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Describing Sequences

Recursive and closed definitions of a sequence, introduced by counting the tilings of a strip by dominoes and unit squares.

Discrete Mathematics: An Open Introduction, Describing Sequences

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7 exercises

Exercise 3

Consider the sequence (an)n≥1(a_n)_{n \ge 1} that starts 1,3,5,7,9,…1, 3, 5, 7, 9, \ldots (i.e., the odd numbers in order).

  1. Give a recursive definition and closed formula for the sequence.

  2. Write out the sequence (bn)n≥2(b_n)_{n \ge 2} of partial sums of (an)(a_n). Write down the recursive definition for (bn)(b_n) and guess at the closed formula.

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Exercise 5

Consider the three sequences below. For each, find a recursive definition. How are these sequences related?

  1. 2,4,6,10,16,26,42,…2, 4, 6, 10, 16, 26, 42, \ldots.
  2. 5,6,11,17,28,45,73,…5, 6, 11, 17, 28, 45, 73, \ldots.
  3. 0,0,0,0,0,0,0,…0, 0 , 0 , 0 , 0 , 0 , 0 ,\ldots.
2.5

Exercise 6

Write out the first few terms of the sequence given by a1=3a_1 = 3; an=2an−1+4a_n = 2a_{n-1} + 4. Then find a recursive definition for the sequence 10,24,52,108,…10, 24, 52, 108, \ldots.

2.5

Exercise 7

Write out the first few terms of the sequence given by an=n2−3n+1a_n = n^2 - 3n + 1. Then find a closed formula for the sequence (starting with a1a_1) 0,2,6,12,20,…0, 2, 6, 12, 20, \ldots.

2.5

Exercise 8

Show that an=3⋅2n+7⋅5na_n = 3\cdot 2^n + 7\cdot 5^n is a solution to the recurrence relation an=7an−1−10an−2a_n = 7a_{n-1} - 10a_{n-2}. What would the initial conditions need to be for this to be the closed formula for the sequence?

2.5

Exercise 12

Use summation (∑\sum) or product (∏\prod) notation to rewrite the following.

  1. 2+4+6+8+⋯+2n2 + 4 + 6 + 8 + \cdots + 2n.
  2. 1+5+9+13+⋯+4251 + 5 + 9 + 13 + \cdots + 425.
  3. 1+12+13+14+⋯+1501 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{50}.
  4. 2⋅4⋅6⋅⋯⋅2n2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2n.
  5. (12)(23)(34)⋯(100101)(\frac{1}{2})(\frac{2}{3})(\frac{3}{4})\cdots(\frac{100}{101}).
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Exercise 13

Expand the following sums and products. That is, write them out the long way.

  1. ∑k=1100(3+4k)\d\sum_{k=1}^{100} (3+4k).
  2. ∑k=0n2k\d\sum_{k=0}^n 2^k.
  3. ∑k=2501(k2−1)\d\sum_{k=2}^{50}\frac{1}{(k^2 - 1)}.
  4. ∏k=2100k2(k2−1)\d\prod_{k=2}^{100}\frac{k^2}{(k^2-1)}.
  5. ∏k=0n(2+3k)\d\prod_{k=0}^n (2+3k).
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Adapted from Discrete Mathematics: An Open Introduction © 2013–2019 Oscar Levin. Licensed under CC BY-SA 4.0. Source: Discrete Mathematics: An Open Introduction. Imported at revision 1e2e26b.