Chapter 8 Answers. Lesson a. This a bipartite graph. b. 1, 4, 9 c. Number of Couples

Similar documents
Lesson 26: Problem Set Sample Solutions

Write each expression as a sum or difference of logarithms. All variables are positive. 4) log ( ) 843 6) Solve for x: 8 2x+3 = 467

Review of Exponential Relations

The given pattern continues. Write down the nth term of the sequence {an} suggested by the pattern. 3) 4, 10, 16, 22, 28,... 3)

Modeling with Exponential Functions

Will Murray s Differential Equations, IV. Applications, modeling, and word problems1

Small Investment, Big Reward

M112 Short Course In Calculus V. J. Motto Spring 2013 Applications of Derivatives Worksheet

Lesson 12: Systems of Linear Equations

Materials: Hw #9-6 answers handout; Do Now and answers overhead; Special note-taking template; Pair Work and answers overhead; hw #9-7

y = b x Exponential and Logarithmic Functions LESSON ONE - Exponential Functions Lesson Notes Example 1 Set-Builder Notation

CHAPTER 6. Exponential Functions

Page Points Score Total: 100

Letter STUDENT NUMBER FURTHER MATHEMATICS. Written examination 2. Day Date

Final Review. Non-calculator problems are indicated. 1. (No calculator) Graph the function: y = x 3 + 2

Ch1 Algebra and functions. Ch 2 Sine and Cosine rule. Ch 10 Integration. Ch 9. Ch 3 Exponentials and Logarithms. Trigonometric.

JULY EXAMINATION 2015 MATHEMATICS GRADE 12 PAPER 1: LO 1, LO 2 PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY

Algebra II. Table of Contents. Sequences and Series. Slide 1 / 153. Slide 2 / 153. Slide 3 /

Lesson 8.2. Finite Differences

2.5 Powerful Tens. Table Puzzles. A Practice Understanding Task. 1. Use the tables to find the missing values of x:

Solve the equation. 1) x + 1 = 2 x. Use a method of your choice to solve the equation. 2) x2 + 3x - 28 = 0

Chapter 14: Basics of Functions

4.1 All in a Row (BC)

Lesson 3 - Linear Functions

Mathematics. Pre-Leaving Certificate Examination, Paper 1 Higher Level Time: 2 hours, 30 minutes. 300 marks L.17 NAME SCHOOL TEACHER

7-2 Solving Exponential Equations and Inequalities

10-1 Sequences as Functions. Determine whether each sequence is arithmetic. Write yes or no. 1. 8, 2, 12, 22

Name: Linear and Exponential Functions 4.1H

Math 0312 Intermediate Algebra Chapter 1 and 2 Test Review

The Method of Substitution. Linear and Nonlinear Systems of Equations. The Method of Substitution. The Method of Substitution. Example 2.

MATH NUMBER SENSE 7 Performance Objective Task Analysis Benchmarks/Assessment Students:

Algebra 2 Standards. Essential Standards:

GRADE 12 SEPTEMBER 2012 MATHEMATICS P1

Name Date Period AFM Final Exam Review: Part 2 (Standards 2.01, 2.02, 2.03, 2.04, and 2.05)

O5C1: Graphing Exponential Functions

Homework 2 Solution Section 2.2 and 2.3.

2. a b c d e 13. a b c d e. 3. a b c d e 14. a b c d e. 4. a b c d e 15. a b c d e. 5. a b c d e 16. a b c d e. 6. a b c d e 17.

Chapter 8. Sequences, Series, and Probability. Selected Applications

Precalculus Lesson 5.7: Financial Models Mrs. Snow, Instructor

Geometric Sequences and Series

Name Advanced Math Functions & Statistics. Non- Graphing Calculator Section A. B. C.

notes.notebook April 08, 2014

Sequences, Series, and Probability

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA II. Student Name:

MATH 112 Final Exam Study Questions

MATH 081. Diagnostic Review Materials PART 2. Chapters 5 to 7 YOU WILL NOT BE GIVEN A DIAGNOSTIC TEST UNTIL THIS MATERIAL IS RETURNED.

Chapter Two B: Linear Expressions, Equations, and Inequalities

Algebra 2 Mock Regents Exam

FURTHER MATHEMATICS. Written examination 1. Wednesday 30 May 2018

Inverse Variation Read 7.1 Examples 1-4

Chapter 8 Sequences, Series, and Probability

College Algebra Joysheet 1 MAT 140, Fall 2015 D. Ivanšić. Name: Simplify and write the answer so all exponents are positive:

Chapter Four: Linear Expressions, Equations, and Inequalities

Unit 2: Functions and Patterns

SABIS School Network

FURTHER MATHEMATICS Units 3 & 4 - Written Examination 2

MATH 415, WEEKS 14 & 15: 1 Recurrence Relations / Difference Equations

CHAPTER 4: Polynomial and Rational Functions

K&U = / 68. Appl = / 34. Comm = a 2 = b 2 + c 2 2ab cosa. MAP4C Final Exam Name:

( ). Switch x and y and solve for y:

Modelling Data Using Exponential Functions

Tools. What do you notice? understand the meaning of negative and zero exponents? multiply each power to get the next result? power of 2.

UNLV University of Nevada, Las Vegas

Algebra 2 Level 2. Final Exam June 2003

Identify the graph of a function, and obtain information from or about the graph of a function.

UNLV University of Nevada, Las Vegas

Recurrence Relations

ALGEBRA I SEMESTER EXAMS PRACTICE MATERIALS KEY SEMESTER 1. Selected Response Key

Complete Week 18 Package

Rational Expressions

4.2 write equations from 1 point and slope ink.notebook. November 14, page write equation from slope and a point. page 142.

( ) = 2 x + 3 B. f ( x) = x 2 25

Index No: Supervising Examiner s/ Invigilator s initial:

Directions: This is a final exam review which covers all of the topics of the course. Please use this as a guide to assist you in your studies.

MATH 1710 College Algebra Final Exam Review

Page: Total Points: Score:

Page Points Score Total: 100

Warm Up Lesson Presentation Lesson Quiz. Holt McDougal Algebra 2

Intermediate Algebra Chapter 12 Review

3 Inequalities Absolute Values Inequalities and Intervals... 4

Section 4.5. Using Exponential Functions to Model Data

Math 101 Final Exam Review Solutions. Eric Schmutz

Exponential and Logarithmic Functions

GEOMETRY (and Honors)

BEAULIEU COLLEGE PRELIMINARY EXAMINATIONS 2018 MATHEMATICS GRADE 12 PAPER 1

ALGEBRA I SEMESTER EXAMS PRACTICE MATERIALS SEMESTER Use the diagram below. 9.3 cm. A = (9.3 cm) (6.2 cm) = cm 2. 6.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA II. Thursday, August 16, :30 to 3:30 p.m., only.

Exponential and Logarithmic Functions

Lesson 18 - Solving & Applying Exponential Equations Using Logarithms

FUNCTIONS Families of Functions Common Core Standards F-LE.A.1 Distinguish between situations that can be

College Algebra. Word Problems

Sequences and Series. College Algebra

Pre-Test Unit 4: Exponential Functions KEY

MCF3M1 Exam Review. 1. Which relation is not a function? a. c. b. d. 2. What is the range of the function?

Review for Final Exam Show your work. Answer in exact form (no rounded decimals) unless otherwise instructed.

Polynomials and Rational Functions (2.1) The shape of the graph of a polynomial function is related to the degree of the polynomial

Algebra I EOC Review (Part 3)

MATHEMATICS: PAPER I (LO 1 AND LO 2) PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY

Chapter 11 Logarithms

Transcription:

Chapter 8 Answers Lesson 8.1 1. a. 2. a. This a bipartite graph. b. 1, 4, 9 c. Number of Couples d. 2n 1 e. H n = H n 1 + 2n 1 Number of Handshakes Recurrence Relation 1 1 2 4 H 2 = H 1 + 3 3 9 H 3 = H 2 + 5 4 16 H 4 = H 3 + 7 5 25 H 5 = H 4 + 9 Couples b. 2n 2. c. H n = H n 1 + 2n 2 3. a. i. H n = H n 1 + 3 Handshakes 1 0 2 2 3 6 4 12 ii. H n = (H n 1 ) 2 iii. H n = H n 1 + n iv. H n = (H n 1 ) n b. 16, 64, 36, 5040 Chapter 8 Answers 1

4. a. A given number of the sequence is the sum of the previous two numbers. The last number should be 5 + 8 = 13. b. t n = t n 1 + t n 2 5. a. 4 b. 1 c. 0 d. None 6. a. 6, 10 b. A n = A n 1 + n 1 7. a. 8. a. c. A n = d. 45 n(n! 1) 2 Term Number Number of First Differences Second Differences Handshakes 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 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 2

9. a. Levels Vertices 0 1 1 3 2 7 3 15 4 31 5 63 b. V n = V n 1 + 2 n 10. a. Term Number Number of Bees 0 5,000 1 5,600 2 6,272 3 7,024.64 4 7,867.60 b. B n = 1.12B n 1 c. After 27 years, in 2014 Chapter 8 Answers 3

11. 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. M n = (M n 1 ) 1.05 12. a. It is one-third the length of a segment in the previous figure. b. There are four times as many. c. L n = 4 3 L n!1 Chapter 8 Answers 4

Lesson 8.2 1. 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). 2. a. H n = 2n 2 5n b. H n = 0.29 + 0.23(n 1) c. H n = n 2 5n + 4 d. H n = 3 n 1. 3. a. T n = T n 1 + n 1 b. 0 n(n! 1) c. T n = 2 4. a. 1, 2, 11, 38, 119, 362 b. 2.5 5. a. D n = D n 1 + n 2 b. D n = n2 2! 3n 2 6. a. Row Number Number of Seats Total Seats b. S n = S n 1 + 2 c. S n = 24 + 2n 2 d. 37 1 24 24 2 26 50 3 28 78 4 30 108 5 32 140 6 34 174 e. See the table in the answer to part a. f. T n = T n 1 + 24 + 2n 2 g. T n = n 2 + 23n Chapter 8 Answers 5

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 8. a. b. V n = (V n 1 ) 1.05 c. The differences never become constant. Year b. In 2058 Deer 2010 50 2011 52 2012 54.08 2013 56.24 c. T n = 1.04T n 1 d. The differences never become constant. 9. A recurrence relation is S n = S n 1 + n. A closed-form formula is S n = 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 6

Lesson 8.3 1. a. i. Arithmetic 2. 0 ii. Geometric iii. Geometric iv. Neither v. Geometric vi. Neither b. i. H n = H n 1 + 3 ii. H n = H n!1 2 iii. H n = (H n 1 ) 1.2 iv. H n = H n 1 + H n 2 v. H n = (H n 1 ) 0.1 vi. H n = H n 1 + n 1 c. i. H n = 2 + 3(n 1) ii. H n = 64(0.5 n 1 ) iii. H n = 10(1.2 n 1 ) v. H n = 0.3(0.1 n 1 ) 3. There is no fixed point. 4. a. C n = C n 1 + 0.95 b. C n = 2 + 0.95(n 1) c. $76,500, $61,500 5. a. M n = (M n 1 ) 1.048 b. M n = 5000(1.048 n ) c. $5,755.11 d. 0.4% e. M n = (M n 1 ) 1.004 f. M n = 5000(1.004 n ) 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 7

6. 7. a. 74; 585 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 b. 63.25; 817.5 c. 3272, 51915 15 d. " 5n! 1 = 585 ;! 1.25n + 44.5 = 817.5 ; "!27n + 3677 = 51,915 1 8. After 19 years 9. a. C n = C n 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, 16 15 15. a. 16500(1.05 30 ), 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 15 1 15 1 Chapter 8 Answers 8

18. a. S n = 4 3 S n 1 b. Assuming that "now" means that sales are currently $8 million, predicted sales in 5 years are 8 4 5! $ " # 3% &, 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(4.85 2 ) 4.7 billion Chapter 8 Answers 9

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. 2 64 1, or approximately 1.8447 10 19 b. Approximately 5.845 10 11 years. 3. a. You must show that a one-disk puzzle can be completed in 2 1 1 = 1 moves, which is obvious. 4. a. 5. a. b. Assume that a k-disk puzzle can be completed in 2 k 1 moves and prove that a (k+1)-disk puzzle can be completed in 2 k+1 1 moves. c. The recurrence relation states that the number of moves in a (k+1)-disk puzzle is 2(2 k 1) + 1 = 2 2 k 2 + 1 = 2 k+1 1. N (in months) T n 0 2,000 1 2,110 2 2,220.55 3 2,331.65 b. M n = (M n 1 ) 1.005 + 100 c. 2,443.31; 2,555.53 T(in months) T n 0 12,000 1 11,804 2 11,606.63 3 11,407.87 b. A n = (A n 1 ) 1.007 286 c. It takes 52 months. d. 40,000; 40,000; 40,000 the fixed point Chapter 8 Answers 10

6. a. t n t n 1 = 0.08(t n 1 70) b. t n = 0.92 (t n 1 ) + 5.6 c. 154.64 d. 70 degrees. It is the temperature of the room. 7. a. M n = (M n 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. T n = (T n 1 ) 0.6 + 450 b. 1076; 83 10. a. The values of k and P are 0.4 and 1125, respectively. T n T n 1 = 0.4(1125 T n 1 ), or T n T n 1 = 450 0.4T n 1, or T n = 450 + 0.6T n 1. 11. a. b. 1125. It is the total population. Term 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 6 7 8 9 10 Term number b. You can see that the terms are approaching 2. Chapter 8 Answers 11

Lesson 8.5 1. a. 3; T n = 4(2 n 1 ) 3; 2.5353 10 30 b. 3.5, T n = 1.5(3 n 1 ) + 3.5, 2.5769 10 47 c. 5 3 ; T n = 1 3 (4 n 1 ) + 5 ; 1.3391 1059 3 d. No fixed point; T n = 2n 6; 194! 2. a. B n = (B n 1 ) 1 + 0.08 $ " # 12 % & + 150 b. c. 22,500 Month Balance 0 150 1 301 2 453.01 3 606.03 4 760.07! d. 22,801 1 + 0.08 $ " # 12 % & n'1 e. $225,194.28 f. 472 months g. Approximately $333 3. a. B n = (B n 1 ) 1.01 260 b.! 22,500, or 22,650 1 + 0.08 $ " # 12 % & n 22,500 n B n 0 12,000 1 11,860 2 11,718.60 3 11,575.79 4 11,431.54 c. 26000 d. B n = 14,140(1.01 n 1 ) + 26,000, or B n = 14,000(1.01 n )+ 26,000 e. $8223.71 f. 62 months Chapter 8 Answers 12

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. (t 1 p)(a 1 1 ) + p = t 1 p + p = t 1 b. t n = (t 1 p)(a n 1 ) + p; t n+1 = (t 1 p)(a n ) + p c. t n+1 = t n (a) + b d. t n+1 = a((t 1 p)(a n 1 ) + p) + b = a(t 1 p)(a n 1 ) + ap + b = at 1 a n 1 pa a n 1 + ap + b = t 1 a n pa n + ap + b = (t 1 p)a n + a b 1! a + b ab + b! ab = (t 1 p)a n + 1! a = (t 1 p)a n + p 5. a. T n = 0.92(T n 1 ) + 5.6; The fixed point is 70. b. T n = 100(0.92 n ) + 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. P n = 0.5(P n 1 ) + 200 b. 400 c. P n = 320(0.5 n 1 ) + 400 d. 399 people e. As with the previous disease problem, the fixed point of the recurrence relation equals the population. Chapter 8 Answers 13

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 Chapter 8 Answers 14

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 A n = 1.07(A n 1 ) + 1000. The final balance is approximately what is claimed in the article. Chapter 8 Answers 15

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 Chapter 8 Answers 16

The amounts are generated with the mixed recurrence relation A n = 1.07(A n 1 ) + 1,000. The 20-year balance is about what is claimed in the article.! c. 214,000 0.08 $ " # 12 % &, or about $1,427. Chapter 8 Answers 17

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 Chapter 8 Answers 18

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 A n = 1.09(A n 1 ) + 2,000; the remaining years with the geometric recurrence relation A n = 1.09(A n 1 ). Chapter 8 Answers 19

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 A n = 1.09(A n 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. Chapter 8 Answers 20

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 Chapter 8 Answers 21

4. a. 1 b. 3 c. n t n 1 1 2 5 3 13 d. t n = 2t n 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 P n = 1.04P n 1 + b, b where b is the annual harvest. The fixed point is = 25b. Thus, the 1! 1.04 closed form is (12000 + 25b)1.04 n 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. 6. a. 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. n t n 1 0 2 4 3 4 4 4 5 4 6 4 Chapter 8 Answers 22

b. c. The behavior is unpredictable. d. 2 is a fixed point. e. The terms diverge. f. 2, 4 g. Attracted when t 1 = 0 or 2; repelled when t 1 = 5; unpredictable when t 1 = 1 " " 7. a. t n = 1 + 0.1 1! t n!1 % % # $ # $ 10 & ' & ' t n!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 Answers 23

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. H n = H n 1 + 4; H n = 2 + 4(n 1); 398 b. H n = 3H n 1 1; H n = 5 2 3n 1 + 1 ; 4.29 1047 2 c. H n = (H n 1 ) 2; H n = 3(2 n 1 ); 1.9014759 10 30 d. H n = 2H n 1 + 3; H n = 2 n 3; 1.2676506 10 30 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. H n = 2(5 n 1 ); 3.1554436 10 69 b. H n = 2.75(5 n 1 ) 0.75; 4.338735 10 69 c. H n = 2 + ( 3)(n 1); 295 d. H n = ( 2) n 1 + 1; -6.338253 10 29 6. a. Geometric b. Neither c. Arithmetic d. Neither 7. a. O n = 1.03 O n 1 b. C n = 0.982 C n 1 Chapter 8 Answers 24

8. a. c. n 3 N S n 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 3 + n2 2 + n 6 ; 204 9. Second degree; H n = n 2 6 10. a. b. G n = G n 1 + n Day Gifts That Day Total Gifts 1 1 1 2 3 4 3 6 10 4 10 20 5 15 35 6 21 56 T n = T n 1 + n2 2 + n 2 c. G n = n2 2 + n 2 T n = n3 6 + n2 2 + n 3 11. a. P n = P n 1 + 0.21 12. a. b. P n = 0.49 + 0.21(n 1) Month b. B n = 1.004(B n 1 ) Balance 0 $1,000 1 $1,004 2 $1,008.02 3 $1,012.05 Chapter 8 Answers 25

c. B n = 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. B n = (B n!1 ) 1 + 0.064 % # $ 12 & ' + 100! c. B n = 23,750 1 + 0.064 $ " # 12 % & d. $13,928.99 e. 200 months 14. a. C n = C n 1 + 3.50 b. C n = 50 + 3.5n c. 50 + 3.5! 100 100 n = 4 or $4.00 15. a. B n = 1.008(B n 1 ) 230 ' 18,750 b. B n = 17,750(1.008) n + 28,750 c. 61 months d. $3,030 e. $352.88 16. a. R n = 0.9R n 1 b. R n = 1000(0.9 n ) c. About 6.58 minutes or 6 minutes, 35 seconds 17. a. V n = 1.5V n 1 + 4,000 b. V n = 16,000(1.5) n 1 8000 Chapter 8 Answers 26

18. 20 10 10 20 30 19. a. A n = 0.4A n 1 + 500 b. The amount of medication in the body stabilizes at 833 mg. Chapter 8 Answers 27

1000 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. 800 600 400 200 0 0 200 400 600 800 1000 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. C n = 0.8C n 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 C n = 0.8C n 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. Chapter 8 Answers 28