If `n ge 2`, and `(1 + x + x^(2))^(n) = a_(0) + a_(1)x + a_(2)x^(2) + .. . +a_(2n) x^(2n)`, then `a_(0) - 2a_(1)+ 3a_(2) + ... + (2n + 1)a_(n)` = ______

Answer

To solve the problem, we need to evaluate the expression \( a_0 - 2a_1 + 3a_2 + \ldots + (2n + 1)a_{2n} \) given that \( (1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \ldots + a_{2n}x^{2n} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We want to find the value of \( S = a_0 - 2a_1 + 3a_2 + \ldots + (2n + 1)a_{2n} \).

2. **Using Differentiation**:
We start with the function \( f(x) = (1 + x + x^2)^n \). We will differentiate this function with respect to \( x \).

\[ f'(x) = n(1 + x + x^2)^{n-1} \cdot (1 + 2x) \]

This gives us the derivative of \( f(x) \).

3. **Evaluating at \( x = -1 \)**:
We substitute \( x = -1 \) into our original function and its derivative:

\[ f(-1) = (1 - 1 + 1)^n = 1^n = 1 \]

For the derivative:

\[ f'(-1) = n(1 - 1 + 1)^{n-1} \cdot (1 - 2) = n \cdot 1^{n-1} \cdot (-1) = -n \]

4. **Setting Up the Equation**:
The expression for \( f'(x) \) can also be expressed in terms of the coefficients:

\[ f'(x) = a_1 + 2a_2x + 3a_3x^2 + \ldots + (2n)a_{2n}x^{2n-1} \]

Evaluating this at \( x = -1 \):

\[ f'(-1) = a_1 - 2a_2 + 3a_3 - 4a_4 + \ldots + (2n)a_{2n}(-1)^{2n-1} \]

5. **Combining Results**:
From our evaluations, we have:

\[ S = f'(-1) = -n \]

6. **Final Result**:
Therefore, we conclude that:

\[ a_0 - 2a_1 + 3a_2 + \ldots + (2n + 1)a_{2n} = 1 - n \]

### Final Answer:
\[ \boxed{1 - n} \]