if `(1+x+x^2)^(10)(1-x+x^2)^(10)=a_0+a_x+a_2x^2+a_3x^3+...........+a_(40)x^(40)` then `a_1+a_3+.....+a_(39)` is equal to

Options

A. 0
B. `(3^(10)-1)/(2)`
C. `3^(10)`
D. `4^(10)`

Explanation

To solve the problem, we need to find the sum of the coefficients of the odd powers of \(x\) in the expansion of \((1+x+x^2)^{10}(1-x+x^2)^{10}\).

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ (1+x+x^2)^{10}(1-x+x^2)^{10} \]
We can denote this as \(f(x)\).

2. **Finding \(f(1)\)**:
To find the sum of all coefficients \(a_0 + a_1 + a_2 + \ldots + a_{40}\), we evaluate \(f(1)\):
\[ f(1) = (1+1+1)^{10}(1-1+1)^{10} = 3^{10} \cdot 1^{10} = 3^{10} \]

3. **Finding \(f(-1)\)**:
Next, we find \(f(-1)\) to help us isolate the odd coefficients:
\[ f(-1) = (1-1+1)^{10}(1+1+1)^{10} = 1^{10} \cdot 3^{10} = 3^{10} \]

4. **Using the Results**:
The coefficients of \(f(x)\) can be expressed as:
\[ f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_{40} x^{40} \]
From our evaluations:
- \(f(1) = a_0 + a_1 + a_2 + a_3 + \ldots + a_{40} = 3^{10}\)
- \(f(-1) = a_0 - a_1 + a_2 - a_3 + \ldots + a_{40} = 3^{10}\)

5. **Setting Up the Equations**:
We have two equations:
\[ a_0 + a_1 + a_2 + a_3 + \ldots + a_{40} = 3^{10} \quad \text{(1)} \]
\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{40} = 3^{10} \quad \text{(2)} \]

6. **Adding and Subtracting the Equations**:
- Adding (1) and (2):
\[ (a_0 + a_1 + a_2 + a_3 + \ldots + a_{40}) + (a_0 - a_1 + a_2 - a_3 + \ldots + a_{40}) = 3^{10} + 3^{10} \]
This simplifies to:
\[ 2a_0 + 2a_2 + 2a_4 + \ldots + 2a_{40} = 2 \cdot 3^{10} \]
Dividing by 2 gives:
\[ a_0 + a_2 + a_4 + \ldots + a_{40} = 3^{10} \]

- Subtracting (2) from (1):
\[ (a_0 + a_1 + a_2 + a_3 + \ldots + a_{40}) - (a_0 - a_1 + a_2 - a_3 + \ldots + a_{40}) = 3^{10} - 3^{10} \]
This simplifies to:
\[ 2a_1 + 2a_3 + 2a_5 + \ldots + 2a_{39} = 0 \]
Dividing by 2 gives:
\[ a_1 + a_3 + a_5 + \ldots + a_{39} = 0 \]

7. **Final Result**:
Therefore, the sum of the coefficients of the odd powers of \(x\) is:
\[ a_1 + a_3 + a_5 + \ldots + a_{39} = 0 \]

### Conclusion:
The value of \(a_1 + a_3 + \ldots + a_{39}\) is \(0\).