If `a_r` is the coefficient of `x^r` in the expansion of `(1+x+x^2)^n`, then `a_1-2a_2 + 3a_3.....-2na_(2n)=`

Options

A. 0
B. n
C. `-n`
D. 2n

Explanation

To solve the problem, we need to find the expression \( a_1 - 2a_2 + 3a_3 - 2na_{2n} \) where \( a_r \) is the coefficient of \( x^r \) in the expansion of \( (1 + x + x^2)^n \).

### Step-by-Step Solution:

1. **Understanding the Expansion**:
The expression \( (1 + x + x^2)^n \) can be expanded using the multinomial theorem. The coefficients \( a_r \) represent the number of ways to choose terms from \( 1, x, x^2 \) such that the total degree of \( x \) is \( r \).

2. **Finding the Coefficients**:
We can express \( a_r \) as the number of non-negative integer solutions to the equation:
\[ k_1 + k_2 + k_3 = n \]
where \( k_1 \) is the number of times we choose \( 1 \), \( k_2 \) is the number of times we choose \( x \), and \( k_3 \) is the number of times we choose \( x^2 \). The total degree of \( x \) is given by \( k_2 + 2k_3 = r \).

3. **Differentiating the Function**:
To find the required expression, we differentiate \( (1 + x + x^2)^n \) with respect to \( x \):
\[ \frac{d}{dx}[(1 + x + x^2)^n] = n(1 + x + x^2)^{n-1}(1 + 2x) \]

4. **Substituting \( x = -x \)**:
We replace \( x \) with \( -x \) in the differentiated equation:
\[ n(1 - x + x^2)^{n-1}(1 - 2x) \]

5. **Finding Coefficients**:
The coefficients of \( x^r \) in the expansion can be expressed using the derivatives. The coefficients of \( x^{2n-1} \) will give us the required alternating sum:
\[ a_1 - 2a_2 + 3a_3 - \ldots - 2na_{2n} \]

6. **Evaluating at \( x = 1 \)**:
Now, we evaluate the expression at \( x = 1 \):
\[ n(1 - 1 + 1)^{n-1}(1 - 2) = n \cdot 1^{n-1} \cdot (-1) = -n \]

7. **Final Result**:
Therefore, the final result for \( a_1 - 2a_2 + 3a_3 - 2na_{2n} \) is:
\[ -n \]

### Conclusion:
The value of \( a_1 - 2a_2 + 3a_3 - 2na_{2n} \) is \( -n \).