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IB Mathematics: Applications and Interpretation HL · 12.1: Sequences & Series · 12.1.4
Sigma notation Problem set B One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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Evaluate each sum.
(a) \(\displaystyle\sum_{r=1}^{7}(3r + 2)\)
- 5 + 8 + 11 + … + 23 (7 terms, arithmetic)
- \(\frac{7}{2}(5 + 23) = 98\)
(b) \(\displaystyle\sum_{r=4}^{12}(5r - 3)\)
- 17 + 22 + … + 57 (9 terms, arithmetic)
- \(\frac{9}{2}(17 + 57) = 333\)
(c) \(\displaystyle\sum_{r=1}^{6} 3 \times 2^r\)
- 6 + 12 + 24 + … + 192 (6 terms, geometric)
- \(\frac{6(2^6 - 1)}{2 - 1} = 378\)
(d) \(\displaystyle\sum_{k=0}^{5} 4 \times 3^k\)
- 4 + 12 + 36 + … + 972 (6 terms, geometric)
- \(\frac{4(3^6 - 1)}{3 - 1} = 1456\)
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How many terms are there in \(\displaystyle\sum_{r=7}^{52} u_r\)?
- The terms are \(u_7, u_8, \dots, u_{52}\).
- \(52 - 7 + 1 = 46\)
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Write each series in sigma notation.
(a) 4 + 7 + 10 + … + 97
- \(d = 3,\ u_r = 4 + 3(r-1) = 3r + 1\)
- \(3r + 1 = 97,\ r = 32\)
- \(\displaystyle\sum_{r=1}^{32}(3r + 1)\)
(b) 5 + 15 + 45 + … + 3645
- \(r = 3,\ u_r = 5 \times 3^{r-1}\)
- \(5 \times 3^{r-1} = 3645,\ 3^{r-1} = 729,\ r = 7\)
- \(\displaystyle\sum_{r=1}^{7} 5 \times 3^{r-1}\)
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Say whether \(\displaystyle\sum_{r=1}^{26}(120 - 4r)\) is arithmetic or geometric, and evaluate it.
- The terms are 116, 112, …, 16: the difference is always \(-4\), so it is arithmetic.
- \(S_{26} = 13(116 + 16)\)
- \(S_{26} = 1716\)
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Evaluate each sum.
(a) \(\displaystyle\sum_{r=1}^{25} 8\)
- 25 terms, each equal to 8
- \(25 \times 8 = 200\)
(b) \(\displaystyle\sum_{r=0}^{4} 3\times(-2)^r\)
- \(3,\ -6,\ 12,\ -24,\ \ldots\)
- \(\text{Geometric: } u_1 = 3,\ r = -2,\ 5\text{ terms}\)
- \(S_{5} = \dfrac{3((-2)^{5} - 1)}{-2 - 1} = 33\)
Stretch
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Given that \(\displaystyle\sum_{r=1}^{n}(2r + 5) = 315\), find \(n\).
- \(S_n = \frac{n}{2}(7 + (2n + 5)) = n^2 + 6n\)
- \(n^2 + 6n - 315 = 0\)
- \((n - 15)(n + 21) = 0,\ n = 15\)
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\(\text{Find the smallest } n \text{ for which }\displaystyle\sum_{r=1}^{n} 4\times 3^{r-1} > 5000\).
- \(u_1 = 4,\ r = 3\)
- \(S_n = \dfrac{4(3^n - 1)}{3 - 1}\)
- \(S_{7} = 4372\) and \(S_{8} = 13120\)
- The smallest value is n = 8.
jmaths.xyz · AI HL · 12.1.4 Series & Sigma Notation