If `(1+x+x^(2))^(20) = a_(0) + a_(1)x^(2) "……" + a_(40)x^(40)`, then following questions.
The value of `a_(0) + 3a_(1) + 5a_(2) + "……" + 81a_(40)` is

Options

A. (a) `161 xx 3^(20)`
B. (b) `41 xx 3^(40)`
C. (c) `41 xx 3^(20)`
D. (d) none of these

Explanation

To solve the problem, we need to find the value of \( a_0 + 3a_1 + 5a_2 + \ldots + 81a_{40} \) given that \( (1 + x + x^2)^{20} = a_0 + a_1 x^2 + a_2 x^4 + \ldots + a_{40} x^{40} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 + x + x^2)^{20} \). The coefficients \( a_k \) represent the coefficients of \( x^{2k} \) in this expansion.

2. **Substituting \( x \) with \( \frac{1}{x} \)**:
We substitute \( x \) with \( \frac{1}{x} \) in the original expression:
\[ (1 + \frac{1}{x} + \frac{1}{x^2})^{20} = (x^2 + x + 1)^{20} \cdot \frac{1}{x^{40}} \]
This gives us:
\[ (x^2 + x + 1)^{20} = a_0 x^{40} + a_1 x^{38} + a_2 x^{36} + \ldots + a_{40} \]

3. **Equating Both Expressions**:
Since both expressions are equal, we can compare coefficients. From the left-hand side:
\[ (1 + x + x^2)^{20} = a_0 + a_1 x^2 + a_2 x^4 + \ldots + a_{40} x^{40} \]
From the right-hand side:
\[ (x^2 + x + 1)^{20} = a_0 x^{40} + a_1 x^{38} + a_2 x^{36} + \ldots + a_{40} \]

4. **Finding Relationships Between Coefficients**:
From the comparison, we can establish:
- \( a_{40} = a_0 \)
- \( a_{39} = a_1 \)
- \( a_{38} = a_2 \)
- Continuing this pattern, we find that \( a_k = a_{40-k} \) for \( k = 0, 1, \ldots, 20 \).

5. **Evaluating the Sum**:
We need to evaluate:
\[ S = a_0 + 3a_1 + 5a_2 + \ldots + 81a_{40} \]
Notice that the coefficients can be expressed as \( 2n + 1 \) where \( n \) is the index of \( a_n \). Thus, we can group terms:
\[ S = (a_0 + 81a_{40}) + (3a_1 + 79a_{39}) + (5a_2 + 77a_{38}) + \ldots + (39a_{19} + 43a_{21}) + 41a_{20} \]

6. **Simplifying the Expression**:
Each pair sums to \( 82a_k \):
\[ S = 82(a_0 + a_1 + a_2 + \ldots + a_{19}) + 41a_{20} \]

7. **Finding \( a_0 + a_1 + \ldots + a_{20} \)**:
To find \( a_0 + a_1 + \ldots + a_{20} \), we substitute \( x = 1 \) in the original expression:
\[ (1 + 1 + 1)^{20} = 3^{20} \]
Thus:
\[ a_0 + a_1 + a_2 + \ldots + a_{20} = 3^{20} \]

8. **Final Calculation**:
Therefore:
\[ a_0 + a_1 + \ldots + a_{19} = 3^{20} - a_{20} \]
Substitute back into \( S \):
\[ S = 82 \left( \frac{3^{20} - a_{20}}{2} \right) + 41a_{20} \]
Simplifying gives:
\[ S = 41 \cdot 3^{20} \]

### Final Result:
Thus, the value of \( a_0 + 3a_1 + 5a_2 + \ldots + 81a_{40} \) is:
\[ \boxed{41 \cdot 3^{20}} \]