The coefficients of `x^(11)` in the expansion of `(2x^(2)+x-3)^(6)` is

Options

A. 186
B. 190
C. 192
D. 196

Answer

To find the coefficient of \( x^{11} \) in the expansion of \( (2x^2 + x - 3)^6 \), we can follow these steps:

### Step 1: Identify the General Term
The general term \( T \) in the expansion of \( (a + b + c)^n \) is given by:
\[ T = \frac{n!}{\alpha! \beta! \gamma!} a^\alpha b^\beta c^\gamma \]
where \( n = 6 \), \( a = 2x^2 \), \( b = x \), and \( c = -3 \). Here, \( \alpha + \beta + \gamma = n \).

### Step 2: Write the General Term for Our Case
For our case, the general term becomes:
\[ T = \frac{6!}{\alpha! \beta! \gamma!} (2x^2)^\alpha (x)^\beta (-3)^\gamma \]
This simplifies to:
\[ T = \frac{6!}{\alpha! \beta! \gamma!} 2^\alpha x^{2\alpha} x^\beta (-3)^\gamma \]
or
\[ T = \frac{6!}{\alpha! \beta! \gamma!} 2^\alpha (-3)^\gamma x^{2\alpha + \beta} \]

### Step 3: Set Up the Equations
We need to find the values of \( \alpha \), \( \beta \), and \( \gamma \) such that:
1. \( 2\alpha + \beta = 11 \) (to get the coefficient of \( x^{11} \))
2. \( \alpha + \beta + \gamma = 6 \)

### Step 4: Solve the Equations
From the second equation, we can express \( \gamma \) as:
\[ \gamma = 6 - \alpha - \beta \]
Substituting this into the first equation gives:
\[ 2\alpha + \beta = 11 \]
Now we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = 11 - 2\alpha \]
Substituting this into the equation for \( \gamma \):
\[ \gamma = 6 - \alpha - (11 - 2\alpha) = 6 - \alpha - 11 + 2\alpha = \alpha - 5 \]

### Step 5: Determine Valid Values
Since \( \alpha \), \( \beta \), and \( \gamma \) must be non-negative integers, we have:
1. \( \alpha - 5 \geq 0 \) implies \( \alpha \geq 5 \)
2. \( 11 - 2\alpha \geq 0 \) implies \( \alpha \leq 5.5 \)

Thus, the only integer solution is \( \alpha = 5 \).

### Step 6: Calculate \( \beta \) and \( \gamma \)
Substituting \( \alpha = 5 \):
\[ \beta = 11 - 2(5) = 1 \]
\[ \gamma = 6 - 5 - 1 = 0 \]

### Step 7: Substitute Back into the General Term
Now substituting \( \alpha = 5 \), \( \beta = 1 \), and \( \gamma = 0 \) into the general term:
\[ T = \frac{6!}{5!1!0!} (2x^2)^5 (x)^1 (-3)^0 \]
This simplifies to:
\[ T = \frac{6!}{5!1!} \cdot 2^5 \cdot x^{10} \cdot x = \frac{6 \cdot 32}{1} x^{11} = 192 x^{11} \]

### Step 8: Coefficient of \( x^{11} \)
Thus, the coefficient of \( x^{11} \) is:
\[ \text{Coefficient} = 192 \]

### Final Answer
The coefficient of \( x^{11} \) in the expansion of \( (2x^2 + x - 3)^6 \) is \( \boxed{192} \).