefficients for each combination
1. For \( (1, 5, 0) \):
\[ \text{Coefficient} = \frac{6!}{1!5!0!} (2^1)(-1)^5(3^0) = \frac{720}{1 \cdot 120 \cdot 1} \cdot 2 \cdot (-1) = 6 \cdot 2 \cdot (-1) = -12 \]

2. For \( (2, 3, 1) \):
\[ \text{Coefficient} = \frac{6!}{2!3!1!} (2^2)(-1)^3(3^1) = \frac{720}{2 \cdot 6 \cdot 1} \cdot 4 \cdot (-1) \cdot 3 = 60 \cdot 4 \cdot (-3) = -720 \]

3. For \( (3, 1, 2) \):
\[ \text{Coefficient} = \frac{6!}{3!1!2!} (2^3)(-1)^1(3^2) = \frac{720}{6 \cdot 1 \cdot 2} \cdot 8 \cdot (-1) \cdot 9 = 60 \cdot 8 \cdot (-9) = -4320 \]

### Step 6: Sum the coefficients
Now, we sum the coefficients from all combinations:
\[ -12 - 720 - 4320 = -5052 \]

### Final Answer
The coefficient of \( x^5 \) in the expansion of \( (2 - x + 3x^2)^6 \) is \( -5052 \).

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