If `(1+x-2x^2)^(6) = 1 + a_(1) x + a_(2) x^(2) + ... + a_(12) x^(12)`, then find ` a_2+ a_4 + ... + a_12`

Answer

To solve the problem, we need to find the sum of the coefficients \( a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} \) in the expansion of \( (1 + x - 2x^2)^6 \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 + x - 2x^2)^6 \).

2. **Using the Binomial Theorem**:
The Binomial Theorem states that \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \). Here, we can treat \( a = 1 + x \) and \( b = -2x^2 \).

3. **Setting Up the Coefficients**:
We want to find the coefficients of \( x^2, x^4, x^6, x^8, x^{10}, x^{12} \). To do this, we can evaluate the expression at specific values of \( x \).

4. **Evaluating at \( x = 1 \)**:
First, we substitute \( x = 1 \):
\[ (1 + 1 - 2 \cdot 1^2)^6 = (1 + 1 - 2)^6 = 0^6 = 0 \]
This gives us:
\[ 1 + a_1 + a_2 + a_3 + \ldots + a_{12} = 0 \quad \text{(Equation 1)} \]

5. **Evaluating at \( x = -1 \)**:
Next, we substitute \( x = -1 \):
\[ (1 - 1 - 2 \cdot (-1)^2)^6 = (1 - 1 - 2)^6 = (-2)^6 = 64 \]
This gives us:
\[ 1 - a_1 + a_2 - a_3 + a_4 - a_5 + a_6 - a_7 + a_8 - a_9 + a_{10} - a_{11} + a_{12} = 64 \quad \text{(Equation 2)} \]

6. **Adding Equations**:
Now, we add Equation 1 and Equation 2:
\[ (1 + a_1 + a_2 + a_3 + \ldots + a_{12}) + (1 - a_1 + a_2 - a_3 + \ldots + a_{12}) = 0 + 64 \]
This simplifies to:
\[ 2 + 2(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}) = 64 \]

7. **Solving for the Sum**:
Rearranging gives:
\[ 2(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}) = 64 - 2 = 62 \]
Dividing by 2:
\[ a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = \frac{62}{2} = 31 \]

### Final Answer:
Thus, the value of \( a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} \) is \( \boxed{31} \).