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

To solve the inequality: \[ \sum_{k=1}^{m} (6k - 8) < 2600 \] **Step 1: Simplify the Sum** First, calculate the sum: \[ \sum_{k=1}^{m} (6k - 8) = 6\sum_{k=1}^{m} k - 8m \] Using the formula for the sum of the first \( m \) positive integers: \[ \sum_{k=1}^{m} k = \frac{m(m + 1)}{2} \] Substitute back into the original sum: \[ 6 \cdot \frac{m(m + 1)}{2} - 8m = 3m(m + 1) - 8m = 3m^2 + 3m - 8m = 3m^2 - 5m \] **Step 2: Formulate the Inequality** Now, set up the inequality: \[ 3m^2 - 5m < 2600 \] Rearrange it into a standard quadratic inequality: \[ 3m^2 - 5m - 2600 < 0 \] **Step 3: Find the Roots of the Quadratic Equation** Solve the quadratic equation \( 3m^2 - 5m - 2600 = 0 \) using the quadratic formula: \[ m = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2600)}}{2 \cdot 3} = \frac{5 \pm \sqrt{25 + 31200}}{6} = \frac{5 \pm \sqrt{31225}}{6} = \frac{5 \pm 177}{6} \] This gives two solutions: \[ m = \frac{182}{6} \approx 30.333 \quad \text{and} \quad m = \frac{-172}{6} \approx -28.666 \] **Step 4: Determine the Valid Range for \( m \)** Since \( m \) must be a positive integer, the inequality \( 3m^2 - 5m - 2600 < 0 \) holds for: \[ m < 30.333 \] Therefore, the largest integer \( m \) that satisfies the inequality is \( m = 30 \). **Verification:** - For \( m = 30 \): \[ 3(30)^2 - 5(30) = 2700 - 150 = 2550 < 2600 \] - For \( m = 31 \): \[ 3(31)^2 - 5(31) = 2883 - 155 = 2728 > 2600 \] Thus, the solution is confirmed. **Answer:** The largest integer \( m \) satisfying the inequality is \( m = 30 \).