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 + ………+2n.a_(2n) =`

Options

A. `-a_0`
B. `3^n`
C. `n.3^n`
D. `-n.3^n`

Answer

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

### Step-by-Step Solution:

1. **Write the Expression**:
We start with the expression we need to evaluate:
\[ S = a_1 + 2a_2 + 3a_3 + \ldots + 2n \cdot a_{2n} \]

2. **Identify the Generating Function**:
The generating function for the coefficients \( a_k \) is given by:
\[ (1 + x + x^2)^n \]

3. **Differentiate the Generating Function**:
To find the weighted sum \( S \), we differentiate the generating function with respect to \( x \):
\[ \frac{d}{dx}((1 + x + x^2)^n) = n(1 + x + x^2)^{n-1} \cdot (1 + 2x) \]
This differentiation brings down the powers of \( x \) and gives us the coefficients multiplied by their respective powers.

4. **Evaluate at \( x = 1 \)**:
Now, we substitute \( x = 1 \) into the differentiated function:
\[ \frac{d}{dx}((1 + x + x^2)^n) \bigg|_{x=1} = n(1 + 1 + 1)^{n-1} \cdot (1 + 2 \cdot 1) \]
Simplifying this gives:
\[ = n \cdot 3^{n-1} \cdot 3 = 3n \cdot 3^{n-1} = 3^n \cdot n \]

5. **Conclusion**:
Thus, the value of \( S \) is:
\[ S = 3^n \cdot n \]

### Final Answer:
\[ \boxed{3^n \cdot n} \]