\( \sum _ { k = 1 } ^ { m } ( 6 k - 8 ) < 2600 \)

To solve the inequality \( \sum_{k=1}^{m} (6k - 8) < 2600 \), we start by finding the sum. The expression inside the summation, \( 6k - 8 \), can be split into two parts: \( 6k \) and \(-8\). First, we calculate the sum of \( 6k \): \[ \sum_{k=1}^{m} 6k = 6 \sum_{k=1}^{m} k = 6 \cdot \frac{m(m+1)}{2} = 3m(m+1). \] Next, we calculate the sum of \(-8\): \[ \sum_{k=1}^{m} -8 = -8m. \] Now, combining these results gives us: \[ \sum_{k=1}^{m} (6k - 8) = 3m(m+1) - 8m = 3m^2 + 3m - 8m = 3m^2 - 5m. \] Now, we need to set up the inequality: \[ 3m^2 - 5m < 2600. \] Rearranging this, we have: \[ 3m^2 - 5m - 2600 < 0. \] Next, we can solve the quadratic equation \( 3m^2 - 5m - 2600 = 0 \) using the quadratic formula, \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = -5, c = -2600 \): First, calculate the discriminant: \[ D = b^2 - 4ac = (-5)^2 - 4 \cdot 3 \cdot (-2600) = 25 + 31200 = 31225. \] Now, calculate \( m \): \[ m = \frac{-(-5) \pm \sqrt{31225}}{2 \cdot 3} = \frac{5 \pm 176.5}{6}. \] Calculating the two potential results: 1. \( m_1 = \frac{5 + 176.5}{6} = \frac{181.5}{6} \approx 30.25 \) (we only take the integer part) 2. \( m_2 = \frac{5 - 176.5}{6} = \frac{-171.5}{6} \) (not a valid solution since \( m \) must be positive) Thus, the largest integer \( m \) less than 30.25 is 30. To verify: \[ 3(30)^2 - 5(30) = 2700 - 150 = 2550. \] Since \( 2550 < 2600 \), it satisfies the inequality. Thus, the solution is: \[ m \leq 30. \]