Consider `(1+x+x^(2)) ^(n) = sum _(r=0)^(2n) a_(r) x^(r) , "where " a_(0),a_(1), `
`a_(2),…a_(2n)` are real numbers and n is a positive integer.
The value of `a_(2)` is

Options

A. `""^(4n+1)C_(2)`
B. `""^(3n+1)C_(2)`
C. `""^(2n+1)C_(2)`
D. `""^(n+1)C_(2)`

Explanation

To find the value of \( a_2 \) in the expansion of \( (1 + x + x^2)^n \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + x + x^2)^n \]
We can rewrite this expression as:
\[ (1 + x + x^2)^n = (1 + x^2 + x)^n \]

### Step 2: Use the Binomial Theorem
We know from the Binomial Theorem that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we can treat \( a = 1 \) and \( b = x + x^2 \):
\[ (1 + (x + x^2))^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} (x + x^2)^k \]

### Step 3: Expand \( (x + x^2)^k \)
Next, we need to expand \( (x + x^2)^k \) using the Binomial Theorem again:
\[ (x + x^2)^k = \sum_{j=0}^{k} \binom{k}{j} x^j (x^2)^{k-j} = \sum_{j=0}^{k} \binom{k}{j} x^{j + 2(k-j)} = \sum_{j=0}^{k} \binom{k}{j} x^{2k - j} \]

### Step 4: Collect terms
Now, we want to find the coefficient of \( x^2 \) in the expansion of \( (1 + x + x^2)^n \). The term \( x^2 \) can arise in two ways:
1. From \( (x + x^2)^k \) when \( j = 2 \) and \( k = 0 \) (i.e., \( k = 2 \)).
2. From \( (x + x^2)^k \) when \( j = 0 \) and \( k = 1 \) (i.e., \( k = 1 \)).

### Step 5: Calculate \( a_2 \)
To find \( a_2 \), we need to consider the contributions from the above cases:
- For \( k = 2 \):
\[ \text{Coefficient of } x^2 = \binom{2}{2} = 1 \]
- For \( k = 1 \):
\[ \text{Coefficient of } x^2 = \binom{1}{0} = 1 \]

Thus, the total coefficient \( a_2 \) is:
\[ a_2 = \sum_{k=0}^{n} \binom{n}{k} \cdot \text{(coefficient of } x^2 \text{ from } (x + x^2)^k) \]

### Final Result
The value of \( a_2 \) is:
\[ a_2 = \binom{n}{2} \]