Consider `(1+sin theta+sin^(2)theta)^(n)=sum_(r=0)^(2n) a_(r)(sin theta)^( r ), theta in R`.
The value of `a_(0)+2a_(1)+3a_(2)+…+(2n+1)a_(2n)` is

Answer

To solve the problem, we need to find the value of \( a_0 + 2a_1 + 3a_2 + \ldots + (2n+1)a_{2n} \) 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, and the coefficients \( a_r \) represent the coefficients of \( (\sin \theta)^r \) in this expansion.

2. **Identifying the Required Sum**:
We need to find \( S = a_0 + 2a_1 + 3a_2 + \ldots + (2n+1)a_{2n} \). This can be rewritten using summation notation:
\[ S = \sum_{r=0}^{2n} (r + 1) a_r \]

3. **Using a Generating Function**:
To find \( S \), we can relate it to the derivative of the generating function. We can consider the function:
\[ f(x) = (1 + x + x^2)^n \]
where \( x = \sin \theta \). The coefficients \( a_r \) are the coefficients of \( x^r \) in this expansion.

4. **Differentiating the Function**:
We differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = n(1 + x + x^2)^{n-1}(1 + 2x) \]
This gives us the sum \( S \) when evaluated at \( x = \sin \theta \):
\[ S = f'(1) = n(1 + 1 + 1)^{n-1}(1 + 2 \cdot 1) = n \cdot 3^{n-1} \cdot 3 = 3n \cdot 3^{n-1} \]

5. **Final Calculation**:
Simplifying gives:
\[ S = 3n \cdot 3^{n-1} = 3^n(n + 1) \]

Thus, the value of \( a_0 + 2a_1 + 3a_2 + \ldots + (2n+1)a_{2n} \) is:
\[ \boxed{(n + 1)3^n} \]