If `(1-x-x^(2))^(20) = sum_(r=0)^(40)a_(r).x^(r )`, then value of `a_(1) + 3a_(3) + 5a_(5) + "….." + 39a_(39)` is

Explanation

To solve the problem, we need to find the value of \( a_1 + 3a_3 + 5a_5 + \ldots + 39a_{39} \) from the expansion of \( (1 - x - x^2)^{20} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( (1 - x - x^2)^{20} \). This can be expanded using the binomial theorem, which states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 1 \) and \( b = -x - x^2 \).

2. **Expanding the Expression**:
We can express the expansion as:
\[ (1 - x - x^2)^{20} = \sum_{r=0}^{40} a_r x^r \]
where \( a_r \) are the coefficients of \( x^r \).

3. **Differentiating the Expression**:
To find the coefficients \( a_1, a_3, a_5, \ldots \), we differentiate the expression:
\[ \frac{d}{dx}[(1 - x - x^2)^{20}] = 20(1 - x - x^2)^{19}(-1 - 2x) \]
This gives us:
\[ 20(1 - x - x^2)^{19}(-1 - 2x) = \sum_{r=0}^{39} ra_r x^{r-1} \]

4. **Evaluating at Specific Points**:
We evaluate the differentiated expression at \( x = 1 \) and \( x = -1 \):
- For \( x = 1 \):
\[ (1 - 1 - 1)^{20} = 0 \implies 20(0)(-3) = 0 \]
This gives us:
\[ a_1 + 2a_2 + 3a_3 + \ldots + 40a_{40} = 0 \quad \text{(Equation 1)} \]

- For \( x = -1 \):
\[ (1 + 1 - 1)^{20} = 1 \implies 20(1)(-1) = -20 \]
This gives us:
\[ a_1 - 2a_2 + 3a_3 - 4a_4 + \ldots - 40a_{40} = -20 \quad \text{(Equation 2)} \]

5. **Adding the Two Equations**:
Now, we add Equation 1 and Equation 2:
\[ (a_1 + 2a_2 + 3a_3 + \ldots + 40a_{40}) + (a_1 - 2a_2 + 3a_3 - 4a_4 + \ldots - 40a_{40}) = 0 - 20 \]
This simplifies to:
\[ 2a_1 + 6a_3 + 10a_5 + \ldots + 80a_{39} = -20 \]

6. **Finding the Required Sum**:
We can factor out 2 from the left side:
\[ 2(a_1 + 3a_3 + 5a_5 + \ldots + 39a_{39}) = -20 \]
Thus:
\[ a_1 + 3a_3 + 5a_5 + \ldots + 39a_{39} = -10 \]

### Final Answer:
The value of \( a_1 + 3a_3 + 5a_5 + \ldots + 39a_{39} \) is \( -10 \).