1. The problem asks to show that \(\sum_{r=1}^3 (3r + 4) = \sum_{r=1}^3 3r + \sum_{r=1}^3 4\) and \(\sum_{r=1}^4 (4r) = 4 \sum_{r=1}^4 r\). 2. Start with the first equality: \[\sum_{r=1}^3 (3r + 4) = (3 \times 1 + 4) + (3 \times 2 + 4) + (3 \times 3 + 4)\] \[= (3 + 4) + (6 + 4) + (9 + 4) = 7 + 10 + 13 = 30\] 3. Compute the right-hand side separately: \[\sum_{r=1}^3 3r + \sum_{r=1}^3 4 = (3 \times 1 + 3 \times 2 + 3 \times 3) + (4 + 4 + 4)\] \[= (3 + 6 + 9) + 12 = 18 + 12 = 30\] 4. Both sides equal 30, confirming the first statement is correct. 5. Now, consider the second equality: \[\sum_{r=1}^4 (4r) = (4 \times 1) + (4 \times 2) + (4 \times 3) + (4 \times 4) = 4 + 8 + 12 + 16 = 40\] 6. Compute the right-hand side: \[4 \sum_{r=1}^4 r = 4 (1 + 2 + 3 + 4) = 4 \times 10 = 40\] 7. Both sides equal 40, confirming the second statement is correct. 8. Hence, the sum operation distributes over addition and multiplication as shown, which is consistent with properties of summation. Final answers: \[\sum_{r=1}^3 (3r + 4) = 30\] \[\sum_{r=1}^3 3r + \sum_{r=1}^3 4 = 30\] \[\sum_{r=1}^4 (4r) = 40\] \[4 \sum_{r=1}^4 r = 40\]