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

Options

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

Answer

To solve the problem, we need to find the value of \( a_0^2 - a_1^2 + a_2^2 - a_3^2 + \ldots - a_{2n}^2 \) given that

\[ (1 + \sin \theta + \sin^2 \theta)^n = \sum_{r=0}^{2n} a_r (\sin \theta)^r. \]

### 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 gives us a polynomial in \( \sin \theta \).

2. **Substituting \( \sin \theta \)**:
To analyze the coefficients \( a_r \), we can substitute \( \sin \theta = x \) for simplicity:
\[ (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r. \]

3. **Finding Coefficients**:
The coefficients \( a_r \) represent the coefficients of \( x^r \) in the expansion of \( (1 + x + x^2)^n \).

4. **Using Symmetry**:
We can observe that if we replace \( x \) with \( \frac{1}{x} \), we get:
\[ (1 + \frac{1}{x} + \frac{1}{x^2})^n = x^{-2n} (x^2 + x + 1)^n. \]
This shows that \( a_r = a_{2n - r} \) for \( r = 0, 1, \ldots, 2n \).

5. **Calculating the Desired Sum**:
We need to compute:
\[ a_0^2 - a_1^2 + a_2^2 - a_3^2 + \ldots - a_{2n}^2. \]
Using the symmetry \( a_r = a_{2n - r} \), we can pair terms:
\[ (a_0^2 - a_{2n}^2) + (a_1^2 - a_{2n-1}^2) + \ldots + (a_n^2 - a_n^2). \]
Since \( a_n^2 \) cancels out, we are left with:
\[ a_0^2 - a_{2n}^2 + a_1^2 - a_{2n-1}^2 + \ldots. \]

6. **Final Value**:
The expression simplifies to \( a_n \), which is the coefficient of \( \sin^{2n} \theta \) in the original polynomial.

Thus, the final result is:
\[ \boxed{a_n}. \]