If `(1+px+x^(2))^(n)=1+a_(1)x+a_(2)x^(2)+…+a_(2n)x^(2n)`.
The remainder obtained when `a_(1)+5a_(2)+9a_(3)+13a_(4)+…+(8n-3)a_(2n)` is divided by `(p+2)` is (a) 1 (b) 2 (c) 3 (d) 0

Options

A. `1`
B. `2`
C. `3`
D. `0`

Explanation

To solve the problem, we need to find the remainder when \( S = a_1 + 5a_2 + 9a_3 + 13a_4 + \ldots + (8n - 3)a_{2n} \) is divided by \( (p + 2) \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression given in the problem:
\[ (1 + px + x^2)^n = 1 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n} \]
This indicates that the coefficients \( a_r \) are derived from the expansion of \( (1 + px + x^2)^n \).

2. **Finding the Coefficients**:
The coefficients \( a_r \) can be found using the binomial theorem and combinatorial methods, but we will focus on the expression we need to evaluate.

3. **Rewriting the Sum**:
The sum \( S \) can be rewritten as:
\[ S = \sum_{r=1}^{2n} (4r - 3) a_r = \sum_{r=1}^{2n} 4r a_r - 3 \sum_{r=1}^{2n} a_r \]
Let \( T = \sum_{r=1}^{2n} a_r \).

4. **Finding \( T \)**:
To find \( T \), we can substitute \( x = 1 \) in the original polynomial:
\[ (1 + p + 1)^n = (p + 2)^n \]
Therefore, we have:
\[ T = (p + 2)^n - 1 \]

5. **Finding \( \sum_{r=1}^{2n} r a_r \)**:
To find \( \sum_{r=1}^{2n} r a_r \), we differentiate the original expression:
\[ \frac{d}{dx}((1 + px + x^2)^n) = n(1 + px + x^2)^{n-1}(p + 2x) \]
Evaluating at \( x = 1 \):
\[ n(1 + p + 1)^{n-1}(p + 2) = n(p + 2)^{n-1}(p + 2) = n(p + 2)^n \]
Thus:
\[ \sum_{r=1}^{2n} r a_r = n(p + 2)^n \]

6. **Substituting Back**:
Now substituting back into the expression for \( S \):
\[ S = 4(n(p + 2)^n) - 3((p + 2)^n - 1) \]
Simplifying this gives:
\[ S = 4n(p + 2)^n - 3(p + 2)^n + 3 = (4n - 3)(p + 2)^n + 3 \]

7. **Finding the Remainder**:
We need to find the remainder when \( S \) is divided by \( (p + 2) \):
\[ S \equiv 3 \mod (p + 2) \]

### Final Answer:
Thus, the remainder obtained when \( S \) is divided by \( (p + 2) \) is **3**.