Consider `(1 + x + x^(2))^(n) = sum_(r=0)^(n) a_(r) x^(r)` , where ` a_(0), a_(1), a_(2),…, a_(2n)` are
real number and n is positive integer.
If n is even, the value of ` sum_(r=0)^(n//2-1) a_(2r) ` is

Options

A. `(3^(n) - 1 + a_(n))/(2)`
B. `(3^(n) -1 - a_(n))/(4)`
C. `(3^(n) + 1 + a_(n))/(2)`
D. `(3^(n) + 1 - 2a_(n))/(4)`

Explanation

To solve the problem, we need to find the value of the sum \( \sum_{r=0}^{n/2 - 1} a_{2r} \) where \( (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r \) and \( n \) is an even positive integer.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 + x + x^2)^n \). This can be expanded using the binomial theorem, and it will yield coefficients \( a_r \) for each power of \( x \).

2. **Substituting \( x = 1 \)**:
We substitute \( x = 1 \) into the expression:
\[ (1 + 1 + 1^2)^n = 3^n \]
This gives us:
\[ a_0 + a_1 + a_2 + \ldots + a_{2n} = 3^n \quad \text{(Equation 1)} \]

3. **Substituting \( x = -1 \)**:
Next, we substitute \( x = -1 \):
\[ (1 - 1 + (-1)^2)^n = 1^n = 1 \]
This gives us:
\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} = 1 \quad \text{(Equation 2)} \]

4. **Adding Equations 1 and 2**:
We add Equation 1 and Equation 2:
\[ (a_0 + a_1 + a_2 + \ldots + a_{2n}) + (a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}) = 3^n + 1 \]
This simplifies to:
\[ 2a_0 + 2a_2 + 2a_4 + \ldots + 2a_{2n} = 3^n + 1 \]
Thus:
\[ a_0 + a_2 + a_4 + \ldots + a_{2n} = \frac{3^n + 1}{2} \]

5. **Finding the Required Sum**:
Since \( n \) is even, we can denote \( n = 2k \) for some integer \( k \). The terms \( a_{2r} \) for \( r = 0, 1, \ldots, k-1 \) correspond to the even-indexed coefficients up to \( a_{2(n/2 - 1)} \). Therefore:
\[ \sum_{r=0}^{n/2 - 1} a_{2r} = \frac{3^n + 1}{4} - \frac{a_n}{2} \]
where \( a_n \) is the middle term in the expansion.

6. **Final Result**:
Since \( n \) is even, we can conclude that:
\[ \sum_{r=0}^{n/2 - 1} a_{2r} = \frac{3^n + 1}{4} \]

### Conclusion:
Thus, the value of \( \sum_{r=0}^{n/2 - 1} a_{2r} \) is \( \frac{3^n + 1}{4} \).