If `(1 + x + x^2)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + ……..C_nx^n` then find `C_0 + C_3 + C_6 + ……. `

Answer

To solve the problem, we need to find the sum \( C_0 + C_3 + C_6 + \ldots \) from the expansion of \( (1 + x + x^2)^n \).

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ (1 + x + x^2)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \]
Here, \( C_k \) represents the coefficient of \( x^k \) in the expansion.

2. **Substituting \( x = 1 \)**:
To find the total sum of coefficients, we substitute \( x = 1 \):
\[ (1 + 1 + 1^2)^n = 3^n \]
This gives us:
\[ C_0 + C_1 + C_2 + C_3 + \ldots + C_n = 3^n \tag{1} \]

3. **Substituting \( x = \omega \)**:
Next, we substitute \( x = \omega \), where \( \omega \) is a complex cube root of unity (i.e., \( \omega = e^{2\pi i / 3} \)). We know that:
\[ 1 + \omega + \omega^2 = 0 \]
Therefore,
\[ (1 + \omega + \omega^2)^n = 0^n = 0 \]
This results in:
\[ C_0 + C_1 \omega + C_2 \omega^2 + C_3 \omega^3 + \ldots + C_n \omega^n = 0 \tag{2} \]

4. **Substituting \( x = \omega^2 \)**:
Similarly, substituting \( x = \omega^2 \):
\[ (1 + \omega^2 + (\omega^2)^2)^n = 0 \]
This gives us:
\[ C_0 + C_1 \omega^2 + C_2 \omega + C_3 \omega^3 + \ldots + C_n (\omega^2)^n = 0 \tag{3} \]

5. **Adding Equations (2) and (3)**:
Now, we add equations (2) and (3):
\[ (C_0 + C_1 \omega + C_2 \omega^2 + C_3 \omega^3 + \ldots) + (C_0 + C_1 \omega^2 + C_2 \omega + C_3 \omega^3 + \ldots) = 0 + 0 \]
This simplifies to:
\[ 2C_0 + (C_1 + C_2)(\omega + \omega^2) + C_3(1 + 1) + \ldots = 0 \]
Since \( \omega + \omega^2 = -1 \), we can express this as:
\[ 2C_0 - C_1 - C_2 + 2C_3 + \ldots = 0 \]

6. **Using the Symmetry**:
We can observe that the coefficients \( C_k \) for \( k \equiv 0 \mod 3 \) will sum up to a specific value. The coefficients can be grouped into three sets based on their indices modulo 3.

7. **Final Calculation**:
From the earlier results, we can conclude that:
\[ 3^n = 3(C_0 + C_3 + C_6 + \ldots) \]
Thus,
\[ C_0 + C_3 + C_6 + \ldots = \frac{3^n}{3} = 3^{n-1} \]

### Final Answer:
\[ C_0 + C_3 + C_6 + \ldots = 3^{n-1} \]