Consider `(1+sin theta+sin^(2)theta)^(n)=sum_(r=0)^(2n)a_(r) (sin theta)^( r ), theta in R`.
`a_(n+1)+a_(n+2)+…+a_(2n-1)` equals

Options

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

Answer

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

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 + \sin \theta + \sin^2 \theta)^n \). This can be expanded using the binomial theorem, which will give us coefficients \( a_r \) for each power of \( \sin \theta \).

2. **Substituting \( \sin \theta = 1 \)**:
To find the total sum of coefficients, we can substitute \( \sin \theta = 1 \):
\[ (1 + 1 + 1^2)^n = 3^n \]
This means:
\[ \sum_{r=0}^{2n} a_r = 3^n \]

3. **Using Symmetry**:
The coefficients \( a_r \) have a symmetry property. Specifically, \( a_r \) for \( r \) and \( 2n - r \) will be equal due to the structure of the polynomial. This means:
\[ a_r = a_{2n - r} \]

4. **Finding \( a_{n+1} + a_{n+2} + \ldots + a_{2n-1} \)**:
The sum \( a_{n+1} + a_{n+2} + \ldots + a_{2n-1} \) can be expressed in terms of the total sum:
\[ a_0 + a_1 + \ldots + a_{2n} = 3^n \]
Since \( a_r = a_{2n - r} \), we can pair the coefficients:
\[ a_0 + a_{2n} + a_1 + a_{2n-1} + \ldots + a_n \]
This gives us:
\[ a_0 + a_{2n} + a_1 + a_{2n-1} + a_2 + a_{2n-2} + \ldots + a_n = 3^n \]
The middle term \( a_n \) is counted once, and the rest can be paired.

5. **Calculating the Required Sum**:
The sum \( a_{n+1} + a_{n+2} + \ldots + a_{2n-1} \) is half of the remaining terms after excluding \( a_0, a_1, \ldots, a_n \):
\[ a_{n+1} + a_{n+2} + \ldots + a_{2n-1} = \frac{3^n - a_0 - a_1 - \ldots - a_n}{2} \]
Since \( a_0 = 1 \) (the constant term when \( \sin \theta = 0 \)), we can find:
\[ a_0 + a_1 + \ldots + a_n = 3^n - \text{(sum of the first half)} \]

6. **Final Calculation**:
After careful analysis, we find that:
\[ a_{n+1} + a_{n+2} + \ldots + a_{2n-1} = \frac{3^n - 1}{2} \]

### Conclusion:
Thus, the value of \( a_{n+1} + a_{n+2} + \ldots + a_{2n-1} \) equals \( \frac{3^n - 1}{2} \).