4. \( \left[\sum_{m=2}^{7}(4 m-13)\right]\left[\sum_{m=1}^{9}(2 m-2)\right] \) 5. \( \sum_{s=5}^{11}(7 s-4)-6 \sum_{s=6}^{13}(-3 s-2) \)

Let's solve each expression step by step. ### Problem 4: \[ \left[\sum_{m=2}^{7}(4 m-13)\right]\left[\sum_{m=1}^{9}(2 m-2)\right] \] **Step 1: Calculate the first sum \(\sum_{m=2}^{7}(4m-13)\)** We can evaluate this sum by substituting the values of \(m\) from 2 to 7: \[ \begin{align*} m = 2: & \quad 4(2) - 13 = 8 - 13 = -5 \\ m = 3: & \quad 4(3) - 13 = 12 - 13 = -1 \\ m = 4: & \quad 4(4) - 13 = 16 - 13 = 3 \\ m = 5: & \quad 4(5) - 13 = 20 - 13 = 7 \\ m = 6: & \quad 4(6) - 13 = 24 - 13 = 11 \\ m = 7: & \quad 4(7) - 13 = 28 - 13 = 15 \\ \end{align*} \] Now, summing these results: \[ -5 + (-1) + 3 + 7 + 11 + 15 = 30 \] **Step 2: Calculate the second sum \(\sum_{m=1}^{9}(2m-2)\)** We can factor out the 2: \[ \sum_{m=1}^{9}(2m-2) = 2\sum_{m=1}^{9}(m-1) = 2\sum_{m=0}^{8}m = 2\left(\frac{8(8+1)}{2}\right) = 2 \cdot 36 = 72 \] **Step 3: Multiply the two sums** Now we multiply the results of the two sums: \[ 30 \cdot 72 = 2160 \] ### Final Result for Problem 4: \[ \boxed{2160} \] --- ### Problem 5: \[ \sum_{s=5}^{11}(7s-4) - 6 \sum_{s=6}^{13}(-3s-2) \] **Step 1: Calculate the first sum \(\sum_{s=5}^{11}(7s-4)\)** Calculating each term: \[ \begin{align*} s = 5: & \quad 7(5) - 4 = 35 - 4 = 31 \\ s = 6: & \quad 7(6) - 4 = 42 - 4 = 38 \\ s = 7: & \quad 7(7) - 4 = 49 - 4 = 45 \\ s = 8: & \quad 7(8) - 4 = 56 - 4 = 52 \\ s = 9: & \quad 7(9) - 4 = 63 - 4 = 59 \\ s = 10: & \quad 7(10) - 4 = 70 - 4 = 66 \\ s = 11: & \quad 7(11) - 4 = 77 - 4 = 73 \\ \end{align*} \] Now, summing these results: \[ 31 + 38 + 45 + 52 + 59 + 66 + 73 = 364 \] **Step 2: Calculate the second sum \(\sum_{s=6}^{13}(-3s-2)\)** Calculating each term: \[ \begin{align*} s = 6: & \quad -3(6) - 2 = -18 - 2 = -20 \\ s = 7: & \quad -3(7) - 2 = -21 - 2 = -23 \\ s = 8: & \quad -3(8) - 2 = -24 - 2 = -26 \\ s = 9: & \quad -3(9) - 2 = -27 - 2 = -29 \\ s = 10: & \quad -3(10) - 2 = -30 - 2 = -32 \\ s = 11: & \quad -3(11) - 2 = -33 - 2 = -35 \\ s = 12: & \quad -