If `(1+x+x^(2))^(6n)=a_(0)+a_(1)x+a_(2)x^(2)+……,` then
`a_(2)+a_(4)+a_(6)+` ….. = …..

Answer

To solve the problem, we need to find the sum of the coefficients of even powers of \( x \) in the expansion of \( (1 + x + x^2)^{6n} \). Let's denote this sum as \( S = a_0 + a_2 + a_4 + \ldots \).

### Step-by-step Solution:

1. **Identify the expression**: We start with the expression \( (1 + x + x^2)^{6n} \).

2. **Evaluate at \( x = 1 \)**:
\[ (1 + 1 + 1^2)^{6n} = 3^{6n} \]
This gives us the sum of all coefficients:
\[ S_{\text{total}} = a_0 + a_1 + a_2 + a_3 + \ldots = 3^{6n} \]

3. **Evaluate at \( x = -1 \)**:
\[ (1 - 1 + (-1)^2)^{6n} = (1)^{6n} = 1 \]
This gives us the sum of coefficients of even powers minus the sum of coefficients of odd powers:
\[ S_{\text{even}} - S_{\text{odd}} = 1 \]

4. **Set up the equations**:
Let \( S_{\text{even}} = a_0 + a_2 + a_4 + \ldots \) and \( S_{\text{odd}} = a_1 + a_3 + a_5 + \ldots \). We have:
\[ S_{\text{total}} = S_{\text{even}} + S_{\text{odd}} = 3^{6n} \quad (1) \]
\[ S_{\text{even}} - S_{\text{odd}} = 1 \quad (2) \]

5. **Solve the equations**:
From equation (1), we can express \( S_{\text{odd}} \) in terms of \( S_{\text{even}} \):
\[ S_{\text{odd}} = 3^{6n} - S_{\text{even}} \]
Substitute this into equation (2):
\[ S_{\text{even}} - (3^{6n} - S_{\text{even}}) = 1 \]
Simplifying gives:
\[ 2S_{\text{even}} - 3^{6n} = 1 \]
\[ 2S_{\text{even}} = 3^{6n} + 1 \]
\[ S_{\text{even}} = \frac{3^{6n} + 1}{2} \]

6. **Final result**: Thus, the sum of the coefficients of even powers of \( x \) is:
\[ a_0 + a_2 + a_4 + \ldots = \frac{3^{6n} + 1}{2} \]