Let n be positive integer such that, `(1+x+x^(2))^(n)=a_(0)+a_(1)x+a_(2)x^(2)+….+a_(2n)x^(2n)`, then `a_(r)` is :

Options

A. `(r+1)a_(r+1)`
B. `(r-1)a_(r+1)`
C. `(2r+1)a_(r+1)`
D. None of these

Answer

To solve the problem, we need to find the coefficients \( a_r \) in the expansion of \( (1 + x + x^2)^n \).

### 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, but we will analyze it in a different way to find the coefficients.

2. **Substituting \( x \)**:
We will substitute \( x \) with \( \frac{1}{x} \) in the original expression:
\[ (1 + x + x^2)^n = (1 + \frac{1}{x} + \frac{1}{x^2})^n \]

3. **Rewriting the Expression**:
This gives us:
\[ (1 + \frac{1}{x} + \frac{1}{x^2})^n = \left(\frac{x^2 + x + 1}{x^2}\right)^n = \frac{(x^2 + x + 1)^n}{x^{2n}} \]

4. **Equating the Two Forms**:
We can express the expansion of \( (1 + x + x^2)^n \) in terms of its coefficients:
\[ (1 + x + x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n} \]
and from our substitution:
\[ (1 + \frac{1}{x} + \frac{1}{x^2})^n = a_0 + a_1 \frac{1}{x} + a_2 \frac{1}{x^2} + \ldots + a_{2n} \frac{1}{x^{2n}} \]

5. **Multiplying by \( x^{2n} \)**:
Now, multiply both sides by \( x^{2n} \):
\[ (1 + x + x^2)^n \cdot x^{2n} = a_0 x^{2n} + a_1 x^{2n-1} + a_2 x^{2n-2} + \ldots + a_{2n} \]

6. **Setting Up the Equations**:
From the two expansions, we can equate the coefficients:
\[ a_r = a_{2n - r} \]
for \( r = 0, 1, 2, \ldots, 2n \).

7. **Conclusion**:
This means that the coefficients \( a_r \) are symmetric around \( n \). Specifically, \( a_r \) is equal to \( a_{2n - r} \).

### Final Result:
Thus, the value of \( a_r \) is given by:
\[ a_r = a_{2n - r} \]