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

**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 \( \sum_{n=1}^{20}(3n+5) \).** Separate 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) Evaluate \( \sum_{n=1}^{16}(30-2n) \).** Split the sum: \[ \sum_{n=1}^{16}(30-2n)=30\cdot16-2\sum_{n=1}^{16}n. \] Since: \[ \sum_{n=1}^{16}n=\frac{16\cdot 17}{2}=136, \] we have: \[ 30\cdot 16-2\cdot136=480-272=208. \] --- **d) Evaluate \( \sum_{n=5}^{12}(6n+1) \).** One way is to subtract the sum from \( n=1 \) to 4 from the sum from \( n=1 \) to 12: \[ \sum_{n=5}^{12}(6n+1)=\sum_{n=1}^{12}(6n+1)-\sum_{n=1}^{4}(6n+1). \] Calculate for \( n=1 \) to 12: \[ \sum_{n=1}^{12}(6n+1)=6\left(\frac{12\cdot 13}{2}\right)+12=6\cdot78+12=468+12=480. \] And for \( n=1 \) to 4: \[ \sum_{n=1}^{4}(6n+1)=6\left(\frac{4\cdot 5}{2}\right)+4=6\cdot10+4=60+4=64. \] Subtract: \[ 480-64=416. \] --- **e) Evaluate \( \sum_{n=10}^{30}(15-n) \).** First determine the number of terms: \[ \text{Number of terms} = 30-10+1=21. \] Write the sum: \[ \sum_{n=10}^{30}(15-n)=15\cdot 21-\sum_{n=10}^{30}n. \] Find \(\sum_{n=10}^{30}n\) by writing: \[ \sum_{n=10}^{30}n=\sum_{n=1}^{30}n-\sum_{n=1}^{9}n. \] Now, \[ \sum_{n=1}^{30}n=\frac{30\cdot 31}{2}=465,\quad \sum_{n=1}^{9}n=\frac{9\cdot 10}{2}=45. \] Then, \[ \sum_{n=10}^{30}n=465-45=420. \] Thus, \[ 15\cdot 21-420=315-420=-105. \] --- **f) Evaluate \( \sum_{n=4}^{20}(2n-6) \).** Separate the sum: \[ \sum_{n=4}^{20}(2n-6)=2\sum_{n=4}^{20}n-6\cdot(\text{number of terms}). \] The number of terms is: \[ 20-4+1=17. \] Now, compute: \[ \sum_{n=4}^{20}n=\sum_{n=1}^{20}n-\sum_{n=1}^{3}n. \] With, \[ \sum_{n=1}^{20}n=\frac{20\cdot 21}{2}=210,\quad \sum_{n=1}^{3}n=\frac{3\cdot 4}{2}=6, \] we get: \[ \sum_{n=4}^{20}n=210-6=204. \] Thus, \[ 2\cdot204-6\cdot17=408-102=306. \] --- **g) Evaluate \( \sum_{n=4}^{14}(8-n) \).** Determine the number of terms: \[ 14-4+1=11. \] Separate the sum: \[ \sum_{n=4}^{14}(8-n)=8\cdot 11-\sum_{n=4}^{14}n. \] Compute \(\sum_{n=4}^{14}n\) by: \[ \sum_{n=4}^{14}n=\sum_{n=1}^{14}n-\sum_{n=1}^{3}n. \] Now, \[ \sum_{n=1}^{14}n=\frac{14\cdot 15}{2}=105,\quad \sum_{n=1}^{3}n=\frac{3\cdot 4}{2}=6, \] so, \[ \sum_{n=4}^{14}n=105-6=99. \] Thus, \[ 8\cdot 11-99=88-99=-11. \] --- **h) Evaluate \( \sum_{n=1}^{15}(3n) \).** Factor out the 3: \[ \sum_{n=1}^{15}(3n)=3\sum_{n=1}^{15}n. \] Then, \[ \sum_{n=1}^{15}n=\frac{15\cdot 16}{2}=120. \] So the sum is: \[ 3\cdot120=360. \]