If `(1 + x - 2 x^(2))^(6) = 1 + C_(1) x + C_(2) x^(2) + C_(3) x^(3) + …+ C_(12) x^(12)`,
then the value of `C_(2) + C_(4) + C_(6) + …+ C_(12)` is

Options

A. 30
B. 32
C. 31
D. none of these

Answer

To find the value of \( C_2 + C_4 + C_6 + \ldots + C_{12} \) from the expansion of \( (1 + x - 2x^2)^6 \), we can follow these steps:

### Step 1: Write the expression
We start with the expression:
\[ (1 + x - 2x^2)^6 \]

### Step 2: Substitute \( x \) with \( -x \)
Next, we will substitute \( x \) with \( -x \):
\[ (1 - x - 2x^2)^6 \]

### Step 3: Set up the equations
Now we can express both expansions:
1. \( (1 + x - 2x^2)^6 = 1 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_{12} x^{12} \) (Equation 1)
2. \( (1 - x - 2x^2)^6 = 1 - C_1 x + C_2 x^2 - C_3 x^3 + \ldots + C_{12} x^{12} \) (Equation 2)

### Step 4: Add the two equations
Adding Equation 1 and Equation 2:
\[ (1 + x - 2x^2)^6 + (1 - x - 2x^2)^6 = 2 + 2C_2 x^2 + 2C_4 x^4 + \ldots + 2C_{12} x^{12} \]
The terms with odd powers of \( x \) cancel out.

### Step 5: Evaluate at \( x = 1 \)
Now, we evaluate the left-hand side at \( x = 1 \):
\[ (1 + 1 - 2)^6 + (1 - 1 - 2)^6 = 0 + (-2)^6 = 64 \]
Thus, we have:
\[ 64 = 2 + 2(C_2 + C_4 + C_6 + \ldots + C_{12}) \]

### Step 6: Solve for the sum of coefficients
Rearranging gives:
\[ 64 - 2 = 2(C_2 + C_4 + C_6 + \ldots + C_{12}) \]
\[ 62 = 2(C_2 + C_4 + C_6 + \ldots + C_{12}) \]
Dividing by 2:
\[ C_2 + C_4 + C_6 + \ldots + C_{12} = 31 \]

### Final Answer
Thus, the value of \( C_2 + C_4 + C_6 + \ldots + C_{12} \) is:
\[ \boxed{31} \]