If `(1 +x+x^2)^25 = a_0 + a_1x+ a_2x^2 +..... + a_50.x^50` then `a_0 + a_2 + a_4 + ... + a_50` is :

Options

A. even
B. odd and of the form `3n`
C. odd and of the form `(3n-1)`
D. odd and of the form `(3n+1)`

Explanation

To solve the problem, we need to find the sum of the coefficients of the even powers of \( x \) in the expansion of \( (1 + x + x^2)^{25} \). The coefficients of the polynomial can be represented as \( a_0, a_1, a_2, \ldots, a_{50} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We have the expression \( (1 + x + x^2)^{25} \). This can be expanded using the binomial theorem.

2. **Finding the Total Sum of Coefficients**:
To find \( a_0 + a_1 + a_2 + \ldots + a_{50} \), we can substitute \( x = 1 \):
\[ (1 + 1 + 1^2)^{25} = (3)^{25} \]
This gives us the total sum of all coefficients.

3. **Finding the Sum of Even Coefficients**:
To find the sum of the coefficients of the even powers, we can use the substitution \( x = -1 \):
\[ (1 - 1 + (-1)^2)^{25} = (1)^{25} = 1 \]
This gives us the alternating sum of the coefficients, which can be expressed as:
\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{50} = 1 \]

4. **Setting Up the Equations**:
Let \( S_e \) be the sum of the coefficients of even powers and \( S_o \) be the sum of the coefficients of odd powers. From the previous steps, we have:
\[ S_e + S_o = 3^{25} \]
\[ S_e - S_o = 1 \]

5. **Solving the System of Equations**:
We can solve these two equations:
- Adding both equations:
\[ 2S_e = 3^{25} + 1 \implies S_e = \frac{3^{25} + 1}{2} \]
- Subtracting the second equation from the first:
\[ 2S_o = 3^{25} - 1 \implies S_o = \frac{3^{25} - 1}{2} \]

6. **Final Result**:
The sum of the coefficients of the even powers \( a_0 + a_2 + a_4 + \ldots + a_{50} \) is:
\[ S_e = \frac{3^{25} + 1}{2} \]

### Summary:
Thus, the final answer is:
\[ \boxed{\frac{3^{25} + 1}{2}} \]