Cevaluate: \( \begin{array}{ll}\text { a) } \sum_{n=1}^{12}(5 n-1) & \text { b) } \sum_{n=1}^{20}(3 n+5) \\ \text { c) } \sum_{n=1}^{16}(30-2 n) & \text { d) } \sum_{n=5}^{12}(6 n+1) \\ \text { e) } \sum_{n=10}^{30}(15-n) & \text { f) } \sum_{n=4}^{20}(2 n-6) \\ \text { 9) } \sum_{n=4}^{14}(8-n) & \text { n) } \sum_{n=1}^{15}(3 n)\end{array} \)

**a) Evaluation of \(\sum_{n=1}^{12}(5n-1)\):** We split the sum: \[ \sum_{n=1}^{12}(5n-1) = 5\sum_{n=1}^{12} n - \sum_{n=1}^{12} 1 \] where \[ \sum_{n=1}^{12} n = \frac{12 \cdot 13}{2} = 78 \] and \[ \sum_{n=1}^{12} 1 = 12. \] Thus, \[ 5 \cdot 78 - 12 = 390 - 12 = 378. \] --- **b) Evaluation of \(\sum_{n=1}^{20}(3n+5)\):** Break the sum: \[ \sum_{n=1}^{20}(3n+5) = 3\sum_{n=1}^{20} n + 5\sum_{n=1}^{20} 1. \] Calculate: \[ \sum_{n=1}^{20} n = \frac{20 \cdot 21}{2} = 210,\quad \sum_{n=1}^{20} 1 = 20. \] Thus, \[ 3\cdot 210 + 5\cdot 20 = 630 + 100 = 730. \] --- **c) Evaluation of \(\sum_{n=1}^{16}(30-2n)\):** Separate the sum: \[ \sum_{n=1}^{16}(30-2n) = 30\cdot 16 - 2\sum_{n=1}^{16} n. \] With \[ \sum_{n=1}^{16} n = \frac{16 \cdot 17}{2} = 136, \] we have \[ 480 - 2\cdot 136 = 480 - 272 = 208. \] --- **d) Evaluation of \(\sum_{n=5}^{12}(6n+1)\):** First, determine the number of terms from \(n=5\) to \(n=12\): \[ 12-5+1 = 8. \] Then, split the sum: \[ \sum_{n=5}^{12}(6n+1) = 6\sum_{n=5}^{12} n + \sum_{n=5}^{12} 1. \] Compute \[ \sum_{n=5}^{12} n = \sum_{n=1}^{12} n - \sum_{n=1}^{4} n = \frac{12\cdot13}{2} - \frac{4\cdot5}{2} = 78 - 10 = 68. \] Thus, \[ 6\cdot68 + 8 = 408 + 8 = 416. \] --- **e) Evaluation of \(\sum_{n=10}^{30}(15-n)\):** Count the number of terms, from \(n=10\) to \(n=30\): \[ 30-10+1 = 21. \] Express the sum: \[ \sum_{n=10}^{30}(15-n) = 15\cdot21 - \sum_{n=10}^{30} n. \] Find \[ \sum_{n=10}^{30} n = \sum_{n=1}^{30} n - \sum_{n=1}^{9} n = \frac{30\cdot31}{2} - \frac{9\cdot10}{2} = 465 - 45 = 420. \] Thus, \[ 15\cdot21 - 420 = 315 - 420 = -105. \] --- **f) Evaluation of \(\sum_{n=4}^{20}(2n-6)\):** Determine the number of terms from \(n=4\) to \(n=20\): \[ 20-4+1 = 17. \] Separate the sum: \[ \sum_{n=4}^{20}(2n-6) = 2\sum_{n=4}^{20} n - 6\sum_{n=4}^{20} 1. \] Compute \[ \sum_{n=4}^{20} n = \sum_{n=1}^{20} n - \sum_{n=1}^{3} n = \frac{20\cdot21}{2} - \frac{3\cdot4}{2} = 210 - 6 = 204. \] Then, \[ 2\cdot204 - 6\cdot17 = 408 - 102 = 306. \] --- **9) Evaluation of \(\sum_{n=4}^{14}(8-n)\):** Count the terms: \[ 14-4+1 = 11. \] Separate the sum: \[ \sum_{n=4}^{14}(8-n) = 8\cdot11 - \sum_{n=4}^{14} n. \] Compute \[ \sum_{n=4}^{14} n = \sum_{n=1}^{14} n - \sum_{n=1}^{3} n = \frac{14\cdot15}{2} - \frac{3\cdot4}{2} = 105 - 6 = 99. \] Thus, \[ 8\cdot11 - 99 = 88 - 99 = -11. \] --- **n) Evaluation of \(\sum_{n=1}^{15}(3n)\):** Extract the constant: \[ \sum_{n=1}^{15}(3n) = 3\sum_{n=1}^{15} n. \] Compute \[ \sum_{n=1}^{15} n = \frac{15\cdot16}{2} = 120. \] Thus, \[ 3\cdot120 = 360. \]