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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

  1. Evaluate each sum.

    1. (a) \(\displaystyle\sum_{r=1}^{7}(3r + 2)\)

      • 5 + 8 + 11 + … + 23 (7 terms, arithmetic)
      • \(\frac{7}{2}(5 + 23) = 98\)
    2. (b) \(\displaystyle\sum_{r=4}^{12}(5r - 3)\)

      • 17 + 22 + … + 57 (9 terms, arithmetic)
      • \(\frac{9}{2}(17 + 57) = 333\)
    3. (c) \(\displaystyle\sum_{r=1}^{6} 3 \times 2^r\)

      • 6 + 12 + 24 + … + 192 (6 terms, geometric)
      • \(\frac{6(2^6 - 1)}{2 - 1} = 378\)
    4. (d) \(\displaystyle\sum_{k=0}^{5} 4 \times 3^k\)

      • 4 + 12 + 36 + … + 972 (6 terms, geometric)
      • \(\frac{4(3^6 - 1)}{3 - 1} = 1456\)
  2. How many terms are there in \(\displaystyle\sum_{r=7}^{52} u_r\)?

    • The terms are \(u_7, u_8, \dots, u_{52}\).
    • \(52 - 7 + 1 = 46\)
  3. Write each series in sigma notation.

    1. (a) 4 + 7 + 10 + … + 97

      • \(d = 3,\ u_r = 4 + 3(r-1) = 3r + 1\)
      • \(3r + 1 = 97,\ r = 32\)
      • \(\displaystyle\sum_{r=1}^{32}(3r + 1)\)
    2. (b) 5 + 15 + 45 + … + 3645

      • \(r = 3,\ u_r = 5 \times 3^{r-1}\)
      • \(5 \times 3^{r-1} = 3645,\ 3^{r-1} = 729,\ r = 7\)
      • \(\displaystyle\sum_{r=1}^{7} 5 \times 3^{r-1}\)
  4. 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.
    • \(S_{26} = 13(116 + 16)\)
    • \(S_{26} = 1716\)
  5. Evaluate each sum.

    1. (a) \(\displaystyle\sum_{r=1}^{25} 8\)

      • 25 terms, each equal to 8
      • \(25 \times 8 = 200\)
    2. (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

  1. Given that \(\displaystyle\sum_{r=1}^{n}(2r + 5) = 315\), find \(n\).

    • \(S_n = \frac{n}{2}(7 + (2n + 5)) = n^2 + 6n\)
    • \(n^2 + 6n - 315 = 0\)
    • \((n - 15)(n + 21) = 0,\ n = 15\)
  2. \(\text{Find the smallest } n \text{ for which }\displaystyle\sum_{r=1}^{n} 4\times 3^{r-1} > 5000\).

    • \(u_1 = 4,\ r = 3\)
    • \(S_n = \dfrac{4(3^n - 1)}{3 - 1}\)
    • \(S_{7} = 4372\) and \(S_{8} = 13120\)
    • The smallest value is n = 8.

jmaths.xyz · AI HL · 12.1.4 Series & Sigma Notation