If `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+…+a_(2n)x^(2n)`.
The value of `a_(1)+3a_(2)+5a_(3)+7a_(4)+….(4n-1)a_(2n)` when `p=-3` and `n in` even is

Options

A. `n`
B. `2n-1`
C. `2n-2`
D. `2n`

Explanation

To solve the problem, we need to find the value of \( a_1 + 3a_2 + 5a_3 + 7a_4 + \ldots + (4n-1)a_{2n} \) given that \( (1 + px + x^2)^n = 1 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n} \), with \( p = -3 \) and \( n \) being an even number.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 + px + x^2)^n \). Here, we will substitute \( p = -3 \):
\[ (1 - 3x + x^2)^n \]

2. **Using the Binomial Theorem**:
We can expand \( (1 - 3x + x^2)^n \) using the multinomial expansion. The coefficients \( a_k \) represent the coefficients of \( x^k \) in the expansion.

3. **Finding the Required Sum**:
We need to evaluate:
\[ S = a_1 + 3a_2 + 5a_3 + 7a_4 + \ldots + (4n-1)a_{2n} \]
This can be expressed as:
\[ S = \sum_{r=1}^{2n} (2r - 1) a_r \]

4. **Separating the Terms**:
We can separate the sum:
\[ S = 2 \sum_{r=1}^{n} r a_r - \sum_{r=1}^{2n} a_r \]

5. **Finding the Values**:
We know from the binomial expansion that:
\[ \sum_{r=0}^{2n} a_r = (1 - 3 + 1)^n = (-1)^n \]
Since \( n \) is even, \( (-1)^n = 1 \). Thus:
\[ \sum_{r=1}^{2n} a_r = 1 - a_0 = 1 - 1 = 0 \]

6. **Calculating \( \sum_{r=1}^{n} r a_r \)**:
We also have:
\[ S = 2 \sum_{r=1}^{n} r a_r - 0 = 2 \sum_{r=1}^{n} r a_r \]
Now, using the formula for \( S \):
\[ S = 2(n - 1) + 1 = 2n - 1 \]

7. **Final Calculation**:
Substituting \( p = -3 \) and \( n \) being even, we can simplify \( S \):
\[ S = 2n - 1 + 1 = 2 \]

### Conclusion:
Thus, the value of \( a_1 + 3a_2 + 5a_3 + 7a_4 + \ldots + (4n-1)a_{2n} \) when \( p = -3 \) and \( n \) is even is:
\[ \boxed{2} \]