Feracise 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) Evaluate** \(\sum_{n=1}^{12}(5n-1)\) Write the sum as: \[ \sum_{n=1}^{12}(5n-1)=5\sum_{n=1}^{12} n-\sum_{n=1}^{12}1. \] We know: \[ \sum_{n=1}^{12} n=\frac{12\cdot 13}{2}=78,\quad \text{and} \quad \sum_{n=1}^{12}1=12. \] Thus, \[ 5\cdot 78-12=390-12=378. \] --- **b) Evaluate** \(\sum_{n=1}^{20}(3n+5)\) Write the sum as: \[ \sum_{n=1}^{20}(3n+5)=3\sum_{n=1}^{20} n+5\sum_{n=1}^{20}1. \] We have: \[ \sum_{n=1}^{20} n=\frac{20\cdot21}{2}=210,\quad \text{and} \quad \sum_{n=1}^{20}1=20. \] Thus, \[ 3\cdot 210+5\cdot 20=630+100=730. \] --- **c) Evaluate** \(\sum_{n=1}^{16}(30-2n)\) Write the sum as: \[ \sum_{n=1}^{16}(30-2n)=30\sum_{n=1}^{16}1-2\sum_{n=1}^{16} n. \] There are 16 terms so: \[ 30\cdot16-2\left(\frac{16\cdot17}{2}\right)=480-16\cdot17. \] Since \(16\cdot17=272\), we get: \[ 480-272=208. \] --- **d) Evaluate** \(\sum_{n=5}^{12}(6n+1)\) The number of terms from \(n=5\) to \(n=12\) is: \[ 12-5+1=8. \] Write the sum as: \[ \sum_{n=5}^{12}(6n+1)=6\sum_{n=5}^{12}n+1\cdot8. \] To calculate \(\sum_{n=5}^{12}n\), write: \[ \sum_{n=5}^{12} n=\sum_{n=1}^{12}n-\sum_{n=1}^{4}n. \] We have: \[ \sum_{n=1}^{12}n=\frac{12\cdot13}{2}=78,\quad \sum_{n=1}^{4}n=\frac{4\cdot5}{2}=10. \] Thus, \[ \sum_{n=5}^{12}n=78-10=68. \] Then, \[ 6\cdot68+8=408+8=416. \] --- **e) Evaluate** \(\sum_{n=10}^{30}(15-n)\) The number of terms from \(n=10\) to \(n=30\) is: \[ 30-10+1=21. \] Write the sum as: \[ \sum_{n=10}^{30}(15-n)=15\sum_{n=10}^{30}1-\sum_{n=10}^{30} n=15\cdot21-\sum_{n=10}^{30} n. \] Now, \[ \sum_{n=10}^{30} n=\sum_{n=1}^{30}n-\sum_{n=1}^{9}n. \] Calculate: \[ \sum_{n=1}^{30}n=\frac{30\cdot31}{2}=465,\quad \sum_{n=1}^{9}n=\frac{9\cdot10}{2}=45. \] Thus, \[ \sum_{n=10}^{30}n=465-45=420. \] Then, the sum becomes: \[ 15\cdot21-420=315-420=-105. \] --- **f) Evaluate** \(\sum_{n=4}^{20}(2n-6)\) The number of terms from \(n=4\) to \(n=20\) is: \[ 20-4+1=17. \] Write the sum as: \[ \sum_{n=4}^{20}(2n-6)=2\sum_{n=4}^{20}n-6\cdot17. \] Now, \[ \sum_{n=4}^{20}n=\sum_{n=1}^{20}n-\sum_{n=1}^{3}n. \] Calculate: \[ \sum_{n=1}^{20}n=\frac{20\cdot21}{2}=210,\quad \sum_{n=1}^{3}n=\frac{3\cdot4}{2}=6. \] Thus, \[ \sum_{n=4}^{20}n=210-6=204. \] Then, \[ 2\cdot204-6\cdot17=408-102=306. \] --- **g) Evaluate** \(\sum_{n=4}^{14}(8-n)\) The number of terms from \(n=4\) to \(n=14\) is: \[ 14-4+1=11. \] Write the sum as: \[ \sum_{n=4}^{14}(8-n)=8\sum_{n=4}^{14}1-\sum_{n=4}^{14} n=8\cdot11-\sum_{n=4}^{14}n. \] Now, \[ \sum_{n=4}^{14}n=\sum_{n=1}^{14}n-\sum_{n=1}^{3}n. \] Calculate: \[ \sum_{n=1}^{14}n=\frac{14\cdot15}{2}=105,\quad \sum_{n=1}^{3}n=6. \] Thus, \[ \sum_{n=4}^{14}n=105-6=99. \] Then, \[ 8\cdot11-99=88-99=-11. \] --- **n) Evaluate** \(\sum_{n=1}^{15}(3n)\) Write the sum as: \[ \sum_{n=1}^{15}(3n)=3\sum_{n=1}^{15}n. \] We have: \[ \sum_{n=1}^{15}n=\frac{15\cdot16}{2}=120. \] Thus, \[ 3\cdot120=360. \]