(8) \( \sum_{r=2}^{m} 2(3-r)>-800 \)

To solve the inequality \( \sum_{r=2}^{m} 2(3-r) > -800 \), we can first simplify the summation: 1. **Expanding the Sum:** \[ \sum_{r=2}^{m} 2(3-r) = 2 \sum_{r=2}^{m} (3-r) \] This can be further simplified to: \[ \sum_{r=2}^{m} (3-r) = \sum_{r=2}^{m} 3 - \sum_{r=2}^{m} r \] where \( \sum_{r=2}^{m} 3 = 3(m - 1) \) (since there are \( m-1 \) terms from 2 to m) and \( \sum_{r=2}^{m} r = \frac{m(m+1)}{2} - 1 \) (the sum of integers from 1 to m minus 1). Therefore, we can compute: \[ \sum_{r=2}^{m} (3-r) = 3(m-1) - \left( \frac{m(m+1)}{2} - 1 \right) \] Simplifying that gives: \[ = 3m - 3 - \frac{m(m+1)}{2} + 1 = 3m - \frac{m(m+1)}{2} - 2 \] 2. **Now putting back into the inequality:** \[ 2(3m - \frac{m(m+1)}{2} - 2) > -800 \] Simplifying: \[ 6m - m(m+1) - 4 > -800 \] Rearranging yields: \[ -m^2 + 5m + 796 > 0 \] 3. **Rearranging this can yield a standard quadratic inequality:** \[ m^2 - 5m - 796 < 0 \] 4. **Finding the roots of the corresponding equation:** We can use the quadratic formula: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 + 3184}}{2} = \frac{5 \pm \sqrt{3209}}{2} \] Now \( \sqrt{3209} \approx 56.6 \), which gives us: \[ m \approx \frac{5 \pm 56.6}{2} \] This yields two roots: \[ m_1 \approx 30.8, \quad m_2 \approx -25.8 \] 5. **Determine intervals for the quadratic to be less than zero:** Since the parabola opens upwards, it is negative between its roots. Hence, the solution is: \[ -25.8 < m < 30.8 \] So, the integer solution for \( m \) is: \[ m = 2, 3, \ldots, 30 \] In conclusion, \( m \) can take any integer value from 2 to 30, inclusive.