Chapter 8 Answers - Weebly

Chapter 8 Answers 1

Chapter 8 Answers

Lesson 8.1 11.. a.

This a bipartite graph. b. 1, 4, 9c.

Number of Couples

Number of Handshakes

Recurrence Relation

1 1 — 2 4 H2 = H1 + 3 3 9 H3 = H2 + 5 4 16 H4 = H3 + 7 5 25 H5 = H4 + 9

d. 2n – 1e. Hn = Hn–1 + 2n – 1

22.. a.

Couples Handshakes 1 0 2 2 3 6 4 12

b. 2n – 2.c. Hn = Hn–1 + 2n – 2

33.. a. i. Hn = Hn–1 + 3

ii. Hn = (Hn–1) × 2iii. Hn = Hn–1 + n

iv. Hn = (Hn–1) × nb. 16, 64, 36, 5040

Chapter 8 Answers 2

44.. a. A given number of the sequence is the sum of the previous two numbers. The last number should be 5 + 8 = 13.

b. tn = tn–1 + tn–2 55.. a. 4

b. –1 c. 0 d. None

66.. a. 6, 10 b. An = An–1 + n – 1

c. An = n(n!1)

2

d. 45 77.. a.

Term Number Number of Handshakes

First Differences Second Differences

1 0 — — 2 1 1 — 3 3 2 1 4 6 3 1 5 10 4 1 6 15 5 1 7 21 6 1 8 28 7 1

The values in the second difference column are constant. b. Second degree, 2

88.. a. Term Number

Value

First Differences

Second Differences

Differences

1 3 — — — 2 28 25 — — 3 101 73 48 — 4 246 145 72 24 5 487 241 96 24 6 848 361 120 24 7 1,353 505 144 24 8 2,026 673 168 24

b. 3 c. They are the same.

Chapter 8 Answers 3

99.. a.

Levels Vertices 0 1 1 3 2 7 3 15 4 31 5 63

b. Vn = Vn–1 + 2n 1100.. a.

Term Number Number of Bees 0 5,000 1 5,600 2 6,272 3 7,024.64 4 7,867.60

b. Bn = 1.12Bn–1 c. After 27 years, in 2014

Chapter 8 Answers 4

1111.. a. Year Amount

0 800.00 1 840.00 2 882.00 3 926.10 4 972.41 5 1021.03 6 1072.08 7 1125.68 8 1181.96 9 1241.06 10 1303.12 11 1368.27 12 1436.69 13 1508.52 14 1583.95 15 1663.14

It takes 15 years for Susie's money to double.

b. Mn = (Mn–1) × 1.05 1122.. a. It is one-third the length of a segment in the previous figure.

b. There are four times as many.

c. Ln =

43

Ln!1

Chapter 8 Answers 5

Lesson 8.2 11.. a. 2

b. 1 c. 2 d. This sequence was not generated by a polynomial; it was generated by an

exponential function (note that each term is 3 times the previous term). 22.. a. Hn = 2n2 – 5n

b. Hn = 0.29 + 0.23(n – 1) c. Hn = n2 – 5n + 4 d. Hn = 3n–1.

33.. a. Tn = Tn–1 + n – 1 b. 0

c. Tn =

n(n!1)2

44.. a. 1, –2, –11, –38, –119, –362 b. 2.5

55.. a. Dn = Dn–1 + n – 2

b. Dn =

n2

2! 3n

2

6. a.

Row Number Number of Seats Total Seats 1 24 24 2 26 50 3 28 78 4 30 108 5 32 140 6 34 174

b. Sn = Sn–1 + 2 c. Sn = 24 + 2n – 2 d. 37 e. See the table in the answer to part a. f. Tn = Tn–1 + 24 + 2n – 2 g. Tn = n2 + 23n

Chapter 8 Answers 6

7. a.

Year Value 2000 88,000 2001 92,400 2002 97020 2003 101,871 2004 106,964.55 2005 112,31278 2006 117,928.42 2007 123,824.84 2008 130.016.08 2009 136,516.88 2010 143,34273 2011 150,509.86 2012 158,035.36 2013 165,937.12 2014 174,233.98 2015 182,945.68

b. Vn = (Vn–1) × 1.05 c. The differences never become constant.

8. a.

Year Deer 2010 50 2011 52 2012 54.08 2013 56.24

b. In 2058 c. Tn = 1.04Tn–1 d. The differences never become constant.

9. A recurrence relation is Sn = Sn–1 + n. A closed-form formula is Sn = n2 + n

2.

10. For a third-degree polynomial, the third differences are constant. Each third difference is 6 times the leading coefficient.

Chapter 8 Answers 7

Lesson 8.3 11.. a. i. Arithmetic

ii. Geometric iii. Geometric iv. Neither v. Geometric vi. Neither

b. i. Hn = Hn–1 + 3

ii. Hn =

Hn!1

2

iii. Hn = (Hn–1) × 1.2 iv. Hn = Hn–1 + Hn–2

v. Hn = (Hn–1) × 0.1 vi. Hn = Hn–1 + n – 1

c. i. Hn = 2 + 3(n – 1) ii. Hn = 64(0.5n–1) iii. Hn = 10(1.2n–1) v. Hn = 0.3(0.1n–1)

2. 0 3. There is no fixed point. 4. a. Cn = Cn–1 + 0.95

b. Cn = 2 + 0.95(n – 1) c. $76,500, $61,500

5. a. Mn = (Mn–1) × 1.048 b. Mn = 5000(1.048n) c. $5,755.11 d. 0.4%

e. Mn = (Mn–1) × 1.004 f. Mn = 5000(1.004n) g. $5,772.76 h. Yearly = $5,755.11. Monthly = $5,772.76 Monthly compounding produces $17.65 more interest than yearly

compounding.

Chapter 8 Answers 8

6.

Year 4.8% monthly 5% yearly 0 $5,000.00 $5,000.00 1 $5,245.35 $5,250.00 2 $5,502.74 $5,512.50 3 $5,772.76 $5,788.13

7. a. 74; 585 b. 63.25; 817.5 c. 3272, 51915

d.

5n!11

15

" = 585 ;

1.25n+ 44.5 = 817.51

15

! ;

!27n+ 3677 = 51,9151

15

"

8. After 19 years 9. a. Cn = Cn–1 + 14.95

b. Arithmetic, 14.95, 14.95 10. $117,463.15 11. Approximately 6.5% 12. a. 4

b. 67° c. 40°

13. a. 1.4238, or approximately 1.4 feet b. 47.457, or approximately 47.5 feet

14. –5.6, 1615

15. a. 16500(1.0530), or about $71,300 b. A little over $22,200

16. a. 1.14471 b. 2,318 c. After 33 hours

17. a. 3.85 meters b. 12.5 meters

Chapter 8 Answers 9

18. a. Sn = 43

Sn–1

b. Assuming that "now" means that sales are currently $8 million, predicted

sales in 5 years are 8 4

3!"#

$%&

5

, or about $33.7 million. In 10 years, about

$142 million. c. No is a reasonable answer because predictions are unlikely to remain accurate

for 10 years. 19. a. 4.85

b. 200(4.852) ≈ 4.7 billion

Chapter 8 Answers 10

Lesson 8.4

1. a. $93,070.22 b. $48,000 c. $45,070.22 d. $202,107.52, $93,070.22 e. $118,789.44, $93,070.22 f. $139,605.33, $93,070.22

2. a. 264 – 1, or approximately 1.8447 × 1019

b. Approximately 5.845 × 1011 years. 3. a. You must show that a one-disk puzzle can be completed in 21 – 1 = 1 moves,

which is obvious. b. Assume that a k-disk puzzle can be completed in 2k – 1 moves and prove that

a (k+1)-disk puzzle can be completed in 2k+1 – 1 moves. c. The recurrence relation states that the number of moves in a (k+1)-disk puzzle

is 2(2k – 1) + 1 = 2 × 2k – 2 + 1 = 2k+1 – 1. 4. a.

N (in months) Tn 0 2,000 1 2,110 2 2,220.55 3 2,331.65

b. Mn = (Mn–1) × 1.005 + 100 c. 2,443.31; 2,555.53

5. a.

T(in months) Tn 0 12,000 1 11,804 2 11,606.63 3 11,407.87

b. An = (An–1) × 1.007 – 286 c. It takes 52 months. d. 40,000; 40,000; 40,000 the fixed point


6. a. tn – tn–1 = –0.08(tn–1 – 70)

b. tn = 0.92 × (tn–1) + 5.6 c. 154.64 d. 70 degrees. It is the temperature of the room.

7. a. Mn = (Mn–1) × 1.1 – 2000 b. $9,352; $8,287.20 c. 10% d. $20,000. The interest would equal the payments, and the loan amount would

never decrease. 8. 13845(1.05) + 300 = $14,837.25

9. a. Tn = (Tn–1) × 0.6 + 450 b. 1076; 83

10. a. The values of k and P are 0.4 and 1125, respectively. Tn – Tn–1 = 0.4(1125 – Tn–1), or Tn – Tn–1 = 450 – 0.4Tn–1, or Tn = 450 + 0.6Tn–1.

b. 1125. It is the total population. 11. a.

b. You can see that the terms are approaching 2.

Term4.5

4

3.5

3

2.5

2

1.5

1

0.5

00 1 2 3 4 5 6 7 8 9 10

Term number


Lesson 8.5 1. a. –3; Tn = 4(2n–1) – 3; 2.5353 × 1030

b. 3.5, Tn = 1.5(3 n–1) + 3.5, 2.5769 × 1047

c. 53

; Tn = 13

(4 n–1) + 53

; 1.3391 × 1059

d. No fixed point; Tn = 2n – 6; 194

2. a. Bn = (Bn–1)

1+ 0.0812

!"#

$%&

+ 150

b.

Month Balance 0 150 1 301 2 453.01 3 606.03 4 760.07

c. –22,500

d. 22,801

1+ 0.0812

!"#

$%&

n'1

– 22,500, or 22,650

1+ 0.0812

!"#

$%&

n

– 22,500

e. $225,194.28 f. 472 months g. Approximately $333

3. a. Bn = (Bn–1) × 1.01 – 260 b.

n Bn

0 12,000 1 11,860 2 11,718.60 3 11,575.79 4 11,431.54

c. 26000 d. Bn = –14,140(1.01n–1) + 26,000, or Bn = –14,000(1.01n)+ 26,000 e. $8223.71 f. 62 months


g. 62 × 260 = $16,120; $4,120 h. $398.57 i. If Jilian borrowed this amount ($26,000), the interest would equal the

payments, and the loan balance would never decrease. 4. a. (t1 – p)(a1–1) + p = t1 – p + p = t1

b. tn = (t1 – p)(an–1) + p; tn+1 = (t1 – p)(an) + p c. tn+1 = tn(a) + b d. tn+1 = a((t1 – p)(an–1) + p) + b

= a(t1 – p)(an–1) + ap + b

= at1an–1 – pa × an–1 + ap + b = t1an – pan + ap + b

= (t1 – p)an + a b

1! a + b

= (t1 – p)an + ab+ b! ab

1! a

= (t1 – p)an + p 5. a. Tn = 0.92(Tn–1) + 5.6; The fixed point is 70.

b. Tn = 100(0.92n) + 70 c. 135.91 degrees d.

Minutes Temperature 0 170 1 162 2 154.64 3 147.87 4 141.64 5 135.91

e. No, because solving the equation requires taking the log of a negative number.

6. a. Pn = 0.5(Pn–1) + 200 b. 400 c. Pn = –320(0.5n–1) + 400 d. 399 people e. As with the previous disease problem, the fixed point of the recurrence

relation equals the population.


7. a. The following table shows the growth of the account.

Age Amount 25 1000 26 2070 27 3214.9 28 4439.943 29 5750.73901 30 7153.29074 31 8654.02109 32 10259.8026 33 11977.9888 34 13816.448 35 15783.5993 36 17888.4513 37 20140.6429 38 22550.4879 39 25129.022 40 27888.0536 41 30840.2173 42 33999.0325 43 37378.9648 44 40995.4923 45 44865.1768 46 49005.7392 47 53436.1409 48 58176.6708 49 63249.0377 50 68676.4704 51 74483.8233 52 80697.6909 53 87346.5293 54 94460.7863 55 102073.041 56 110218.154 57 118933.425 58 128258.765


59 138236.878 60 148913.46 61 160337.402 62 172561.02 63 185640.292 64 199635.112 65 214609.57

The amounts are generated with the mixed recurrence relation An = 1.07(An–1) + 1000. The final balance is approximately what is claimed in the article.


b. The following table tracks the account for 20 years

Age Amount 35 1000 36 2070 37 3214.9 38 4439.943 39 5750.73901 40 7153.29074 41 8654.02109 42 10259.8026 43 11977.9888 44 13816.448 45 15783.5993 46 17888.4513 47 20140.6429 48 22550.4879 49 25129.022 50 27888.0536 51 30840.2173 52 33999.0325 53 37378.9648 54 40995.4923 55 44865.1768 56 49005.7392 57 53436.1409 58 58176.6708 59 63249.0377 60 68676.4704 61 74483.8233 62 80697.6909 63 87346.5293 64 94460.7863 65 102073.041


The amounts are generated with the mixed recurrence relation An = 1.07(An–1) + 1,000. The 20-year balance is about what is claimed in the article.

c. 214,000

0.0812

!"#

$%&

, or about $1,427.


8. The following table shows the growth in Plan A.

Age Amount 21 $2,000.00 22 $4,180.00 23 $6,556.20 24 $9,146.26 25 $11,969.42 26 $13,046.67 27 $14,220.87 28 $15,500.75 29 $16,895.81 30 $18,416.44 31 $20,073.92 32 $21,880.57 33 $23,849.82 34 $25,996.31 35 $28,335.97 36 $30,886.21 37 $33,665.97 38 $36,695.91 39 $39,998.54 40 $43,598.41 41 $47,522.26 42 $51,799.27 43 $56,461.20 44 $61,542.71 45 $67,081.55 46 $73,118.89 47 $79,699.59 48 $86,872.56 49 $94,691.09 50 $103,213.28 51 $112,502.48 52 $122,627.70


53 $133,664.20 54 $145,693.97 55 $158,806.43 56 $173,099.01 57 $188,677.92 58 $205,658.94 59 $224,168.24 60 $244,343.38 61 $266,334.29 62 $290,304.37 63 $316,431.76 64 $344,910.62 65 $375,952.58

The first five years are generated with the mixed recurrence relation An = 1.09(An–1) + 2,000; the remaining years with the geometric recurrence relation An = 1.09(An–1).


The following table tracks the account in Plan B.

Age Amount 45 $2,000.00 46 $4,180.00 47 $6,556.20 48 $9,146.26 49 $11,969.42 50 $15,046.67 51 $18,400.87 52 $22,056.95 53 $26,042.07 54 $30,385.86 55 $35,120.59 56 $40,281.44 57 $45,906.77 58 $52,038.38 59 $58,721.83 60 $66,006.80 61 $73,947.41 62 $82,602.68 63 $92,036.92 64 $102,320.24 65 $113,529.06

The amounts are generated with the mixed recurrence relation An = 1.09(An–1) + 2,000.

Sasha is wrong. Plan A has over three times the amount that Plan B has. 9. If a = 1, the recurrence relation is arithmetic. An arithmetic sequence does not

have a fixed point (unless the constant difference is 0). The closed form for the recurrence relation can be found by applying previous results for arithmetic sequences.


Lesson 8.6 1.

2. a. 4, repelling

b. 2.5, neither c. 6, attracting d. 12, attracting

3. Attracting when |a| < 1, repelling when |a| > 1


4. a. 1 b. 3 c.

n tn 1 1 2 5 3 13

d. tn = 2tn–1 + 3 e. The fixed point is –3. It can be found algebraically by solving the equation

x = 2x + 3. 5. a. The population can be modeled by the recurrence relation Pn = 1.04Pn–1 + b,

where b is the annual harvest. The fixed point is

b1!1.04

= –25b. Thus, the

closed form is (12000 + 25b)1.04n–1 – 25b. Setting this equal to 10,000, assigning n the value 10, and solving for b gives approximately –669. Thus issuing 670 permits annually for 10 years would bring the population to about 10,000 in 10 years, assuming the each permit is fulfilled.

b. Approximately 400 permits a year would hold the population constant. c. The plans seems reasonable assuming the nearly all permits are fulfilled.

However, the decrease from 670 to 400 permits at the end of 10 years could prove unpopular with hunters.

6. a.

n tn 1 0 2 4 3 –4 4 –4 5 –4 6 –4


b.

c. The behavior is unpredictable. d. 2 is a fixed point. e. The terms diverge. f. 2, –4 g. Attracted when t1 = 0 or 2; repelled when t1 = 5; unpredictable when t1 = 1

7. a. tn = 1+ 0.1 1! tn!1

10"#$

%&'

"#$

%&'

tn!1

b. Approaches 10,000 but never reaches it c. The population approaches 10,000. d. 0 and 10,000 are the fixed points.


Chapter 8 Review 1. A reasonable summary should include the following points:

A description of recurrence relations, perhaps with a few examples, including recursive and closed-form representations.

A discussion of finite difference techniques for finding polynomial closed forms.

Arithmetic and geometric recurrence relations, including closed forms. Mixed recurrence relations, including fixed points and their role in closed

forms. Cobweb diagrams, repelling and attracting fixed points. Important applications, particularly financial applications.

2. a. Hn = Hn–1 + 4; Hn = 2 + 4(n – 1); 398

b. Hn = 3Hn–1 – 1; Hn = 52

3n–1 + 12

; 4.29 × 1047

c. Hn = (Hn–1) × 2; Hn = 3(2n–1); 1.9014759 × 1030

d. Hn = 2Hn–1 + 3; Hn = 2n – 3; 1.2676506 × 1030 3. a. Arithmetic

b. Neither c. Geometric d. Neither

4. a. 0; 0, 0, 0, 0 b. –0.75; –0.75, –0.75, –0.75, –0.75 c. No fixed point d. 1; 1, 1, 1, 1

5. a. Hn = 2(5n–1); 3.1554436 × 1069

b. Hn = 2.75(5n–1) – 0.75; 4.338735 × 1069 c. Hn = 2 + (–3)(n – 1); –295

d. Hn = (–2)n – 1 + 1; -6.338253 × 1029 6. a. Geometric

b. Neither c. Arithmetic d. Neither

7. a. On = 1.03 On–1 b. Cn = 0.982 Cn–1


8. a.

N Sn First Differences Second Differences Third Differences 1 1 — — — 2 5 4 — — 3 14 9 5 — 4 30 16 7 2 5 55 25 9 2

c. n3

3+ n2

2+ n

6; 204

9. Second degree; Hn = n2 – 6 10. a.

Day Gifts That Day Total Gifts 1 1 1 2 3 4 3 6 10 4 10 20 5 15 35 6 21 56

b. Gn = Gn–1 + n

Tn = Tn–1 + n2

2+ n

2

c. Gn = n2

2+ n

2

Tn =

n3

6+ n2

2+ n

3

11. a. Pn = Pn–1 + 0.21 b. Pn = 0.49 + 0.21(n – 1)

12. a.

Month Balance 0 $1,000 1 $1,004 2 $1,008.02 3 $1,012.05

b. Bn = 1.004(Bn–1)


c. Bn = 1000(1.004)n d. Solve the equation 1000(1.004)n = 2000, which gives 14 years, 6 months.

13. a.

Month Balance 0 $5,000 1 $5,126.67 2 $5,254.01 3 $5,382.03

b. Bn = (Bn!1 ) 1+ 0.064

12"#$

%&'+100

c. Bn = 23,750 1+ 0.064

12!"#

$%&

n

'18,750

d. $13,928.99 e. 200 months

14. a. Cn = Cn–1 + 3.50 b. Cn = 50 + 3.5n

c. 50+ 3.5!100

100= 4 or $4.00

15. a. Bn = 1.008(Bn–1) – 230 b. Bn = –17,750(1.008)n + 28,750 c. 61 months d. $3,030 e. $352.88

16. a. Rn = 0.9Rn–1 b. Rn = 1000(0.9n) c. About 6.58 minutes or 6 minutes, 35 seconds

17. a. Vn = 1.5Vn–1 + 4,000 b. Vn = 16,000(1.5)n–1 – 8000


18.

19. a. An = 0.4An–1 + 500

b. The amount of medication in the body stabilizes at 833 mg.

20

20

10

10 30


c. The cobweb would be attracted to the point (833.33, 833.33), which is the intersection of y = x and y = 0.4x + 500, as shown in this figure.

d. The amount in the body reaches the stable value (833 mg, in this case) more

quickly. The stable value is probably near the optimal dosage of the drug. 20. a. Cn = 0.8Cn–1 + 1

b. The daily concentration sequence is 2, 2.6, 3.08, 3.464, … .The concentration quickly exceeds the recommended maximum of 3 ppm and gradually approaches 5 ppm.

c. Sample answer: a daily addition of 0.4 ppm. The recurrence relation Cn = 0.8Cn–1 + 0.4 has a fixed point of 2. Daily additions of 0.4 ppm keep the concentration at a constant 2 ppm, which is the middle of the recommended range.

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