IntroductionA sequence is an ordered list of numbers. The numbers, or terms, in the ordered list are determined by a formula that is a function of the position of the term in the list. So if we have a sequence A determined by a function f, the terms of the sequence will be:

A = a1, a2, a3, …, an, where a1 = f(1), a2 = f(2), a3 = f(3), …, an = f(n)

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3.2.1: Sequences As Functions

Introduction, continuedUnlike a typical function on a variable, there are no fractional terms between the first and second term, between the second and third term, etc. Each term is a whole number. Just like no one can place 1.23rd in a race, there is no 1.23rd term in a sequence. Therefore, the domain of f  is at most {1, 2, 3, …, n}. This sub-set of the integers is called the natural number system. Natural numbers are the numbers we use for counting. Because every element of the domain of a sequence is individually separate and distinct, we say a sequence is a discrete function.

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3.2.1: Sequences As Functions

Key Concepts• Sequences are ordered lists determined by functions.

• The domain of the function that generates a sequence is all natural numbers.

• A sequence is itself a function.

• There are two ways sequences are generally defined—recursively and explicitly.

• An explicit formula is a formula used to find the nth term of a sequence. If a sequence is defined explicitly (that is, with an explicit formula), the function is given.

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3.2.1: Sequences As Functions

Key Concepts, continued• A recursive formula is a formula used to find the next

term of a sequence when the previous term or terms are known. If the sequence is defined recursively (that is, with a recursive formula), the next term is based on the term before it and the commonality between terms.

• For this lesson, sequences have a common difference or a common ratio.

• To determine the common difference, subtract the second term from the first term. Then subtract the third term from the second term and so on.

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3.2.1: Sequences As Functions

Key Concepts, continued• If a common difference exists, an – an – 1 = constant.

So, a4 – a3 = a3 – a2 = a2 – a1.

• If the difference is not constant, then the commonality might be a ratio between terms. To find a common ratio, divide the second term by the first term. Then, divide the third term by the second term, and so on.

• If a common ratio exists, then .

So, .

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3.2.1: Sequences As Functions

Key Concepts, continuedExplicit Sequences• Explicitly defined sequences provide the function that

will generate each term. For example:

an = 2n + 3 or bn = 5(3)n .

• In each case, we simply plug in the n representing the nth term and we get an or bn.

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3.2.1: Sequences As Functions

Key Concepts, continuedRecursive Sequences• The second way a sequence may be defined is

recursively. In a recursive sequence, each term is a function of the term, or terms, that came before it. For example: an = an – 1 + 2 or b n = bn – 1 • 3, where n is the number of the term.

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3.2.1: Sequences As Functions

Key Concepts, continuedGraphing Sequences• Sequences can be graphed with a domain of natural

numbers. Compare the graphs on the following slide. The first graph is of a sequence, an = n – 1, while the second graph is of the line f(x) = x – 1.

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3.2.1: Sequences As Functions

Key Concepts, continued

• Notice the sequence only has values where n = 1, 2, 3, 4, etc. Also notice the labels on the axes of each graph. The sequence is in terms of n and an, while the line is in terms of x and y. 9

3.2.1: Sequences As Functions

Sequence graph Line graph

Common Errors/Misconceptions• not realizing that a sequence is not a function

• not understanding that a sequence is graphed without a line connecting the points

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3.2.1: Sequences As Functions

Guided Practice

Example 2Find the missing terms in the sequence using recursion.

A = {8, 13, 18, 23, a5, a6, a7}

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3.2.1: Sequences As Functions

Guided Practice: Example 2, continued

1. First look for the pattern. Is there a common difference or common ratio?

a2 – a1 = 5

a3 – a2 = 5

a4 – a3 = 5

The terms are separated by a common difference of 5.

From this, we can deduce an = an – 1 + 5.

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3.2.1: Sequences As Functions

Guided Practice: Example 2, continued

2. Use the formula to find the missing terms.an = an – 1 + 5

a5 = a4 + 5 = 23 + 5 = 28

a6 = a5 + 5 = 28 + 5 = 33

a7 = a6 + 5 = 33 + 5 = 38

The missing terms are 28, 33, and 38.

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3.2.1: Sequences As Functions

✔

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3.2.1: Sequences As Functions

Guided Practice: Example 2, continued

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Guided Practice

Example 5Find the seventh term in the sequence given by an = 3 • 2n – 1. Then, graph the first 5 terms in the sequence.

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3.2.1: Sequences As Functions

Guided Practice: Example 5, continued

1. Substitute 7 for n.an = 3 • 2n – 1

a7 = 3 • 2(7) – 1 = 3 • 26 = 3 • 64 = 192

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3.2.1: Sequences As Functions

Guided Practice: Example 5, continued

2. Generate the first 5 terms of the sequence.

an = 3 • 2n – 1

a1 = 3 • 2(1) – 1 = 3 • 20 = 3 • 1 = 3

a2 = 3 • 2(2) – 1 = 3 • 21 = 3 • 2 = 6

a3 = 3 • 2(3) – 1 = 3 • 22 = 3 • 4 = 12

a4 = 3 • 2(4) – 1 = 3 • 23 = 3 • 8 = 24

a5 = 3 • 2(5) – 1 = 3 • 24 = 3 • 16 = 48

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3.2.1: Sequences As Functions

Guided Practice: Example 5, continued

3. Create the ordered pairs from the sequence.n corresponds to x, and an corresponds to y.

(n, an )

(1, 3)

(2, 6)

(3, 12)

(4, 24)

(5, 48)

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3.2.1: Sequences As Functions

Guided Practice: Example 4, continued

4. Plot the ordered pairs. Do not connect the points.

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3.2.1: Sequences As Functions

✔

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3.2.1: Sequences As Functions

Guided Practice: Example 5, continued