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Section 3.2

Section 3.2. Sequences and Summations. Sequence. Function from a subset of Z (usually the set beginning with 1 or 0) to a set S a n denotes the image of n (n  Z) a n is called a term of the sequence The notation {a n } is used to describe the sequence. Example.

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Section 3.2

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  1. Section 3.2 Sequences and Summations

  2. Sequence • Function from a subset of Z (usually the set beginning with 1 or 0) to a set S • an denotes the image of n (n  Z) • an is called a term of the sequence • The notation {an} is used to describe the sequence

  3. Example Find the first 5 terms of sequence {an} where {an} = 2 * (-3)n + 5n a0 = 2 * (-3)0 + 50 = 2 * 1 + 1 = 3 a1 = 2 * (-3)1 + 51 = 2 * (-3) + 5 = -1 a2 = 2 * (-3)2 + 52 = 2 * 9 + 25 = 43 a3 = 2 * (-3)3 + 53 = 2 * (-27) + 125 = 71 a4 = 2 * (-3)4 + 54 = 2 * 81 + 625 = 786

  4. Strings • Strings are finite sequences of the form: a1, a2, a3, … , an • The number of terms in a string is the length of the string • The empty string has 0 terms

  5. Special Integer Sequences • Find the formula or general rule for constructing the terms of a sequence, given a few initial terms • Look for a pattern in the terms you’re given • Determine how a term can be produced from a preceding term

  6. Useful clues for special integer sequences • Runs of a value • Terms obtained by adding to previous term: • the same amount • an amount that depends on the term’s position in the sequence • Terms obtained by multiplying the previous term by some amount • Terms obtained by combining previous terms

  7. Examples 1,0,1,1,0,0,1,1,1,0,0,0,1,… Both 1 and 0 appear exactly n times, alternating 1,2,2,3,4,4,5,6,6,7,8,8,… The positive integers appear in increasing order, with the odd numbers appearing once and the even numbers appearing twice 1,0,2,0,4,0,8,0,16,0,… The even-numbered terms are all 0; the odd-numbered terms are successive powers of 2

  8. Arithmetic Progression • An arithmetic progression is a sequence of the form a, a+d, a+2d, a+3d, … , a+nd • For example, the sequence: 1, 7, 13, 19, 25, 31, 37, 43, 49, 55 … is an arithmetic progression with a = 1 and d = 6 • The next term in this arithmetic progression will be a + d(n-1)

  9. Well-known Sequences • n2 = 1, 4, 9, 16, 25, … • n3 = 1, 8, 27, 64, 125, … • n4 = 1, 16, 81, 256, 625, … • 2n = 2, 4, 8, 16, 32, … • 3n = 3, 9, 27, 81, 243, … • n! = 1, 2, 6, 24, 120, …

  10. Summations This symbol represents the following sum: am + am+1 + … + an where: j: subscript of term (index of summation) m: 1st subscript value (lower limit) n: last subscript value (upper limit) Note that the choice of these letters (a, j, m and n) is arbitrary

  11. Example Express the sum of the first 100 terms of sequence {an} where an = n2 + 1 for n = 1, 2, 3, … Recall the summation denotation: So j (the index of summation) goes from m=1 to n=100, and we want the sum of all (j2 + 1) between m and n Thus the expression is:

  12. Example Find the value of each of the following summations: (1+1) + (2+1) + (3+1) + (4+1) + (5+1) = 2 + 3 + 4 + 5 + 6 = 20 (-2)0 + (-2)1 + (-2)2 + (-2)3 + (-2)4 = 1 + -2 + 4 + -8 + 16 = 11

  13. Geometric Progression • A geometric progression is a sequence of the form ar0, ar1, ar2, ar3, ar4, … , ark where: • a = initial term • r = common ratio • both a & r are real numbers • The summation of the terms of a geometric progression is called a geometric series

  14. S: sum of the first n+1 terms of a geometric series To find S, we could do the problem longhand, but we can derive a formula that provides a significant shortcut by following the steps below: 1. Multiply both sides by r: 2. Shift the index of summation. Suppose k = j + 1; then: 3. Shift back to 0: Since was the original S, we can conclude rS = S + (arn+1 - a) so if r  1, S = (arn+1 - a)/(r-1) and if r = 1, S = (n+1)a

  15. Finding S: Example Find the value of: In this expression, a = 3, r = 2 and n = 8 Applying the formula: S = (arn+1 - a)/(r-1) S = (3 * 29 - 3) / (2 - 1) = 3 * 512 - 3 = 1533 Can confirm this by doing problem longhand: 3*20 + 3*21 + 3*22 + 3*23 + 3*24 + 3*25 + 3*26 + 3*27 +3*28 = 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = 1533

  16. Double Summations A double summation has the form of one sigma after another, followed by the formula involving the two indexes of summation To solve a double summation: 1. Expand the inner summation 2. Compute the outer summation given the expansion

  17. Double Summation Example 1. Expand the inner summation (i*1 + i*2 + i*3) 6i 2. Compute the outer summation given the expansion = (6*1 + 6*2 + 6*3 + 6*4 + 6*5 = 90

  18. Double Summation Example (2i+0) + (2i+3) + (2i+6) + (2i+9) (8i + 18) = 18 + 26 +34 = 78

  19. Using Summation Notation with Sets & Functions S f(s) sS represents the sum of all values f(s) where sS S s = 1 + 2 + 3 = 6 s{1,2,3} S s2 + s = 2 + 6 + 12 = 20 s{1,2,3}

  20. Formulae for commonly-occurring summations n S ark (arn+1 – a) / (r – 1), r  1 k=0 n S k (n(n + 1)) / 2 k=1

  21. Formulae for commonly-occurring summations n S k2 (n(n + 1)(2n + 1)) / 6 k=1 n S k3 (n2(n + 1) 2) / 4 k=1

  22. Using the formulae to solve summation problems Sometimes summation problems don’t start at a convenient index. For example, find: 200 200 99 S k = S k - S k k=100 k=1 k=1 Applying formula: n S k (n(n + 1)) / 2 k=1 (200(201))/2 - (99(100))/2 = 20100 - 4950 = 15150

  23. Section 3.2 Sequences and Summations - ends -

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