Sigma notation Problem set A Worked solutions

Evaluate each sum.

1. (a) \(\displaystyle\sum_{r=1}^{6}(4r - 1)\)

   • \(3, 7, 11, 15, 19, 23\ldots\)
   • \(\text{Arithmetic: } u_1 = 3,\ d = 4,\ n = 6\)
   • \(S_6 = \frac{6}{2}(3 + 23) = 78\)

2. (b) \(\displaystyle\sum_{r=3}^{10}(2r + 7)\)

   • \(13, 15, 17, \ldots, 27\)
   • \(\text{Arithmetic: 8 terms, } u_1 = 13,\ d = 2\)
   • \(S_8 = \frac{8}{2}(13 + 27) = 160\)

3. (c) \(\displaystyle\sum_{r=1}^{5} 2 \times 3^r\)

   • \(6, 18, 54, \ldots\)
   • \(\text{Geometric: } u_1 = 6,\ r = 3,\ n = 5\)
   • \(S_5 = \dfrac{6(3^5 - 1)}{3 - 1} = 726\)

4. (d) \(\displaystyle\sum_{k=0}^{4} 5 \times 2^k\)

   • \(5, 10, 20, \ldots\)
   • \(\text{Geometric: } u_1 = 5,\ r = 2,\ n = 5\)
   • \(S_5 = \dfrac{5(2^5 - 1)}{2 - 1} = 155\)

How many terms are there in \(\displaystyle\sum_{r=5}^{40} u_r\)?

• \(\text{Number of terms} = \text{last} - \text{first} + 1\)
• \(40 - 5 + 1 = 36\)

Write each series in sigma notation.

1. (a) 6 + 11 + 16 + … + 101

   • \(d = 5,\ u_r = 6 + (r - 1)(5) = 5r + 1\)
   • \(5r + 1 = 101 \Rightarrow r = 20\)
   • \(\displaystyle\sum_{r=1}^{20}(5r + 1)\)

2. (b) 3 + 6 + 12 + … + 768

   • \(r = 2,\ u_r = 3 \times 2^{r-1}\)
   • \(3 \times 2^{r-1} = 768 \Rightarrow 2^{r-1} = 256 = 2^8 \Rightarrow r = 9\)
   • \(\displaystyle\sum_{r=1}^{9} 3 \times 2^{r-1}\)

Say whether \(\displaystyle\sum_{r=1}^{30}(100 - 3r)\) is arithmetic or geometric, and evaluate it.

• \(u_r = 100 - 3r:\ u_1 = 97,\ u_2 = 94,\ u_3 = 91\)
• \(\text{Difference } -3 \text{ each time: arithmetic}\)
• \(u_{30} = 100 - 90 = 10\)
• \(S_{30} = \frac{30}{2}(97 + 10) = 1605\)

Evaluate each sum.

1. (a) \(\displaystyle\sum_{r=1}^{20} 6\)

   • 20 terms, each equal to 6
   • \(20 \times 6 = 120\)

2. (b) \(\displaystyle\sum_{r=0}^{3} 5\times(-2)^r\)

   • \(5,\ -10,\ 20,\ -40,\ \ldots\)
   • \(\text{Geometric: } u_1 = 5,\ r = -2,\ 4\text{ terms}\)
   • \(S_{4} = \dfrac{5((-2)^{4} - 1)}{-2 - 1} = -25\)

Given that \(\displaystyle\sum_{r=1}^{n}(2r + 3) = 320\), find \(n\).

• \(u_1 = 5,\ d = 2:\ \sum = \frac{n}{2}(10 + 2(n - 1)) = n^2 + 4n\)
• \(n^2 + 4n = 320 \Rightarrow n^2 + 4n - 320 = 0\)
• \((n - 16)(n + 20) = 0 \Rightarrow n = 16\ (n > 0)\)

\(\text{Find the smallest } n \text{ for which }\displaystyle\sum_{r=1}^{n} 5\times 2^{r-1} > 3000\).

• \(u_1 = 5,\ r = 2\)
• \(S_n = \dfrac{5(2^n - 1)}{2 - 1}\)
• \(S_{9} = 2555\) and \(S_{10} = 5115\)
• The smallest value is n = 10.