If `(1+x-2x^2)^6=1+a_1x+a_2x^(2)+.........+a_(12)x^(12),` then find the value of `a_2+a_4+a_6+................+a_(12)`.

Explanation

To solve the problem, we need to find the sum of the coefficients of the even powers of \( x \) in the expansion of \( (1 + x - 2x^2)^6 \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 + x - 2x^2)^6 \). We want to find the coefficients \( a_2, a_4, a_6, \ldots, a_{12} \) in the expansion.

2. **Substituting \( x = 1 \)**:
Substitute \( x = 1 \) into the expression:
\[ (1 + 1 - 2 \cdot 1^2)^6 = (1 + 1 - 2)^6 = (0)^6 = 0 \]
This gives us:
\[ 1 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} + a_{11} + a_{12} = 0 \quad \text{(Equation 1)} \]

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

4. **Adding Equation 1 and Equation 2**:
Now, we add Equation 1 and Equation 2:
\[ (0) + (64) = (1 + 1) + (a_2 + a_2) + (a_4 + a_4) + (a_6 + a_6) + (a_8 + a_8) + (a_{10} + a_{10}) + (a_{12} + a_{12}) \]
Simplifying this gives:
\[ 64 = 2 + 2(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}) \]
Thus:
\[ 64 - 2 = 2(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}) \]
\[ 62 = 2(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}) \]
Dividing both sides by 2:
\[ 31 = a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} \]

5. **Final Result**:
Therefore, the value of \( a_2 + a_4 + a_6 + \ldots + a_{12} \) is \( \boxed{31} \).