If `a_k` is the coefficient of `x^k` in the expansion of `(1+x+x^2)^n` for `k = 0, 1, 2, ......., 2n` then `a_1+2a_2+3a_3+...+2na_(2n)=`

Options

A. `-a_(0)`
B. `3^(n)`
C. `n*3^(n+1)`
D. `n*3^(n)`

Answer

To solve the problem, we need to find the sum \( S = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n} \), where \( a_k \) is the coefficient of \( x^k \) in the expansion of \( (1 + x + x^2)^n \).

### Step-by-Step Solution:

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

2. **Differentiate the Expansion**:
To find the weighted sum \( S \), we can differentiate the expansion with respect to \( x \):
\[ \frac{d}{dx}[(1 + x + x^2)^n] = n(1 + x + x^2)^{n-1}(1 + 2x) \]
This differentiation brings down the powers of \( x \) in front of the coefficients.

3. **Express the Coefficients**:
After differentiation, we can express the result as:
\[ \frac{d}{dx}[(1 + x + x^2)^n] = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 2n a_{2n} x^{2n-1} \]

4. **Evaluate at \( x = 1 \)**:
To find the sum \( S \), we substitute \( x = 1 \) into the differentiated expression:
\[ \frac{d}{dx}[(1 + x + x^2)^n] \bigg|_{x=1} = n(1 + 1 + 1)^{n-1}(1 + 2 \cdot 1) = n \cdot 3^{n-1} \cdot 3 = 3n \cdot 3^{n-1} \]

5. **Final Result**:
Therefore, we have:
\[ S = 3n \cdot 3^{n-1} = n \cdot 3^n \]

### Conclusion:
The value of \( a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n} \) is \( n \cdot 3^n \).