If `(1+x+2x^(2))^(20) = a_(0) + a_(1)x^(2) "……" + a_(40)x^(40)`, then find the value of `a_(0) + a_(1) + a_(2) + "……" + a_(38)` .

Explanation

To solve the problem, we need to find the sum \( a_0 + a_1 + a_2 + \ldots + a_{38} \) from the expansion of \( (1 + x + 2x^2)^{20} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We have the expression \( (1 + x + 2x^2)^{20} \). This can be expanded using the Binomial Theorem, which states that \( (a + b + c)^n \) can be expanded into a sum of terms involving \( a^i b^j c^k \) where \( i + j + k = n \).

2. **Finding \( a_0 + a_1 + a_2 + \ldots + a_{40} \)**:
To find the sum of all coefficients \( a_0 + a_1 + a_2 + \ldots + a_{40} \), we can substitute \( x = 1 \):
\[ (1 + 1 + 2 \cdot 1^2)^{20} = (1 + 1 + 2)^{20} = 4^{20} \]

3. **Finding \( a_0 + a_2 + a_4 + \ldots + a_{40} \)**:
To find the sum of the coefficients of even powers, we substitute \( x = -1 \):
\[ (1 - 1 + 2 \cdot (-1)^2)^{20} = (1 - 1 + 2)^{20} = 2^{20} \]

4. **Using the Results**:
We have:
- From step 2: \( a_0 + a_1 + a_2 + \ldots + a_{40} = 4^{20} \)
- From step 3: \( a_0 + a_2 + a_4 + \ldots + a_{40} = 2^{20} \)

5. **Finding the Sum of Odd Coefficients**:
The sum of the coefficients of odd powers can be found by subtracting the sum of even coefficients from the total sum:
\[ a_1 + a_3 + a_5 + \ldots + a_{39} = (a_0 + a_1 + a_2 + \ldots + a_{40}) - (a_0 + a_2 + a_4 + \ldots + a_{40}) \]
This gives:
\[ a_1 + a_3 + a_5 + \ldots + a_{39} = 4^{20} - 2^{20} \]

6. **Finding \( a_0 + a_1 + a_2 + \ldots + a_{38} \)**:
To find \( a_0 + a_1 + a_2 + \ldots + a_{38} \), we note that:
\[ a_0 + a_1 + a_2 + \ldots + a_{38} = (a_0 + a_1 + a_2 + \ldots + a_{40}) - (a_{39} + a_{40}) \]
Since \( a_{40} \) is the coefficient of \( (2x^2)^{20} \) which is \( 2^{20} \), we have:
\[ a_{40} = 1 \]
Thus:
\[ a_{39} = 0 \quad \text{(as there is no term with } x^{39} \text{)} \]

7. **Final Calculation**:
Therefore:
\[ a_0 + a_1 + a_2 + \ldots + a_{38} = 4^{20} - 1 \]

### Final Answer:
\[ a_0 + a_1 + a_2 + \ldots + a_{38} = 4^{20} - 1 \]