If `S=sum_(r=1)^(80)(r)/((r^4+r^2+1))` , then the value of `(6481s)/(1000)` is

Answer

To solve the problem, we need to evaluate the sum \( S = \sum_{r=1}^{80} \frac{r}{r^4 + r^2 + 1} \) and then find the value of \( \frac{6481 S}{1000} \).

### Step-by-Step Solution

1. **Rewrite the Denominator**:
We start with the expression for \( S \):
\[ S = \sum_{r=1}^{80} \frac{r}{r^4 + r^2 + 1} \]
Notice that we can rewrite the denominator:
\[ r^4 + r^2 + 1 = (r^2 + 1)^2 - r^2 \]
This allows us to express it as a difference of squares.

2. **Multiply and Divide by 2**:
To simplify the expression, we multiply and divide by 2:
\[ S = \sum_{r=1}^{80} \frac{2r}{2(r^4 + r^2 + 1)} = \frac{1}{2} \sum_{r=1}^{80} \frac{2r}{(r^2 + 1)^2 - r^2} \]

3. **Add and Subtract \( r^2 \)**:
Now, we add and subtract \( r^2 \) in the denominator:
\[ S = \frac{1}{2} \sum_{r=1}^{80} \left( \frac{2r}{(r^2 + r + 1)(r^2 - r + 1)} \right) \]

4. **Split the Fraction**:
We can express the fraction as:
\[ \frac{2r}{(r^2 + r + 1)(r^2 - r + 1)} = \frac{A}{r^2 + r + 1} + \frac{B}{r^2 - r + 1} \]
for some constants \( A \) and \( B \).

5. **Find Constants A and B**:
By equating coefficients, we can determine \( A \) and \( B \). After some algebra, we find:
\[ A = 1, \quad B = -1 \]
Thus:
\[ S = \frac{1}{2} \sum_{r=1}^{80} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) \]

6. **Telescoping Series**:
The series is telescoping:
\[ S = \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \ldots + \left( \frac{1}{6401} - \frac{1}{6481} \right) \right) \]
Most terms cancel out, and we are left with:
\[ S = \frac{1}{2} \left( 1 - \frac{1}{6481} \right) \]

7. **Calculate S**:
Thus, we get:
\[ S = \frac{1}{2} \left( \frac{6480}{6481} \right) = \frac{3240}{6481} \]

8. **Final Calculation**:
Now we need to find:
\[ \frac{6481 S}{1000} = \frac{6481 \cdot \frac{3240}{6481}}{1000} = \frac{3240}{1000} = 3.24 \]

### Final Answer:
\[ \frac{6481 S}{1000} = 3.24 \]