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Section 3.2

Sequences and Summations

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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

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ExampleFind 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

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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

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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

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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

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Examples1,0,1,1,0,0,1,1,1,0,0,0,1,… 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 twice1,0,2,0,4,0,8,0,16,0,… The even-numbered terms are all 0; the odd-numberedterms are successive powers of 2

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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)

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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, …

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SummationsThis 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

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ExampleExpress 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:

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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 =

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(-2)0 + (-2)1 + (-2)2 + (-2)3 + (-2)4 = 1 + -2 + 4 + -8 + 16 =

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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

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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 canconclude rS = S + (arn+1 - a)so if r 1, S = (arn+1 - a)/(r-1) andif r = 1, S = (n+1)a

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Finding S: ExampleFind 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

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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 summation2. Compute the outer summation given the expansion

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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

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Double Summation Example

(2i+0) + (2i+3) + (2i+6) + (2i+9)

(8i + 18) = 18 + 26 +34 = 78

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Using Summation Notation with Sets & Functions

f(s)sS represents the sum of all values f(s) where sS

s = 1 + 2 + 3 = 6s{1,2,3}

s2 + s = 2 + 6 + 12 = 20s{1,2,3}

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Formulae for commonly-occurring summations

n ark (arn+1 – a) / (r – 1), r 1k=0

n k (n(n + 1)) / 2k=1

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Formulae for commonly-occurring summations

n k2 (n(n + 1)(2n + 1)) / 6k=1

n k3 (n2(n + 1) 2) / 4k=1

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Using the formulae to solve summation problems

Sometimes summation problems don’t start at a convenientindex. For example, find:

k = k - kk=100 k=1 k=1

Applying formula:n k (n(n + 1)) / 2k=1

(200(201))/2 - (99(100))/2 =20100 - 4950 = 15150

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Section 3.2

Sequences and Summations- ends -