AQR Unit 4: Using Recursion in Models and Decision Making...

AQR Unit 4: Using Recursion in Models and Decision Making Sequence Notes Name: __________________________________________ Date: _____________________________

Sequences

Sequences (Sucesiones)

A number sequence is a set of numbers, usually separated by commas, arranged in an

order. The first term is referred to as t 1 , the second term as t 2 , the third term as

t 3 and so on. The nth term is referred to as t n .

A sequence may stop at some number, or continue indefinitely. The following are examples of

sequences:

a) 5, 7, 9, 11 t 1=5 ; t 2=7 ; t 3=9 ; t 4=11b) 80, 40, 20, 10, … The three dots, called an ellipsis, indicate that the sequence continues indefinitely.

Exercise

1.- Describe the pattern and write the next three terms.

a) 4, 7, 10, 13, … pattern: _________________________________________________

b) 3, 6, 12, 24, … pattern: _________________________________________________

c) 1, 2, 4, 7, 11, 16, … pattern: _________________________________________________

d) 4, 12, 6, 18, 9, … pattern: _________________________________________________

e) 2, 3, 9, 10, 30, … pattern: _________________________________________________

f) 3, 4, 7, 11, 18, 29, … pattern: _________________________________________________

g) 9, 27 45, 63, … pattern: _________________________________________________

h) 1, 4, 9, 16, 25, … pattern: _________________________________________________

k) 2, 5, 10, 17, 26, … pattern: _________________________________________________

l) 11, 8, 3, 5, -2, 7, … pattern: _________________________________________________

2.- Imagine a pattern to create your own sequence. Write down its first five terms.

Finding the formula for the n th term

Sometimes a pattern can lead to a general rule for finding the terms of a sequence. The rule is

called the formula for the nth term, t n . Given the formula for the nth term, t n , the terms of

a sequence can be written by substituting term numbers for n.

Practise1.- Given the formula for the nth term, state the first five terms of the sequence.

a) an=3n a1=3⋅1=3 ; a2=3⋅2=6 ; a3=¿ ; a4=¿ ; a5=¿ ; … ; a18=¿

b) bn=2 n4

c) cn=5&2 n

d) d n=2n

e) en=n&3n4

1

AQR Unit 4: Using Recursion in Models and Decision Making Sequence Notes

SEQUENCES

ARITHMETIC SEQUENCES

An ordered list of numbers such as: 4, 9, 16, 25, 36… is a sequence. Each number in the

sequence is a term. Usually variables with subscripts are used to label terms. For example,

in the sequence above, the first term is 4 and the third term is 16. This might be written

a1= 4 and a

3= 16 where a is the variable used to label the sequence.

In the sequence 1, 5, 9, 13, …, there is a common difference (d = 4 ) between the successive

terms and this is called an arithmetic sequence. There are two common methods to define a

sequence. An explicit formula tells you exactly how to find any specific term in the sequence.

A recursive formula tells first term and how to get from one term to the next. Formally, for

arithmetic sequences, this is written:

Explicit: an = a1 + (n !1)d where n = term number and d = common difference.

Recursive: a1 = some specific value, an+1 = an + d , and d = common difference.

For the sequence 1, 5, 9, 13, …, the explicit formula is: an= 1+ (n !1)(4) = 4n ! 3 and the

recursive formula is: a1 = 1,!!an+1 = an + 4 . In each case, successively replacing n by

1, 2, 3, … will yield the terms of the sequence. See the examples below.

Examples

List the first five terms of each arithmetic sequence.

Example 1 (An explicit formula)

an= 5n + 2

a1 = 5(1) + 2 = 7

a2 = 5(2) + 2 = 12

a3 = 5(3) + 2 = 17

a4 = 5(4) + 2 = 22

a5 = 5(5) + 2 = 27

The sequence is: 7, 12, 17, 22, 27, …

Example 2 (A recursive formula)

b1 = 3,!!bn+1 = bn ! 5

b1 = 3

b2 = b1 ! 5 = 3! 5 = !2

b3 = b2 ! 5 = !2 ! 5 = !7

b4 = b3 ! 5 = !7 ! 5 = !12

b5 = b4 ! 5 = !12 ! 5 = !17

The sequence is: 3, –2, –7, –12, –17, …

Example 3 Find an explicit and a recursive formula for the sequence: –2, 1, 4, 7, …

Explicit: a1 = !2,!d = 3 so the answer is: an = a1 + (n !1)d = !2 + (n !1)(3) = 3n ! 5

Recursive: a1 = !2,!d = 3 so the answer is: a1 = !2,!!an+1 = an + 3

AQR Unit 4: Using Recursion in Models and Decision Making Sequence Notes Problems: Please complete on another piece of paper.

QUADRATIC SEQUENCES

Problems


1. an= 5n ! 2 2. bn = !3n + 5

3. an= !15 +

1

2n 4. c

n= 5 + 3(n !1)

5. a1 = 5,!!an+1 = an + 3 6. a1 = 5,!!an+1 = an ! 3

7. a1 = !3,!!an+1 = an + 6 8. a1 =

1

3,!!a

n+1 = an +1

2

Find the 30th term of each arithmetic sequence.

9. an= 5n ! 2 10. a

n= !15 +

1

2n

11. a31 = 53,!!d = 5 12. a1 = 25,!!an+1 = an ! 3

For each arithmetic sequence, find an explicit and a recursive formula.

13. 4, 8, 12, 16, 20, … 14. –2, 5, 12, 19, 26, …

15. 27, 15, 3, –9, –21, … 16. 3,!3 13,!3 2

3,!4,!4 1

3, ...

Sequences are graphed using points of the form: (term number, term value).

For example, the sequence 4, 9, 16, 25, 36, … would be graphed by plotting the points (1, 4),

(2, 9), (3, 16), (4, 25), (5, 36), … Sequences are graphed as points and not connected.

17. Graph the sequences from problems (1) and (2) above and determine the slope of each line.

18. How does the slope of the line found in the previous problem relate to the sequence?

MEP Y9 Practice Book B

40

(b) Copy and complete this table by writing expressions:

(c) Write an expression to show the total number of tiles in patternnumber n. Simplify your expression.

(d) A different series of patterns is made with tiles.

patternnumber1 pattern

number2 pattern

number3

The series of patterns continues by adding each time.

For this series of patterns, write an expression to show the totalnumber of tiles in pattern number n.

Show your working and simplify your expression.

(KS3/98/Ma/Tier 5-7/P1)

10.3 Second Differences and Quadratic SequencesIn section 10.2 we dealt with sequences where the differences between the termswas a constant value. In this section we extend this idea to sequences where thedifferences are not constant.

Example 1(a) Calculate the first 6 terms of the sequence defined by the quadratic formula,

u n nn = + −2 1

(b) Calculate the first differences between the terms.

(c) Comment on the results you obtain.

10.2

pattern number expression for expression forthe number of the number of

grey tiles white tiles

n


MEP Y9 Practice Book B

41

Solution(a) Substituting n = 1 gives,

u 1 = 1 1 12 + −

= 1

For n = 2 ,u 2 = 2 2 12 + −

= 5

For n = 3,u 3 = 3 3 12 + −

= 11

For n = 4 ,u 4 = 4 4 12 + −

= 19

For n = 5,u 5 = 5 5 12 + −

= 29

For n = 6,u 6 = 6 6 12 + −

= 41

So the first 6 terms are,

1, 5, 11, 19, 29, 41

(b) The differences can now be calculated,

1 5 11 19 29 41

4 6 8 10 12

(c) Note that the differences between the first differences are constant. They areall equal to 2. These are called the second differences, as shown below.

Sequence 1 5 11 19 29 41

First differences 4 6 8 10 12

Second differences 2 2 2 2MEP Y9 Practice Book B

42

Example 2(a) Calculate the first 5 terms of the sequence defined by the quadratic formula

u n nn = − −3 22

(b) Determine the first and second differences for this sequence.

(c) Comment on your results.

Solution

(a) For n = 1, u 1 = 3 1 1 22× − −= 3 1 2− −= 0

For n = 2 , u 2 = 3 2 2 22× − −= 8

For n = 3, u 3 = 3 3 3 22× − −= 22

For n = 4 , u 4 = 3 4 4 22× − −= 42

For n = 5, u 5 = 3 5 5 22× − −= 68

So the sequence is,

0, 8, 22, 42, 68, ...

(b) The differences are calculated below:

Sequence 0 8 22 42 68 ...

First differences 8 14 20 26

Second differences 6 6 6

(c) Again, the second differences are constant; this time they are all 6.

NoteFor a sequence defined by a quadratic formula, the second differences will beconstant and equal to twice the number of n 2 .For example,

u n = n n2 1+ − Second difference = 2u n = 3 22n n− − Second difference = 6u n = 5 72n n− + Second difference = 10

10.3MEP Y9 Practice Book B

43

Example 3Determine a formula for the general term of the sequence,

2, 9, 20, 35, 54, ...

SolutionConsider the first and second differences of the sequence:

2 9 20 35 54 ...

7 11 15 19

4 4 4

As the second differences are constant and equal to 4, the formula will begin

u nn = +2 2 ...

To determine the rest of the formula, subtract 2 2n from each term of thesequence, as shown below:

Sequence 2 9 20 35 54 ...

2 2n 2 8 18 32 50

New sequence 0 1 2 3 4

1 1 1 1

The new sequence has a constant difference of 1 and begins with 0, so for thissequence the formula is n − 1.Combining this with the 2 2n gives

u n nn = + −2 12

Example 4(a) Calculate the first and second differences for the sequence,

4, 1, 0, 1, 4, 9, ...(b) Use the differences to determine the next 2 terms of the sequence.

(c) Determine a formula for the general term of the sequence.

Solution(a) 4 1 0 1 4 9 ...

– 3 – 1 1 3 5

2 2 2 2


Lesson 2.2A: Worksheet Honors Geometry Name: Quadratic Sequences Mr. Ferwerda

Find the nth term and the 20th term in each sequence: 1.

Term 1 2 3 4 5 6 … n … 20

Value 1 4 9 16 25 36 … …

2.

Term 1 2 3 4 5 6 … n … 20

Value 6 12 20 30 42 56 … …

3.

Term 1 2 3 4 5 6 … n … 20

Value 0 4 12 24 40 60 … …

4.

Term 1 2 3 4 5 6 … n … 20

Value 4 15 32 55 84 119 … …

5.

Term 1 2 3 4 5 6 … n … 20

Value -2 3 8 13 18 23 … …

6.

Term 1 2 3 4 5 6 … n … 20

Value -6 7 24 45 70 99 … …


GEOMETRIC SEQUENCES

In the sequence 2, 6, 18, 54, …, there is a common ratio ( r = 3 ) between the successive terms

and this is called an geometric sequence. As before, there are two common methods to

define a geometric sequence. The explicit formula tells you exactly how to find any specific

term in the sequence. The recursive formula gives first term and how to get from one term to

the next. Formally, for geometric sequences, this is written:

Explicit: an= a1 ! r

n"1 where n = term number and r = common ratio

Recursive: a1 = some specific value and an+1 = an ! r where r = common ratio

For the sequence 2, 6, 18, 54, …, the explicit formula is: an= a1 ! r

n"1= 2 ! 3

n"1 , and the

recursive formula is: a1 = 2,!!an+1 = an ! 3 . In each case, successively replacing n by

1, 2, 3, … will yield the terms of the sequence. See the examples below.

Examples

List the first five terms of each geometric sequence.

Example 1 (An explicit formula)

an= 3 !2

n"1

a1 = 3 !21"1

= 3 !20= 3

a2 = 3 !22"1

= 3 !21= 6

a3 = 3 !23"1

= 3 !22= 12

a4 = 3 !24"1

= 3 !23= 24

a5 = 3 !25"1

= 3 !24 = 48

The sequence is: 3, 6, 12, 24, 48, …

Example 2 (A recursive formula)

b1 = 8,!!bn+1 = bn !1

2

b1 = 8

b2 = b1 !1

2= 8 !

1

2= 4

b3 = b2 !1

2= 4 !

1

2= 2

b4 = b3 !1

2= 2 !

1

2= 1

b5 = b4 !1

2= 1 !

1

2=1

2

The sequence is: 8, 4, 2, 1, 12

, …

Example 3 Find an explicit and a recursive formula for the sequence: 81, 27, 9, 3, …

Explicit: a1 = 81,!r =1

3 so the answer is: an = a1 ! r

n"1 = 81 ! 1

3( )n"1

Recursive: a1 = 81,!r =1

3 so the answer is: a1 = 81,!!an+1 = an !

1

3


Problems


19. an= 5 !2n"1 20. bn = !3 " 3

n!1

21. an= 40 1

2( )n!1

22. cn= 6 !

1

2( )n!1

23. a1 = 5,!!an+1 = an ! 3 24. a1 = 100,!!an+1 = an !1

2

25. a1 = !3,!!an+1 = an " !2( ) 26. a1 =

1

3,!!a

n+1 = an !1

2

Find the 15th term of each geometric sequence.

27. b14 = 232,!r = 2 28. b16 = 32,!r = 2

29. a14 = 9,!r =2

3 30. a16 = 9,!r =

23

Find an explicit and a recursive formula for each geometric sequence.

31. 2, 10, 50, 250, 1250, … 32. 16, 4, 1, 14,! 116, ...

33. 5, 15, 45, 135, 405, … 34. 3, –6, 12, –24, 48, …

35. Graph the sequences from problems (19) and (32). Remember the note before problem

(17) about graphing sequences.

36. How are the graphs of geometric sequences different from arithmetic sequences?


Write an explicit and recursive formula for the following sequences (examples from worksheet).

1. -4, -6, -8, -10, …

Explicit:

Recursive:

2. 19, 13, 7, 1, …

Explicit:

Recursive:

3. 25, 75, 225, …

Explicit:

Recursive:

4. 3, 9, 27, 81, …

Explicit:

Recursive:

AQR Unit 4: Using Recursion in Models and Decision Making...

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