1. 1 Section 3.2 Sequences and Summations

2. 2 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

3. 3 Example Find the first 5 terms of sequence {a n } where {a n } = 2 * (-3) n + 5 n a 0 = 2 * (-3) 0 + 5 0 = 2 * 1 + 1 = 3 a 1 = 2 * (-3) 1 + 5 1 = 2 * (-3) + 5 = -1 a 2 = 2 * (-3) 2 + 5 2 = 2 * 9 + 25 = 43 a 3 = 2 * (-3) 3 + 5 3 = 2 * (-27) + 125 = 71 a 4 = 2 * (-3) 4 + 5 4 = 2 * 81 + 625 = 786

4. 4 Strings Strings are finite sequences of the form: a 1, a 2, a 3, …, a n The number of terms in a string is the length of the string The empty string has 0 terms

5. 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. 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. 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. 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. 9 Well-known Sequences n 2 = 1, 4, 9, 16, 25, … n 3 = 1, 8, 27, 64, 125, … n 4 = 1, 16, 81, 256, 625, … 2 n = 2, 4, 8, 16, 32, … 3 n = 3, 9, 27, 81, 243, … n! = 1, 2, 6, 24, 120, …

10. 10 Summations This symbol represents the following sum: a m + a m+1 + … + a n 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. 11 Example Express the sum of the first 100 terms of sequence {a n } where a n = n 2 + 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 (j 2 + 1) between m and n Thus the expression is:

12. 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. 13 Geometric Progression A geometric progression is a sequence of the form ar 0, ar 1, ar 2, ar 3, ar 4, …, ar k 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. 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:Sincewas the original S, we can conclude rS = S + (ar n+1 - a) so if r  1, S = (ar n+1 - a)/(r-1) and if r = 1, S = (n+1)a

15. 15 Finding S: Example Find the value of: In this expression, a = 3, r = 2 and n = 8 Applying the formula: S = (ar n+1 - a)/(r-1) S = (3 * 2 9 - 3) / (2 - 1) = 3 * 512 - 3 = 1533 Can confirm this by doing problem longhand: 3*2 0 + 3*2 1 + 3*2 2 + 3*2 3 + 3*2 4 + 3*2 5 + 3*2 6 + 3*2 7 +3*2 8 = 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = 1533

16. 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. 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. 18 Double Summation Example (2i+0) + (2i+3) + (2i+6) + (2i+9) (8i + 18) = 18 + 26 +34 = 78

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

20. 20 Formulae for commonly- occurring summations  n  ar k (ar n+1 – a) / (r – 1), r  1 k=0 n  k(n(n + 1)) / 2 k=1 

21. 21 Formulae for commonly- occurring summations n  k 2 (n(n + 1)(2n + 1)) / 6 k=1  n  k 3 (n 2 (n + 1) 2 ) / 4 k=1

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

23. 23 Section 3.2 Sequences and Summations - ends -