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.
The value of ` sum_(r=0)^(n-1) a_(r)` is

Options

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

Explanation

To solve the problem, we need to find the value of \( \sum_{r=0}^{n-1} a_r \) from the expression \( (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r \]
This means that the coefficients \( a_r \) represent the number of ways to choose terms from the expansion.

2. **Substituting \( x = 1 \)**:
To find \( \sum_{r=0}^{n-1} a_r \), we can substitute \( x = 1 \) into the expression:
\[ (1 + 1 + 1)^n = 3^n = \sum_{r=0}^{2n} a_r \]
This gives us the total sum of all coefficients \( a_r \).

3. **Substituting \( x = -1 \)**:
Next, we substitute \( x = -1 \):
\[ (1 - 1 + 1)^n = 1^n = 1 = \sum_{r=0}^{2n} a_r (-1)^r \]
This expression helps us find the alternating sum of the coefficients.

4. **Combining Results**:
From the two substitutions, we have:
- From \( x = 1 \):
\[ \sum_{r=0}^{2n} a_r = 3^n \]
- From \( x = -1 \):
\[ \sum_{r=0}^{2n} a_r (-1)^r = 1 \]

5. **Finding \( \sum_{r=0}^{n-1} a_r \)**:
We can express \( \sum_{r=0}^{2n} a_r \) in terms of even and odd indexed coefficients:
\[ \sum_{r=0}^{2n} a_r = \sum_{k=0}^{n} a_{2k} + \sum_{k=0}^{n-1} a_{2k+1} \]
And for the alternating sum:
\[ \sum_{r=0}^{2n} a_r (-1)^r = \sum_{k=0}^{n} a_{2k} - \sum_{k=0}^{n-1} a_{2k+1} = 1 \]

6. **Setting Up the Equations**:
Let:
\[ S_e = \sum_{k=0}^{n} a_{2k} \quad \text{(sum of even indexed coefficients)} \]
\[ S_o = \sum_{k=0}^{n-1} a_{2k+1} \quad \text{(sum of odd indexed coefficients)} \]
From our previous results:
\[ S_e + S_o = 3^n \]
\[ S_e - S_o = 1 \]

7. **Solving the System of Equations**:
Adding the two equations:
\[ 2S_e = 3^n + 1 \implies S_e = \frac{3^n + 1}{2} \]
Subtracting the second from the first:
\[ 2S_o = 3^n - 1 \implies S_o = \frac{3^n - 1}{2} \]

8. **Finding \( \sum_{r=0}^{n-1} a_r \)**:
The sum \( \sum_{r=0}^{n-1} a_r \) is equal to \( S_e \) since it includes all even indexed coefficients up to \( n-1 \):
\[ \sum_{r=0}^{n-1} a_r = S_e = \frac{3^n + 1}{2} \]

### Final Result:
Thus, the value of \( \sum_{r=0}^{n-1} a_r \) is:
\[ \frac{3^n + 1}{2} \]