mathematical patterns (for help, go to skills handbook page 838.) algebra 2 lesson 11-1 1.1, 3, 5,...

Mathematical PatternsMathematical Patterns

(For help, go to Skills Handbook page 838.)

ALGEBRA 2 LESSON 11-1ALGEBRA 2 LESSON 11-1

1. 1, 3, 5, 7, 9, 11, . . . 2. –2, –4, –6, –8, –10, –12, . . .

3. 0.2, 1, 5, 25, 125, 625, . . . 4. 50, 45, 40, 35, 30, 25, . . .

5. 512, 256, 128, 64, 32, 16, . . . 6. 2, 5, 8, 11, 14, 17, . . .

7. 16, 32, 64, . . . 8. –3, –7, –11, –15, . . .

Find the next two numbers of each pattern. Then write a rule to describe the pattern.

11-1

Mathematical PatternsMathematical PatternsALGEBRA 2 LESSON 11-1ALGEBRA 2 LESSON 11-1

1. 1, 3, 5, 7, 9, 11, 13, 15; add 2

2. –2, –4, –6, –8, –10, –12, –14, –16; subtract 2

3. 0.2, 1, 5, 25, 125, 625, 3125, 15,625; multiply by 5

4. 50, 45, 40, 35, 30, 25, 20, 15; subtract 5

5. 512, 256, 128, 64, 32, 16, 8, 4; divide by 2

6. 2, 5, 8, 11, 14, 17, 20, 23; add 3

7. 16, 32, 64, 128, 256; multiply by 2

8. –3, –7, –11, –15, –19, –23; subtract 4

Solutions

11-1


a. Start with a square with sides 1 unit long. On the right

side, add on a square of the same size. Continue adding one square

at a time in this way. Draw the first four figures of the pattern.

b. Write the number of 1-unit segments in each figure above as a sequence.

c. Predict the next term of the sequence. Explain your choice.

Each term is 3 more than the preceding term.

The next term is 13 + 3, or 16.

There will be 16 segments in the next figure in the pattern.

4, 7, 10, 13, . . .

11-1


Suppose you drop a ball from a height of 100 cm. It

bounces back to 80% of its previous height. How high will it go after

its fifth bounce?

The ball will rebound about 32.8 cm after the fifth bounce.

Original height of ball: 100 cm

After first bounce: 80% of 100 = 0.80(100) = 80

After 2nd bounce: 0.80(80) = 64

After 3rd bounce: 0.80(64) = 51.2

After 4th bounce: 0.80(51.2) = 40.96

After 5th bounce: 0.80(40.96) = 32.768

11-1


a. Describe the pattern that allows you to find the next term

in the sequence 2, 6, 18, 54, 162, . . . . Write a recursive formula for

the sequence.

Multiply a term by 3 to find the next term.

A recursive formula is an = an – 1 • 3, where a1 = 2.

b. Find the sixth and seventh terms in the sequence.

Since a5 = 162

a6 = 162 • 3 = 486

and a7 = 486 • 3 = 1458

11-1


(continued)

c. Find the value of a10 in the sequence.

The term a10 is the tenth term.

a10 = a9 • 3

= (a8 • 3) • 3

= ((a7 • 3) • 3) • 3

= ((1458 • 3) • 3) • 3

= 39,366

11-1


The spreadsheet shows the perimeters of regular pentagons

with sides from 1 to 4 units long. The numbers in each row form a

sequence.

A B C D E1 a1 a2 a3 a4

2 Length of Side 1 2 3 43 Perimeter 5 10 15 20

a. For each sequence, find the next term (a5) and the twentieth term (a20).

In the sequence in row 2, each term is the same as its subscript. Therefore, a5 = 5 and a20 = 20.

In the sequence in row 3, each term is 5 times its subscript. Therefore, a5 = 5(5) = 25 and a20 = 5(20) = 100.

11-1


(continued)

A B C D E1 a1 a2 a3 a4

2 Length of Side 1 2 3 43 Perimeter 5 10 15 20

b. Write an explicit formula for each sequence.

The explicit formula for the sequence in row 2 is an = n. The explicit formula for the sequence in row 3 is an = 5n.

11-1


pages 591–593 Exercises

1. Subtract 3; 65, 62, 59.

2. Multiply by 2; 128, 256, 512.

3. Add one more to each term (add 3, add 4, add 5, etc.); 25, 33, 42.

4. Add 3; 16, 19, 22.

5. Divide by 10; 0.001, 0.0001, 0.00001.

6. Multiply by ; , , .

7. Multiply by –2; –128, 256, –512.

8. Each term is the preceding term multiplied by n; 720, 5040, 40,320.

9. Every odd-numbered term is 0, and

every even-numbered term is ; 0, , 0. 1 n – 1

17

10.

11.

12. an = an–1 + 1, a1 = –2; 3

13. an = an–1 – 2, a1 = 43; 33

14. an = , a1 = 40;

15. an = an–1 – 5, a1 = 6; –14

an–1

254

11-1

12

1 64

1 128

1 256

16. an = , a1 = 144;

17. an = an – 1 • , a1 = ;

18. an = n + 3; 15

19. an = ;

20. an = 3n + 1; 37

21. an = 4n – 1; 47

22. an = ; 3

23. an = n2 + 1; 145

24. recursive; 3, 9, 21, 45, 93

25. explicit; 0, 1, 3, 6, 10

26. explicit; –24, –21, –16, –9, 0


an – 1

4

9 16

1 64

12

12

1 n – 1

1 13

n – 62

27. recursive; –2, 6, –18, 54, –162

28. explicit; –6, –18, –38, –66, –102

29. explicit; 3, 9, 19, 33, 51

30. explicit; 5, 10, 15, 20, 25

31. recursive; 340, 323, 306, 289, 272

32. 15, 26, 40

33. 20, 23; an = 3n + 2; explicit or an = an – 1 + 3, a1 = 5; recursive

34. 96, 192; an = 3 • 2n–1; explicit or an = 2an – 1, a1 = 3; recursive

35. 216, 343; an = n3; explicit

36. 4096, 16,384; an = 4n; explicit or an = 4an – 1; a1 = 4, recursive

11-1


37. 144, 169; an = (n + 6)2; explicit or an+1 = an + 2n + 13; a1 = 49, recursive

38. –1, 1; an = –1(an–1), a1 = –1; recursive or an = ( –1)n; explicit

39. –1,– ; an = , a1 = –16; recursive

or an = ; explicit

40. –47, –40; an = an–1 + 7, a1 = –75; recursive or an = –82 + 7n; explicit

41. –11, –19; an = an–1 – 8, a1 = 21; recursive or an = 29 – 8n; explicit

42. an–2, an+2

43. Answers may vary. Sample: A recursive formula requires that the previous term be known to find a given term. An explicit formula only requires the number of the term.

an–1

212

–322n

44. a–c. Answers may vary. Sample:a. 1, –2, 4, –8, ...b. an = –2(an–1), a1 = 1; an = (–2)n–1

c. –524,288

45. 26,677; 458,330; 210,066,388,901

46. 24, 78, 240, 726

47. 25, 36, 49, 64

48. 54, 128, 250, 432

49. , , ,

50. , , ,

51. a. 25 boxesb. 110 boxesc. 9 levels

165

256

367

498

56

67

78

89

11-1


52. an = 10 • 2n–1

53. an = –n – 4

54. an = –2 •

55. an = 1 + 4(n –1)

56. a. an = an–1 + 5, a1 = 25; an = 20 + 5nb. $40c. an = (an–1 + $20) • 1.005, a1 = $40.20d. 6.5%

57. a. 15, 21b. an = an–1 + n, a1 = 1

c. Yes; the formula yields the same values as the recursive formula.

n–112

58. B

59. G

60. C

61. I

62. [2] an = an–1 + n2, a1 = 1; Each term consists of the previous term added to the square of the number of the term.

[1] incorrect formula OR no explanation OR incorrect explanation

11-1


63. (x + 4)2 + (y + 2)2 = 5

64. + = 1

65. xy = 20

66. xy = 10

67. xy = 117

68. xy = 27

69. xy = 10

70. xy = 72

71. xy = –

72. xy = 100

14

11-1

(x + 1)2

36(y + 2)2

36


Describe each pattern. Find the next three terms.

1. 5, 15, 25, 35, . . . 2. 1, , , , . . .23

49

827

Each new term is 10 more than the preceding term; 45, 55, 65

Each new term is of the

preceding term;

23

1681

32243

64729, ,

3. Write a recursive formula for the sequence 7, –1, –9, –17, . . .. Then find the next term.

an = an – 1 – 8, where a1 = 7; –25

4. Write an explicit formula for the sequence 1, , , , . . .. Then find a15

14

19

116

an = ;1n2

1225

5. A recursive formula for a sequence is an = an – 1 + 2n, where a1 = 1. Write the first five terms of the sequence.

1, 5, 11, 19, 29

11-1

Arithmetic SequencesArithmetic Sequences



Describe the pattern in each sequence. Use at least one of the words add, subtract, or difference.

1. 10, 8, 6, 4, 2, 0, . . . 2. 100, 117, 134, 151, 168, . . .

3. , , , 2, . . . 4. – , – , – , –1, – , – , . . .57

87

117

14

12

34

54

32

11-2

Arithmetic SequencesArithmetic SequencesALGEBRA 2 LESSON 11-2ALGEBRA 2 LESSON 11-2

1. 10, 8, 6, 4, 2, 0, . . .; subtract 2

2. 100, 117, 134, 151, 168, . . .; add 17

3. , , , 2, . . .; add

4. – , – , – , –1, – , – , . . .; subtract

57

87

117

14

12

34

54

32

37

14

Solutions

11-2


Is the given sequence arithmetic?

a. 7, 10, 13, 16, . . .

The common difference is +3. This is an arithmetic sequence.

7, 10, 13, 16

10 – 7 = 3

+3

13 – 10 = 3

+3

16 – 13 = 3

+3

11-2


(continued)

b. The sequence of dots in the “triangles” shown below.

3, 6, 10, 15

There is no common difference. This is not an arithmetic sequence.

10 – 6 = 4

+4

15 – 10 = 5

+5

6 – 3 = 3

+3

11-2


Suppose you have already saved $75 toward the purchase

of a new CD player and speakers. You plan to save at least $12 a

week from money you earn at a part-time job. In all, what is the

minimum amount you will have after 26 weeks?

Find the 27th term of the sequence 75, 87, 99, . . . .

an = a1 + (n – 1)d Use the explicit formula.

a27 = 75 + (27 – 1)(12) Substitute a1 = 75, n = 27, and d = 12.

= 75 + (26)(12) Subtract within parentheses.

= 75 + 312 Multiply.

= 387 Simplify.

After 26 weeks, you will have saved a minimum of $387.

11-2


Find the missing term of the arithmetic sequence 50, , 92.

= 71 Divide.

The missing term is 71.

arithmetic mean = Write the average.92 + 50

2

= Simplify the numerator.142

2

11-2



1. no

2. yes; 10

3. no

4. no

5. yes; 3

6. yes; –11

7. yes; 4

8. no

9. no

10. no

11. 127

12. 0.3

13. 12.5

14. 0.0085

15. 225

16. –159

17. –59

18. 240

19. –146

20. 137

21. –7.5

22. 21

23. 13

24. 16

25. –7

26. 660

27. 7.5

28. 2.5

29. a11 or

30. 82.5

31. 4

32.

33. 13

34. 120

35. –19.5

a10 + a12

2

11-2

12


36. 1.1

37. –1

38.

39.

40.

41. 0

42. 2x + 1

43. The student assumed that the sequence was an = 2n–1. However, a1 = 20 = 1, not 0 as given in the problem.

45

r + s2

2r + s2

44. a. Answers may vary. Sample: 25, 18, 11, 4, –3, –10,...; to find the nth term, multiply n – 1 times (–7) and add to a1.

b. Answers may vary. Sample: Start with the first term and continue to subtract 7 for each term. For each term, you subtract 7x (term number –1) from the first term.

45. Answers may vary. Sample: An advantage of a recursive formula is that only the preceding term must be known to find the next term; a disadvantage is that many calculations may be required to find a term. An advantage of an explicit formula is that it is easy to find any term.

11-2


46. 23

47. 15

48. 18.5

49. 22

50. 6

51. 29

52. an = 2 + 2(n – 1); an = an–1 + 2, a1 = 2

53. an = 0 + 6(n – 1); an = an–1 + 6, a1 = 0

54. an = –5 + 1(n – 1); an = an–1 + 1, a1 = –5

55. an = –4 – 4(n – 1); an = an–1 – 4, a1 = –4

56. an = –2 + 7(n – 1); an = an–1 + 7, a1 = –2

57. an = 27 – 12(n – 1); an = an–1 – 12, a1 = 27

58. an = –5 + 1.5(n – 1); an = an–1 + 1.5, a1 = –5

59. an = –32 + 12(n – 1); an = an–1 + 12, a1 = –32

60. an = 1 + (n – 1); an = an–1 + , a1 = 1

61. an = (n – 1); an = an–1 + , a1 = 0

62. 6 min; 1 min

63. –4, –10, –16

64. 4.6, –0.8, –6.2

65. –8, –17, –-26

66. , 5,

67. 17, 17, 17

11-2

195

315

13

18

18


68. 681, 702, 723

69. –12.5, –8, –3.5

70. a + 5, a + 9, a + 13

71. a. $20, $45, $75, $110, $150, $195, $245, $300, $360, $425, $495

b. an = an – 1 + $20 + $5(n – 1), a1 = $20c. $495

72.

73.

74. 2.46 min or 2 min 28 s

75. 54

76. 8

77. 21st term

78. 43rd term

79. a1 = –1, d = 3

80. a1 = –4, d = 4

81. a1 = 52, d = –10

11-2


82. a1 = –21 , d = 4

83. a1 = –100.5, d = 22

84. a1 = –9, d = 2.2

85. 9k + 32

86. 21k – 43

87. B

88. I

89. B

90. H

91. [2] The third term, 31, is the mean of the first and fifth terms. The second term, 23, is the mean of the first and third terms. The fourth term, 39, is the mean of the third and fifth terms.

[1] incomplete explanation OR incorrect term values

92. [4] 696; a2 – a1 = 10 – 3 = 7, so the common difference is 7. Use a1 = 3, d = 7, and n = 100 in the formula an = a1 + (n – 1)d to find a100.

[3] explanation correct, but computational error

[2] incomplete explanation

[1] correct number, but no explanation

11-2

14

14


93. recursive; –2, –7, –12, –17, –22

94. explicit; 6, 18, 36, 60, 90

95. explicit; 0, 3, 8, 15, 24

96. recursive; –121, –108,–95, –82, –69

97. (0, – 5 ) and (0, 5 )

98. (–4 2, 0) and (4 2, 0)

99. (1 – 21 , 0) and (1 + 21, 0)

100. (1 + 39, 3) and (1 – 39, 3)

101. r =6 2V

2

3

11-2


Is the given sequence arithmetic? If so, identify the common difference.

1. –9, 0, 9, 18, . . . 2. 0.1, 0.01, 0.001, 0.0001, . . .

3. , , , , . . .16

13

12

23

4. Find the 25th term of the arithmetic sequence 26, 13, 0, –13, . . .

Find the missing term of each arithmetic sequence.

5. 8, , 20, . . . 6. 3 , , 12 , . . .12

12

yes; 9 no

yes; 16

–286

14 8

11-2

Geometric SequencesGeometric SequencesALGEBRA 2 LESSON 11-3ALGEBRA 2 LESSON 11-3

(For help, go to Lesson 11-3.)

Find the next term in each sequence.

1. 1, 2, 4, 8, . . . 2. 336, 168, 84, 42, . . .

3. 0.1, 1, 10, 100, . . . 4. 900, 300, 100, . . .

11-3


1. 1, 2, 4, 8, 16 (multiply by 2)

2. 336, 168, 84, 42, 21 (divide by 2)

3. 0.1, 1, 10, 100, 1000 (multiply by 10)

4. 900, 300, 100, (divide by 3)1003

Solutions

11-3


Is the given sequence geometric? If so, identify the

common ratio.

a. 1, –6, 36, –216, . . .

There is a common ratio of –6. This is a geometric sequence.

1, –6, 36, –216

–6 ÷ 1 = –6

–6

36 ÷ –6 = –6

–6

216 ÷ 36 = –6

–6

11-3


(continued)

b. 2, 4, 6, 8, . . .

2, 4, 6, 8

There is no common ratio. This is not a geometric sequence.

4 ÷ 2 = 2

2 43

8 ÷ 6 = 43

32

6 ÷ 4 = 32

11-3


Suppose you have equipment that can enlarge a photo to

120% of its original size. A photo has a length of 10 cm. Find the

length of the photo after 5 enlargements at 120%.

You need to find the 6th term of the geometric sequence 10, 12, 14.4, . . .

an = a1 • r n – 1 Use the explicit formula.

a6 = 10 • 1.206 – 1 Substitute a1 = 10, n = 6, and r = 1.20.

= 10 • 1.205 Simplify the exponent.

After five enlargements of 120%, the photo has a length of about 25 cm.

24.883 Use a calculator.

11-3


A family purchased a home for $150,000. Two years later

the home was valued at $188,160. If the value of the home is

increasing geometrically, how much was the home worth after one

year?

geometric mean = 150,000 • 188, 160 Use the definition.

= 28,224,000,000 Multiply.

= 168,000 Take the square root.

11-3



1. yes; 2; 16, 32

2. no

3. yes; –2; 16, –32

4. yes; –1; –1, 1

5. yes; 0.4; 0.256, 0.1024

6. yes; 0.1; 0.0007, 0.00007

7. yes; – ; , –

8. no

9. yes; 1.5; 50.625, 75.9375

10. yes; –5; 1250, –6250

11. yes; 6; –1296, –7776

13

29

2 27

12. no

13. an = 5 • (–3)n – 1; 5, –15, 45, –135, 405

14. an = 0.0237 • 10n – 1; 0.0237, 0.237,2.37, 23.7, 237

15. an = ; , , , ,

16. an = 0.5n – 1; 1, 0.5, 0.25, 0.125, 0.0625

17. an = 100(–20)n – 1; 100; –2000; 40,000; –800,000; 16,000,000

18. an = 7 • 1n – 1; 7, 7, 7, 7, 7

19. an = 1024(0.5)n – 1; 1024, 512, 256, 128, 64

20. an = 4(0.1)n – 1; 4, 0.4, 0.04, 0.004, 0.0004

21. an = 10(–1)n – 1; 10, –10, 10, –10, 10

22. 67.5

12

23

n – 1 12

13

29

4 27

8 81

11-3


23. 1530

24.

25. 1.5

26. 3.75

27. 6

28. geometric; 720, 1440

29. arithmetic; 125, 150

30. geometric; 3, –3

31. arithmetic; 50, 55

32. geometric; –80, 160

33. geometric; 0.125, 0.0625

34. neither; 20, 26

4 15

35. geometric; 2, 2

36. neither; 25, 36

37. 6561, 2187, 729

38. 7.5, 22.5, 67.5

39. 10, 8, 6.4

40. –6.64, –11.02, –18.30

41. a–d. Answers may vary. Sample:a. 3 and 12; 6b. 3, 6, 12; 2c. 768d. 48; 5th term

42. 768

43. 12,288

44. 786,432

11-3


45. 201,326,592

46. 12,884,901,888

47. 3(4n – 1)

48. 4

49. 16

50. 2.5

51. 10

52. –

53. –

1623

54. Both the common difference and the common ratio are used to find the next term in a sequence, but a common difference is added and a common ratio is multiplied.

55. $142.79, $613.59, $28.62, $58.92, $105.82, $262.94

56. a. d, d, d, d, d, d

b. Yes; the common ratio is .

c. an = an – 1 • , a1 = d

57. Both arithmetic and geometric sequence explicit formulas use the first term a1, n – 1, and a common term. The recursive formulas both use an – 1 and a common term.

11-3

12

14

18

1 16

1 32

12

12


58. a. 5000, 3750, 2812.5, 2109.38, 1582.03

b.

c. 375.42 cm3

d.

e. The common ratio is less than one, so the graph is decreasing.

59. 7

60. 128

34

61. A

62. F

63. A

64. C

65. B

66. [2] geometric mean = 4 • 16

= 64 = 8; arithmetic mean

= = = 10; The

arithmetic mean is greater.

[1] correct answer with no work shown

4 + 162

202

11-3

34


67. [4] Since a4 = a3 • r = a2 • r 2 =

a1• r 3, 192 = 3 • r 3. So r 3 = 64,

and r = 64 = 4. Then a2 = a1• r = 3 • 4 = 12 and a3 = a2 • r = 12 • 4 = 48.

[3] appropriate methods, but with one computational error

[2] correct answer, but with minimal explanation

[1] correct answer, but with no explanation

68. an = –3 + 3(n – 1); an = an – 1 + 3, a1 = –3

69. an = 17 + (–9)(n – 1); an = an – 1 – 9, a1 = 17

3

70. an = –2 + (–11)(n – 1); an = an – 1 – 11, a1 = –2

71. x2 + y2 = 9

72. (x + 3)2 + (y – 1)2 = 25

11-3


73. (x – 1)2 + (y – 1)2 = 4

74. x = –3

75. x = –1

76. x = 0, 1

11-3


Is the given sequence geometric? If so, identify the common ratio and find the next two terms.

4. Write the explicit formula for the geometric sequence for which a1 = 7 and r = . Then generate the first five terms.1

3

1. 1, 2, 6, 12, . . . no

2. 2, 1, 0.5, 0.25, . . .yes; 0.5; 0.125, 0.0625

3. –9, 81, –729, 6561, . . .yes; –9, –59,049, 531,441

5. Find the missing term for the geometric sequence 3, , 48 . . . 12

an = 7 • ; 7, , , , 13

n – 1 73

79

727

781

11-3

Arithmetic SeriesArithmetic Series



Find each sum.

1. 2 + 3.5 + 5 + 6.5 + 8

2. –17 + (–13) + (–9) + (–5) + (–1) + 3

Write an explicit formula for each sequence.

3. 4, 6, 8, 10, 12, . . . 4. 1, 4, 7, 10, 13, 16, . . .

5. –17, –23, –29, –35, . . . 6. 10, 1, –8, –17, . . .

11-4

Arithmetic SeriesArithmetic SeriesALGEBRA 2 LESSON 11-4ALGEBRA 2 LESSON 11-4

1. 2 + 3.5 + 5 + 6.5 + 8 = (2 + 8) + (3.5 + 6.5) + 5 = 10 + 10 + 5 = 25

2. –17 + (–13) + (–9) + (–5) + (–1) + 3 = [–17 + (–13)] + [–9 + (–1)] + [–5 + 3] = –30 + (–10) + (–2) = –42

3. 4, 6, 8, 10, 12, . . .an = a1 + (n – 1)d

= 4 + (n – 1)2= 4 + 2n – 2= 2 + 2n

4. 1, 4, 7, 10, 13, 16, . . .an = a1 + (n – 1)d

= 1 + (n – 1)3= 1 + 3n – 3= –2 + 3n

5. –17, –23, –29, –35, . . .an = a1 + (n – 1)d

= –17 + (n – 1)(–6)= –17 – 6n + 6= –11 – 6n

6. 10, 1, –8, –17, . . .an = a1 + (n – 1)d

= 10 + (n – 1)(–9)= 10 – 9n + 9= 19 – 9n

Solutions

11-4


Use the finite sequence 5, 9, 13, 17, 21, 25, 29. Write the

related series. Evaluate the series.

The sum of the terms of the sequence is 119.

5 + 9 + 13 + 17 + 21 + 25 + 29 = 119

Related series Add to evaluate.

11-4


A staircase uses same-size cement blocks arranged 4

across, as shown below. Find the total number of blocks in the

staircase.

Relate: sum of

the seriesis

number of terms2

timesthe first

termplus

the lastterm

Define: Let Sn = total number of blocks,

and let n = the number of stairs.

Then a1 = the number of blocks in the first stair,

and an = the number of blocks in the last stair.

11-4


(continued)

= 2.5(24) Simplify.

= 60 Multiply.

There are 60 blocks in the stairs.

n2

Write: Sn = ( a1 + an ) Use the formula.

= (4 + 20) Substitute n = 5, a1 = 4, an = 20.52

11-4


Use the summation notation to write the series8 + 16 + 24 + . . . for 50 terms.

8 • 1 = 8, 8 • 2 = 16, 8 • 3 = 24, . . . The explicit formula for the sequence is 8n.

8 + 16 + 24 + . . . + 400 = 8n The lower limit is 1 and the upper limit is 50.

50

n = 1

11-4


Use the series (–2n + 3).4

n = 1

a. Find the number of terms in the series.

Since the values of n are 1, 2, 3, and 4, there are four terms in the series.

b. Find the first and last terms in the series.

The first term of the series –2n + 3 = –2(1) + 3 = 1.

The last term of the series –2n + 3 = –2(4) + 3 = –5.

11-4


(continued)

c. Evaluate the series.

4

n = 1

(–2n + 3) = (–2(1) + 3) + (–2(2) + 3) + (–2(3) + 3) +

(–2(4) + 3) Substitute.

= 1 + (–1) + (–3) + (–5) Simplify within parentheses.

= –8 Add.

The sum of the series is –8.

11-4



1. 21 + 18 + 15 + 12 + 9 + 6 + 3; 84

2. (–5) + (–15) + (–25) + (–35) + (–45); –125

3. 100 + 99 + 98 + 97 + 96 + 95; 585

4. 0.5 + 0.25 + 0 + (–0.25) + (–0.5) + (–0.75); –0.75

5. 17.3 + 19.6 + 21.9 + 24.2 + 26.5; 109.5

6. 4.5 + 5.6 + 6.7 + 7.8 + 8.9 + 10 + 11.1; 54.6

7. 32

8. –48

9. 264

10. 35

11. 4292

12. –146

13. 2n

14. (n + 7)

15. (n + 4)

16. (3n – 2)

17. 7n

18. –3n

4

n = 18

n = 17

n = 111

n = 115

n = 15

n = 1

11-4


19. 5, 1, 9; 25

20. 5, –3, –11; –35

21. 6, 4, –1; 9

22. 5, 0, 0.8; 2

23. 9, , ;72

24. 6, 15, 10; 75

25. sequence; infinite

26. sequence; finite

27. series; finite

28. series; infinite

29. sequence; infinite

30. series; finite

83

403

31. a. 270 chairs on each side, 390 chairs middle, 930 chairs total

b. each side: (n + 3); middle: (n + 9)

c. $46,950

32. a. 8; the formula for the corresponding sequence is 3n + 7. Solving 3n + 7 = 31 for n shows that n = 8.

b. 164

33. a. 91b. 83

20

n = 1

20

n = 1

11-4


34. a. an = n + 1

b. (n + 1)

c. 18 cansd. No; no; 13 rows have 104 cans,

14 rows have 119 cans, 15 rows have 135 cans, and 16 rows have 152 cans. The number of rows would not be an integer for 110 cans or 140 cans.

35. 110

36. –765

37. 5150

38. –22

9

n = 1

39. –0.6

40. 1,000,500

41. a. No; 75 + 25(6) = 225, which is less than 500.

b. Answers may vary. Sample: Pro: spreading the cost over several years, con: calculators purchased first may be outdated by the time 500 calculators have been purchased; check students’ work.

42. a–d. Check students’ work.

43. 300

44. –200

45. 34

11-4

Arithmetic SeriesArithmetic Series

58. an = 3 ; 3, ,

59. an = –7(0.1)n – 1; –7, –0.7, –0.07

60. an = 20(–0.5)n – 1; 20, –10, 5

61.

ellipse: x- and y-axes; domain: –6 x 6; range: –2 3 y 2 3


11-4

46. 12,884,901,888

47. 10x + 45y

48. 45x – 210y

49. B

50. G

51. C

52. A

53. C

54. B

55. an = 1(2)n – 1; 1, 2, 4

56. an = 1(5)n – 1; 1, 5, 25

57. an = –1(–1)n – 1; –1, 1, –1

n – 132

92

274

<– <–<– <–


11-4

64.

65.

66.

x + 3x – 4

c – 2c – 5

z 2 + 12z + 20z – 1

63.

circle: all lines through the origin; domain: –2 x 2; range: –2 y 2<– <– <– <–

62.

hyperbola: x- and y-axes; domain: x –5 or x 5; range: all real numbers

>–<–

6

n = 1

(100 + 2n); 642

6

n = 1

3n; 15, 150

6

n = 1

(15 – 5n); –15


1. Write an addition expression of the finite series represented by 8

n = 110n.

Write each finite arithmetic series using summation notation. Then find the sum of each series.

2. 102 + 104 + 106 + 108 + 110 + 112

3. 10 + 5 + 0 + (–5) + (–10) + (–15)

4. Write the arithmetic series to the given term using summation notation. Then find the sum of the series.

3 + 6 + 9 + 12 + . . .; 100th

10 + 20 + 30 + 40 + 50 + 60 + 70 + 80

11-4

Geometric SeriesGeometric Series



Find each sum or difference.

1. 100 + 50 + 25 + + 2. 3 + 9 + 27 + 81

3. –2 + 4 – 8 + 16 – 32 4. –5 – 10 – 20 – 40

Simplify each fraction.

5. 6. 7. 8.

252

254

1 – 15

13

1

141 –

12

13–

14

1162 +

13

11-5

Geometric SeriesGeometric SeriesALGEBRA 2 LESSON 11-5ALGEBRA 2 LESSON 11-5

1. 100 + 50 + 25 + + = 175 + + = 175 + = 175 + 18 = 193

2. 3 + 9 + 27 + 81 = (3 + 27) + (9 + 81) = 30 + 90 = 120

3. –2 + 4 – 8 + 16 – 32 = (4 + 16) – (2 + 8 + 32) = 20 – 42 = –22

4. –5 – 10 – 20 – 40 = –(5 + 20) – (10 + 40) = –25 – 50 = –75

252

254

504

254

754

34

34

Solutions

11-5


Solutions (continued)

5. = = ÷ = • = or 2

6. = = 1 ÷ = 1 • = or 1

7. = = = = ÷ = • = =

8. = = ÷ = • = or 6

1 – 15

13

1141 –

12

13–

14

1162 +

13

4513

45

13

45

31

125

25

134

34

43

13

1 • 32 • 3

1 • 23 • 2

–

14

36

26–

14

1614

16

14

16

41

46

23

331613

3316

13

3316

31

9916

316

11-5


Use the formula to evaluate the series

5 + 15 + 45 + 135 + 405 + 1215.

The first term is 5, and there are six terms in the series.

The sum of the series is 1820.

The common ratio is = = = = = 3155

4515

13545

405135

1215405

So a1 = 5, r = 3, and n = 6.

Sn = Write the formula.a1 (1 – r n)

1 – r

= Substitute a1 = 5, r = 3, and n = 6.5 (1 – 36)

1 – 3

= = 1820 Simplify.–3640

–2

11-5


The Floyd family starts saving for a vacation that is one year

away. They start with $125. Each month they save 8% more than the

previous month. How much money will they have saved 12 months later?

Relate: Sn = a1 (1 – r n)

1 – rWrite the formula for the sum of a geometric series.

The amount of money the Floyd’s will have saved will be $2372.14.

Define: Sn = total amount saved

a1 = 125 Initial amount.

r = 1.08 Common ratio.

n = 12 Number of months.

11-5

Write: S12 =125 ( 1 – 1.08 12)

1 – 1.08

2372.14 Use a calculator.

Substitute.


Decide whether each infinite geometric series diverges or

converges. Then determine whether the series has a sum.

n = 1

23

n

a. b. 2 + 6 + 18 + . . .

a1 = = , a2 = =23

1 23

23

2 49 a1 = 2, a2 = 6

r = ÷ =49

23

23 r = 6 ÷ 2 = 3

Since | r | < 1, the series converges, and the series has a sum.

Since | r | 1, the series diverges, and the series does not have a sum.

>

11-5


The weight at the end of a pendulum swings through an arc

of 30 inches on its first swing. After that, each successive swing is

85% of the length of the previous swing. What is the total distance the

weight will swing by the time it comes to rest? The largest arc the

pendulum swings through is on the first swing of 30 in., so a1 = 30.

= 200 Simplify.

The total distance that the pendulum swings through is 200 in.

S = Use the formula.a1

1 – r

= Substitute.30

1 – 0.85

11-5



1. 255

2. 1456

3. 381

4. 3647

5. –10,235

6. –

7.

8. –1640

9. converges; has a sum



11116

255256

12. diverges; no sum






18. 1.2

19. 1

20.

21.

22. 9

5692

23.

24. geometric; 2046

25. arithmetic; 420

26. geometric; –1,627,605

27. geometric; 96.47

28. arithmetic; 500,500

29. geometric; 121.5

95

11-5


30. a.

4, 16, 64b. 4 + 16 + 64 + 256 + 1024 + 4096c. 5460

31. a. 20, 18, 16.2, 14.58

b. 198.59

c. S = = 200

d. Check students’ work.

20 1 – 0.9

32.

33. 4

34.

35. no sum

36. 0.83

37. no sum

38. a. 2b. 25 + 20 + 16 + 12.8 + 10.24 + . . .c. 20 + 16 + 12.8 + 10.24 + . . .d. 225 cm

39. Check students’ work.

54

34

11-5


40. a. No; the sum of a series of positive numbers will be positive.

b. Your classmate did not check if |r | were less than 1.

41.

42. 10

43. (b); (a) yields $26,000; using the formula for finding the sum of a finite geometric series, (b) yields $1,342,177.26.

44. a. Answers may vary. Sample: The student used r – 1 instead of 1 – r in the formula for the sum of an infinite geometric series.

b.

45. converges; 1

46. diverges

47. converges; 2.718

48. a. 70th swingb. 10,000 cm

49. a. S = = =

b.

50. a. all integers greater than or equal to 1

b. 10; 18; 24.4; 29.52; 33.62; 36.89; 39.51; 41.61; 43.29; 44.63;

c. 50

78

12

0.142857 1 – 0.000001

0.1428570.999999

17

37

11-5


51.

52.

53.

54. 5

55. 0.424

56. 1543

57. 140

58. –825

59. 1375

60. (0, 4), y = –4;

472323

61. (–1, 0), x = 1;

62. 0, – , y = ;

63.

94

94

64.

65.

66.

7c – 42c2

10(2y + 3) (x + 3)(x – 3)x2 + 6x + 4

(x + 6)(x – 6)

–3(5d + 16) (d + 3)(d – 3)

11-5


5. Find the sum of the infinite geometric series.

15 + 5 + + + . . . 53

59

1. Find the sum of the geometric series.3 – 30 + 300 – 3000 + 30,000 27,273

2. Find the sum of the first twelve terms of the geometric series. Round your answer to the nearest thousandth.16 + 8 + 4 + . . . 31.992

3. How can you tell whether an infinite geometric series converges or diverges?Check the common ratio r. If | r | < 1, the series converges. Otherwise it diverges.

4. Find the sum of the infinite geometric series.16 + 8 + 4 + . . . 32

22.5

11-5

Area Under a CurveArea Under a Curve



Find the area of the rectangle with the given length and width.

1. = 4 ft, w = 1 ft 2. = 5.5 m, w = 0.5 m

3. = 6.2 cm, w = 0.1 cm 4. = 9 in., w = 3 in.12

58

11-6

Area Under a CurveArea Under a CurveALGEBRA 2 LESSON 11-6ALGEBRA 2 LESSON 11-6

Solutions

1. A = w = (4 ft)(1 ft) = 4 ft2

2. A = w = (5.5 m)(0.5 m) = 2.75 m2

3. A = w = (6.2 cm)(0.1 cm) = 0.62 cm2

4. A = w = (9 in.)(3 in.) = = = 34 in.212

58

192

298

55116

716

11-6


The curve shown below approximates the speed of a car

during a 12-minute drive.

a. What does the area under the curve represent?

= feet Simplify.

The area under the curve approximates the total distance traveled by the car.

Area = • minutes Use dimensional analysis.feet

minutes

11-6


(continued)

b. Use the inscribed rectangle 2 units wide to estimate the area under the curve.

The indicated area is 35,000 units2.

Total area

2(2000) + 2(3000) + 2(3500) + 2(4000) + 2(3000) + 2(2000) = 35,000

Value of curve at upper edge of each rectangle.

Width of each rectangle

11-6


Estimate the area under the curve

ƒ(x) = –0.5x2 + 6 for the domain

0 x 2 by evaluating the sum A.

A = (0.5)ƒ(an), where a1 = 0.5, a2 = 1, a3 = 1.5, a4 = 2.4

n = 1

A = 0.5ƒ(0.5) + 0.5ƒ(1) + 0.5ƒ(1.5) + 0.5ƒ(2) Add the areas of the rectangles.

= 0.5(5.875 + 5.5 + 4.875 + 4) total area = width of each rectangle • sum of the heights

= 0.5(20.25) Add within parentheses.

= 10.125 Simplify.

The indicated area is about 10.125 units2.

11-6

<– <–


Use a graphing calculator to graphƒ(x) = –x2 + 4x + 5. Find the area under the curve for the domain 1 x 4.

Step 1: Input the equation. Adjust the window values.

Step 3: Use the lower limit of x = 1.

Step 4: Use the upper limit of x = 4.

The area under the curve between x = 1 and x = 4 is 24 units2.

Step 2: Access the ƒ(x)dx feature from the CALC menu.

11-6

<– <–



1. total produced

2. amount of growth

3. miles

4. distance traveled

5. total price

6. 270 units2

7. 110 units2

8. 480 units2

9. A = 1ƒ(an )

a. 0.5 units2

b. 2.5 units2

10. A = 1ƒ(an )

a. 5 units2

b. 9 units2

11. A = 1g (an )

a. 3 units2

b. 7 units2

12. A = 1ƒ(an )

a. 3 unit2

b. 7 units2

2

n = 1

2

n = 1

2

n = 1

2

n = 1

13. A = 1ƒ(an )

a. 10 units2

b. 13 units2

14. A = 1h (an )

a. 5 units2

b. 25 units2

15. A = 1ƒ(an )

a. 6.75 unit2

b. 7.75 units2

2

n = 1

2

n = 1

2

n = 1

2313

11-6


16. A = 1h (an )

a. 5 units2

b. 9 units2

17. A = 1ƒ(an )

a. 5 units2

b. 9 units2

18. 1.6 units2

19. 2.5 units2

20. 0.25 units2

21. 3.3 units2

22. 2.3 units2

23. 3.25 units2

2

n = 1

2

n = 1

24.

8.75 units2

25.

43 units2

26.

28.75 units2

27.

50 units2

28.

16.875 units2

29.

19.5 units2

11-6


30. a.

b. 3 units2

c.

12 units2

d. 7.5 units2; the mean best approximates the area because it is between the other measures known to be larger and smaller than the actual value.

31. a. Answers may vary. Sample:

0.37 milesb. Answers may vary. Sample: Using

inscribed rectangles will result in an estimate smaller than the actual number.

32. 9 units2

33. 7.5 units2

34. 15 units2

35. 3.46 units2

11-6


36. 8.05 units2

37. 9.75 units2

38. 10 units2

39. 2.5 units2

40. 9.9 units2


42. a. 17.5 units2

b. 14 units2

c. The estimate in (a) is closer to the actual area because there is less area between the curve and the rectangles when more rectangles are used.

43. a.

b. The area under both curves is the same, 3 units2, over the interval–1.5 x 1.5; this is true because

the amount of area above y = x3 + 1

and below y = 1 on the left side of the y-axis is equal to the area below

y = x3 + 1and above y = 1 on the

right side of the y-axis.

44. 6 units2

14

14

11-6

<– <–


45. a. y = 9 –

b.

23.56 units2

c. 47.12 units2

d. Check students’ work.

46. A

47. G

48. B

49. G

9x2

25 50. [2] The mean might be a better estimate since it is between the estimate using inscribed rectangles, which is too low, and the estimate using circumscribed rectangles, which is too high.

[1] incomplete explanation

51. exists

52. exists

53. – = 1;x2

16y 2

9

11-6

54. – = 1

55. – = 1

56. –3

57.

58.


x2

25y 2

1

x2

10y 2

16

3 147 ± 2 73

9

11-6


2. Use inscribed rectangles to approximate the area under the graph of the

function ƒ(x) = 3 + x, for –3 x 0. Use rectangles 0.5 units wide.

1. Use circumscribed rectangles to approximate the area under the graph

of the function ƒ(x) = 5 – x2, for 0 x 3. Use rectangles 1 unit wide.14

3. Evaluate a sum to approximate the area under the graph of the function

ƒ(x) = 3x2, for 0 x 10. Use inscribed rectangles 2 units wide.

4. Use a graphing calculator to find the area under the graph of

y = x3 – 2x2 + 3, 0 x 3.

about 13.75 units2

about 2.96 units2

about 720 units2

11.25 units2

11-6

<– <–

<– <–

<– <–

<– <–

Sequences and SeriesSequences and SeriesALGEBRA 2 CHAPTER 11ALGEBRA 2 CHAPTER 11

1. an = an – 1 + 6, a1 = 7, an = 1 + 6n; 73

2. an = an – 1 • 2, a1 = 10, an = 10 • 2n–1; 20,480

3. a. savings = 50 + 5(months – 1)

b. $75

4. arithmetic; 59

5. arithmetic; 51

6. geometric; 98,415

7. 8

8. 6


10. geometric; r =

11. arithmetic; d = –3

12. geometric; r = 2

13

13. 2, –4, 8, –16, 32

14. 3, 10, 17, 24, 31

15. –100, –20, –4, – ,–

16. 19, 15, 11, 7, 3

17. 1

18. 4

19.

20.

21. 36

22. arithmetic; 156

23. geometric; 6250

24. arithmetic; –4.9

45

4 25

5923

11-A

Page 632

Sequences and SeriesSequences and SeriesALGEBRA 2 CHAPTER 11ALGEBRA 2 CHAPTER 11

25. 5, 4, 16; 50

26. 8, , ; 24

27. 7, 2.8, 7.6; 36.4

28. 5, –2, –32; –22

29. If the absolute value of the common ratio is less than 1, then it will converge; check students’ work.

30. $3693.64

31. 40 cm, 36 cm, 32.4 cm, 29.16 cm

32. miles

33. pounds

34. dollars

23

163

35.

2 units2

36.

4 units2

37.

9 units2

38. Answers may vary. Sample: You could find the sum of the circumscribed rectangles and the sum of the inscribed rectangles and average the numbers.

11-A

mathematical patterns (for help, go to skills handbook page 838.) algebra 2 lesson 11-1 1.1, 3, 5,...

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