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

Question

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

UpStudy AI · Accepted Answer

**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,
$$
3\cdot120 = 360.
$$
**a) 378** **b) 730** **c) 208** **d) 416** **e) -105** **f) 306** **9) -11** **n) 360**

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