IB Diploma Program Mathematics Course Companion Higher Level [1 ed.] 0199129347, 9780199129348, 9780198390121

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OXFORD UNIVERSITY PRESS

Great Clarendon Street, Oxford OX2 6DP Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide in Oxford New York Auckland Cape Town Dar es Salaam Hong Kong Karachi Kuala Lumpur Madrid Melbourne Mexico City Nairobi New Delhi Shanghai Taipei Toronto With offices in Argentina Ausn·ia Brazil Chile Czech Republic France Greece Guatemala Hungary Italy Japan Poland Portugal Singapore South Korea Switzerland Thailand Turkey Ukraine Vietnam © Oxford University Press 2012 The moral rights of the author have been asserted Database right Oxford University Press (maker) First published 2012 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any meam, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, or under terms agreed with the approprate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above You must not circulate this book in any other binding or cover and you must impose this san1e condition on any acquirer British Library Cataloguing in Publication Data Data available ISBN: 978-0-19-912934-8 10 9 8 7 6 5 4 3 2 Printed in Italy by Rotolito Lombarda

Acknowledgments We are grateful to the following for permission to reprint copyright material: Guardian News and Media Ltd for '618 is the Magic Number' by Marcus Chown, The Guardian, 16.1.2003, copyright© Guardian News & Media Ltd 2003. Professor Alan Tennant from the Helmholtz Association of German Research Centres for extract from press release 'Golden ratio discovered in a quantum world', 7 January 2010. Internet Systems Consortium, hlc (!SC) for figure: 'The growth of internet domains on the world wide web since 1994' from www.isc.org. Although we have made every effort to trace and contact copyright holders before publication this has not been possible in all cases. If notified, the publisher will rectify any errors or omissions at the earliest opportunity. The publishers would like to thank the following for permission to reproduce photographs: P3: Mary Evans Picture Library/Alamy; P4: Professor Peter Goddard/Science Photo Library; P12: Photo Researchers, Inc./ Science Photo Library; P24: Science Photo Library; P46: Associated Sports Photography/Alamy; P46: Trinity Mirror/ Mirrorpix/Alamy; P46: Sheila Terry/Science Photo Libraiy; P47: Dvmsimages/Dreamstime.Com: P47: Science Photo Library; P19: Taily_Sindariel/Dreamstime; PS1: New York Public Library Picture Collection/Science Photo Library; P94: Af Archive/

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Course Companion definition The IB Diploma Programme Course Companions are resource materials designed to provide students with support through their two-year course of study. These books will help students gain an understanding of what is expected from the study of an IB Diploma Programme subject. The Course Companions reflect the philosophy and approach of the IB Diploma Programme and present content in a way that illustrates the purpose and aims of the IB. They encourage a deep understanding of each subject by making connections to wider issues and providing opportunities for critical thinking. The books mirror the IB philosophy of viewing the curriculum in terms of a whole-course approach; the

use of a wide range of resources; international­ mindedness; the IB learner profile and the IB Diploma Programme core requirements; theory of knowledge, the extended essay, and creativity, action, service (CAS). Each book can be used in conjunction with other materials and indeed, students of the IB are required and encouraged to draw conclusions from a variety of resources. Suggestions for additional and further reading are given in each book and suggestions for how to extend research are provided. In addition, the Course Companions provide advice and guidance on the specific course assessment requirements and also on academic honesty protocol.

1B mission statement The International Baccalaureate aims to develop inquiring, knowledgable and caring young people who help to create a better and more peaceful world through intercultural understanding and respect. To this end the IB works with schools, governments and international organizations to develop challenging

programmes of international education and rigorous assessment. These programmes encourage students across the world to become active, compassionate, and lifelong learners who understand that other people, with their differences, can also be right.

The 1B Learner Profile The aim of all IB programmes is to develop internationally minded people who, recognizing their common humanity and shared guardianship of the planet, help to create a better and more peaceful world. IB learners strive to be: Inquirers They develop their natural curiosity. They

acquire the skills necessary to conduct inquiry and research and show independence in learning. They actively enjoy learning and this love of learning will be sustained throughout their lives. Knowledgable They explore concepts, ideas, and

issues that have local and global significance. In so doing, they acquire in-depth knowledge and develop understanding across a broad and balanced range of disciplines.

dignity of the individual, groups, and communities. They take responsibility for their own actions and the consequences that accompany them. Open-minded They understand and appreciate their

own cultures and personal histories, and are open to the perspectives, values, and traditions of other individuals and communities. They are accustomed to seeking and evaluating a range of points of view, and are willing to grow from the experience. Caring They show empathy, compassion, and respect

towards the needs and feelings of others. They have a personal commitment to service, and act to make a positive difference to the lives of others and to the environment.

Risk-takers They approach unfamiliar situations and

Thinkers They exercise initiative in applying thinking skills critically and creatively to recognize and approach complex problems, and make reasoned, ethical decisions.

uncertainty with courage and forethought, and have the independence of spirit to explore new roles, ideas, and strategies. They are brave and articulate in defending their beliefs.

Communicators They understand and express ideas and information confidently and creatively in more than one language and in a variety of modes of communication. They work effectively and willingly in collaboration with others.

Balanced They understand the importance of intellectual, physical, and emotional balance to achieve personal well-being for themselves and others.

Principled They act with integrity and honesty, with a

strong sense of fairness, justice, and respect for the

Reflective They give thoughtful consideration to their

own learning and experience. They are able to assess and understand their strengths and limitations in order to support their learning and personal development.

A note on academic honesty It is of vital importance to acknowledge and appropriately credit the owners of information when that information is used in your work. After all, owners of ideas (intellectual property) have property rights. To have an authentic piece of work, it must be based on your individual and original ideas with the work of others fully acknowledged. Therefore, all assignments, written or oral, completed for assessment must use your own language and expression. Where sources are used or referred to, whether in the form of direct quotation or paraphrase, such sources must be appropriately acknowledged.

How do I acknowledge the work of others? The way that you acknowledge that you have used the ideas of other people is through the use of footnotes and bibliographies. Footnotes (placed at the bottom of a page) or endnotes (placed at the end of a document) are to be provided when you quote or paraphrase from another document, or closely summarize the information provided in another document. You do not need to provide a footnote for information that is part of a "body of knowledge". That is, definitions do not need to be footnoted as they are part of the assumed knowledge. Bibliographies should include a formal list of

the resources that you used in your work. "Formal" means that you should use one of the several accepted forms of presentation. This usually involves separating the resources that you use into different categories (e.g. books, magazines, newspaper articles, Internet-based resources, CDs and works of art) and providing full information as to how a reader or viewer of your work can find the same information. A bibliography is compulsory in the extended essay.

What constitutes malpractice? Malpractice is behavior that results in, or may result in, you or any student gaining an unfair advantage in one or more assessment component. Malpractice includes plagiarism and collusion. Plagiarism is defined as the representation of the

ideas or work of another person as your own. The following are some of the ways to avoid plagiarism: • Words and ideas of another person used to support one's arguments must be acknowledged. • Passages that are quoted verbatim must be enclosed within quotation marks and acknowledged. • CD-ROMs, email messages, web sites on the Internet, and any other electronic media must be treated in the same way as books and journals. • The sources of all photographs, maps, illustrations, computer programs, data, graphs, audio-visual, and similar material must be acknowledged if they are not your own work. • Works of art, whether music, film, dance, theatre arts, or visual arts, and where the creative use of a part of a work takes place, must be acknowledged. Collusion is defined as supporting malpractice by another student. This includes: • allowing your work to he copied or submitted for assessment by another student • duplicating work for different assessment components and/ or diploma requirements. Other forms of malpractice include any action that gives you an unfair advantage or affects the results of another student. Examples include, taking unauthorized material into an examination room, misconduct during an examination, and falsifying a CAS record.

About the book The new syllabus for Mathematics Higher Level is thoroughly covered in this book. Each chapter is divided into lesson-size sections with the following features: Investigations

Exploration suggestions

f Examiner's tip ._

[ _o id_yo_ _ u_k_no_ w_?__ ) __.

[ Theory of Knowledge ) Historical exploration

The Course Companion will guide you through the latest curriculum with full coverage of all topics and the new internal assessment. The emphasis is placed on the development and improved understanding of mathematical concepts and their real life application as well as proficiency in problem solving and critical thinking. The Course Companion denotes questions that would be suitable for examination practice and those where a GDC may be used. Questions are designed to increase in difficulty, strengthen analytical skills and build confidence through

understanding. Internationalism, ethics and applications are clearly integrated into every section and there is a TOK application page that concludes each chapter. It is possible for the teacher and student to work through the book in sequence but there is also the flexibility to follow a different order. Where appropriate the solutions to examples using the TI-Nspire calculator are shown. Similar solutions using the TI-84 Plus and Casio fx-9860GII are included on the accompanying interactive CD which includes a complete ebook of the text, extension material, GDC support, a glossary, sample examination papers, and worked solution presentations. Mathematics education is a growing, ever changing entity. The contextual, technology integrated approach enables students to become adaptable, lifelong learners. Note: US spelling has been used, with IB style for mathematical terms.

About the authors Josip Harcet has been teaching the IB programme for 20 years. After teaching for 11 years at different international schools he returned to teach in Zagreb. He has served as a curriculum review member, deputy chief examiner for Further Mathematics, assistant and senior examiner, as well as a workshop leader. Lorraine Heinrichs has been teaching IB mathematics for the past 12 years at Bonn International School. She has been the IB DP coordinator since 2002. During this time she has also been senior moderator for HL Internal Assessment and workshop leader for the IB. She was also a member of the curriculum review team. Palmira Mariz Seiler has been teaching mathematics for 22 years. She joined the IB community 11 years ago and since then has worked as Internal Assessment moderator, in curriculum review working groups, and as a

workshop leader and deputy chief examiner for HL mathematics. Marlene Torres-Skoumal has been teaching IB mathematics for the past 30 years. She has enjoyed various roles with the IB over this time, including deputy chief examiner for HL, senior moderator for Internal Assessment, calculator forum moderator, and workshop leader. A special thanks to Jim Fensom for the GDC chapters and contribution to the Prior Learning chapter.

Contents Chapter 1 Mathematics as the science of patterns

1.1 1.2 1.3 1.4 1.5 1.6 1.7

2

Number patterns: sequences, series and sigma notation Arithmetic sequences and series Geometric sequences and series Conjectures and proofs Mathematical induction Counting methods The binomial theorem

Chapter 2 Mathematics as a language

2.1 2.2 2.3 2.4

Relations and functions Special functions and their graphs Operations with functions Transformations of graphs of functions

Chapter 3 The long journey of mathematics

3.1 3.2 3.3 3.4 3.5 3.6

Introduction to complex numbers Operations with complex numbers Polynomial functions: graphs and operations Polynomial functions: zeros, sum and product Polynomial equations and inequalities Solving systems of equations

5 10 15 24 25 31 38

5.7

48

6.1

50 54 70 79 96

97 109 118 131 140 153

Chapter 4 Modeling the real world

166

4.1 4.2 4.3 4.4

168 180 189

4.5 4.6 4.7 4.8 4.9

Limits, continuity and convergence The derivative of a function Differentiation rules Exploring relationships between JJ' andf" Applications of differential calculus: kinematics Applications of differential calculus: economics Optimization and modeling Differentiation of implicit functions Related rates

205 208 211 215 218 221

Chapter 5 Aesthetics in mathematics 232

5.1 5.2 5.3

Recursive functions Properties of exponents and logarithms Euler's number and exponential functions

5.4

234

5.5 5.6

5.8

Invariance and the exponential function - a different approach to Euler's number 248 Logarithms and bases 249 Logarithmic functions and their behavior 258 Derivatives of exponential and 261 logarithmic functions Angles, arcs and areas 267

Chapter 6 Exploring randomness

278

Classification and representation of statistical data 6.2 Measures of central tendency 6.3 Measures of dispersion 6.4 Theoretical probability 6.5 Probability properties 6.6 Experimental probability 6.7 Conditional probability 6.8 Independent events 6.9 Probability tree diagrams 6.10 Bayes' theorem

280 288 291 299 306 308 312 318 321 326

Chapter 7 The evolution of calculus

342

7.1 7.2 7.3

Integration as anti-differentiation 344 Definite integration 352 Geometric significance of the definite 355 integral

Chapter 8 Ancient mathematics and modern methods

8.1

The right-angled triangle and trigonometric ratios 8.2 The unit circle and trigonometric ratios 8.3 Compound angle identities 8.4 Double angle identities 8.5 Graphs of trigonometric functions 8.6 The inverse trigonometric functions 8.7 Solving trigonometric equations 8.8 The cosine rule 8.9 The sine rule 8.10 Area of a triangle Chapter 9 The power of calculus

238

9.1 9.2

243

9.3

382

384 389 398 401 403 409 412 415 418 423 434

Derivatives of trigonometric functions 436 Related rates of change with trigonometric expressions 450 Integration of trigonometric functions 455

Integration by substitution Integration by parts Special substitutions Applications and modeling

461 466 472 480

Chapter 10 Modeling randomness

494

9.4 9.5 9.6 9.7

10.1 Discrete random variables and distributions 10.2 Binomial distribution 10.3 Poisson distribution 10.4 Continuous random variables 10.5 Normal distribution 10.6 Modeling and problem solving

Chapter 12 Multiple perspectives in mathematics

496 503 513 520 532 544

556 563 571 583 592 596 599 613 628

630 12.1 Complex numbers as vectors 12.2 Complex plane and polar form 633 12.3 Operations with complex numbers in 638 modulus-argument form 12.4 Powers and roots of complex numbers: De Moivre's theorem and applications 643 12.5 Mathematical connections 650 Chapter 13 Exploration

13.1 13.2 13.3 13.4 13.5 13.6 13.7

About the exploration Internal assessment criteria How the exploration is marked Academic honesty Record keeping Choosing a topic Getting started

1 2 3 4

Number Algebra Geometry Statistics

Chapter 15 Practice Papers

Chapter 11 Inspiration and formalism 554

11.1 Geometric vectors and basic operations 11.2 Introduction to vector algebra 11.3 Vectors, points and equations of lines 11.4 Scalar product 11.5 Vector (cross) product and properties 11.6 Vectors and equations of planes 11.7 Angles, distances and intersections 11.8 Modeling and problem solving

Chapter 14 Prior learning

660

660 661 666 666 667 668 669

Practice paper 1 Practice paper 2

672

673 697 719 745 754

754 757

Answers

760

Index

811

The material on your CD-ROM includes the entire student book as an eBook, as well as a wealth of other resources specifically written to support your learning. On these two pages you can see what you will find and how it will help you to succeed in your Mathematics Higher course. The whole print text is presented as a user-friendly eBook for use in class and at home. Extra content can be found in the Contents

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written

as

1

+

+

1

+ 3

2

1

+…+ 4

+

…

r

1

where

is

the

general

term.

r

The

general

is

3r

is

finite.

The

is

but

how

( ∑)

you

2

of

only

the

take

notation

write

1

+

+

term

can

sigma

1

1

r

the

1

the

is

a

series

values

+

6

+

1

9

to

form

+

7

to

12

+

15

because

represent

+

the

a

18

+

21

series

series.

Here

series

1

+…+ 4

3

from

compact

1

+ 3

finite

+

…

+

using

sigma

notation:

largest

20

r

r

can

take

20

1

∑ r

=1

r

The

r

smallest

can

value

take

1

Read

this

as

‘The

summation

of

from

r

equals

1

to

20.’

r

∑

4

1

So

r

=1

1

1

=

∑

r

+

1

1

1

=

3

3

3

1

+

2

1

3

1

+

+

+ 27

9

3

3

the

capital

4

3

is

series

4

+

8

+

12

+

16

+

….

+

48

can

letter

sigma,

81

which

The

Greek

1

+

be

written

as

is

used

represent

‘the

to

sum

12

of ’.

4

×

1

+

4

×

2

+

4

×

3

+

4

×

4

+

…

+

4

×

12

=

∑

Leonhard

Euler

4r (1706–83)

was

the

r =1

rst

use

mathematician

this

to

notation.

Chapter

1

5

Example

Write

the



next

three

terms

and

1 a

2,

4,

8,

16,

…

,

b

the

1

,

12

1

1

,

6

term:

1

,

2

general

1

…

c

1,

,

4

2

20

1

,

,

…

8

Answers

a

For

the

sequence

the

next

are

32,

The

64,

2

,

2

in

3

,

2

2,

the

8,

in

16,

the

...

sequence

sequence

4

,

4,

terms

128.

terms

1

2

three

2

can

be

written

r

,

…

,

2 ,

…

r

General

where

term

r

can

is

2

take

the

values

1 b

For

the

1

,

sequence

1

terms

1

are

42

1

×

be

2

×

, ...

12

the

next

Look

for

three

20

written

1

1

,

2

...

56

can

1

,

ter m.

,

sequence

1

3,

general

1

,

30

The

,

6

2,

the

1

,

2

1,

1

Find

1

,

3

3

×

,

4

4

×

... ,

5

,

r

×

... the

patter ns

in

the

denominators.

+ 1)

(r

1

So

the

general

term

is

r

where

r

can

take

the

For

the

1

1 ,

sequence

1

1

are

1

,

1

32

1

,

1

2

,

,

1

1

next

three

1

1

...

64

1

1

,

2

1

,

3

2

,

32

,

1

2

the

8

,

16

,

0

2

…

written

1

,

1

,

=

, ...

4

be

1

8

3,

64

can

,

4

2,

,

sequence

1

1,

1

,

16

The

+ 1)

1

,

2

terms

(r

values

1 c

×

1

,

4

2

,

5

2

...

,

,

6

2

r

2

... 0

1

Since

the

sequence

has

general

term

the

is r

general

ter m

r

Example

Write

the

can

take

the

r

values

1,

2,

3,

(3r

first

three

terms

of

each

∞

series.

+

b

6)

∑

=1

r

(

1)

2

r

=1

Answers

7

a

∑ r

(3r

+

6)

=

9

+ 12

+ 15

+

...

Substitute

r

=

1

for

r

r

∑ r

(–1)

2

=

for

r

=1

2

=

(–1)

×

1

2

+

(–1)

2

×

2

4

–

3

+

(–1)

2

×

6

the

first

ter m,

=1

2

for

the

second

∞

b

3

+

1



r

r

ter m,

…

∞

∑

first

2

7

a

the

is

1

2

where

in

1

1

The

2

2

…

Mathematics

=

–1

as

+

the

9

+

science

…

of

patterns

the

third

ter m.

ter m,

and

r

=

3

represents

innity.

Example



F ind

Write

these

series

using

sigma

and a

2

+

b

–8

12

+

22

+

32

+

…

+

the

general

term

notation.

the

value

of

r

that

102

gives

the

last

term.

1

+

4

–

2

+

1

–

+

…

2

Answers

a

2

+

12

+

22

+

32

+

…

+

102

The

general

ter m

is

10r

–

8

11

For =

(10r

–

∑ r

the

ter m:

102

8)

10r

=1

8

=

102

10r

–

=

110

r

=

11

1 b

–8

+

4

–

2

+

1

–

+

…

Alter nating

signs

suggest

that

you

2 8

need

8

1

=

to

multiply

each

ter m

by

–1.

2

(–1)

×

+

(–1)

×

0

1

2

2

n

8 3

+

(–1)

8

is

positive

when

n

is

even

×

+

(–1)

×

2

3

2

negative

2

when

n

is

odd.

8

8 r

5

+

×

(–1)

and

4

(–1)

+

…

The

general

ter m

( −1)

is

×

4

r

2

1

2

∞

8 r

=

∑

(

1)

1

r

2

=1

r

Exercise

1

×

Write

1A

the

next

three

terms

for

each

1 a

–6,

–4.5,

–3,

–1.5,

…

3 ,

b

2

2

Write

the

general

term

for

1 a

2,

6,

12,

20,

…

1 ,

b

3

Given

terms

that

of

r

can

the

take

the

sequence

5

a

1 ,

6

...

c

3

4r

–

3

1 ,

1 ,

15

35

,

...

63

1 ,

,

17

1,

...

–1,

c

1,

3,

5,

…

26

2,

3,

general

…,

write

the

first

four

term:

r a

1 ,

8

1 ,

10

values

with

7 ,

sequence.

1 ,

2

5 ,

4

each

sequence.

1

b

c 2

2r

4

Expand

these

series

in

+1

r

full.

4

5

5

r a

∑ r

5

r (r

+ 1)

b

∑ r

=1

Write

these

a

–1

+

3

+

b

–1

+

1

–

c

6

–

12

+

series

7

1

+

+

24

using

11

1

–

–

48

+

1

r

c

=1

2r

∑

( − 1)

2

r

+1 r

sigma

=1

notation.

…

+

+

1

96

–

–

1

+

1

–

1

+

1

192

Chapter

1

7

Investigation

–

Use

at

your

GDC

to

look

quadratic

the

numbers

2

T

Enter

Use

this

3

+

data

table

second

by

2n

–

the

a

generated

by

+

n

=

sequences

in

to

n

where

the

list

.

GDC:

the

differences

quadratic



∈

rst

of

the

differences

numbers

and

the

generated

formula.

F irst

Second

difference

difference

2

n

–

2n

+

3

2 F irst

}

3

are

obtained

2

}

}

subtracting

2

}

consecutive

}

the

original

2

} 11

38

2

}

}

Second

differences

three

consecutive

integers

are

p

−

p

1,

and

p

+

values

to

to

calculate

justify

that

the

n

−

2n

second

+

3

for

each

difference

is

1.

the

one.

always

see

what

happens

if

you

repeat

the

task

rst

subtracting

terms

of

differences.

constant.

A

Now

by

consecutive

2

algebra

are

13

obtained

Use

sequence.

9

}

these

of

2

}

27

Use

terms

7

18

general

two

5

}

In

by

3

11

51

2

}

}

6

differences

1

general

form

for

any

using: quadratic

equation

2

T

=

2n

+

2n

+

1

2

is

T

=

an

+

bn

+

c

2

T

=

–n

+

3n

–

4 where

T ry

to

generalize

your

a,

b

and

c

are

results. constants.

Investigation

The

patterns

Enter

GDC

these

or

in

of

dots

triangular

represent

numbers

a

–

in

the

rst

numbers

ve

triangular

your

spreadsheet

and

then

T riangle work

out

the

rst

and

numbers.

numbers

F irst

difference

differences.

can

continue

consecutive

in

the

2

numbers

as

}

6

shown

ndings

sequences

from

the

}

a

investigation

formula

that

}

}

} }

to

6

the 21

triangular

8

numbers.

Mathematics

as

the

science

of

patterns

=

15

+

6

1

5

quadratic

generates

1

4

21

nd

1

3

}

10

table.

your

}

3

generating

15 Use

difference

1

} Y ou

Second

second

6

=

5

+

1

1

Investigation

The

the

diagrams

show

pentagonal,

lists

and

these

the

the

more

rst

method

of

terms

and

differences

of

the

to

patterns

the

square,

heptagonal

obtain

numbers.

formulae

that

Use

generate

numbers

Pentagonal

Hexagonal

your

numbers

numbers

Heptagonal

numbers

results

from

the

T erm

T riangular

Square

numbers

numbers

Pentagonal

Hexagonal

numbers

numbers

Heptagonal

Octagonal

Nonagonal

Use

ve

hexagonal

number

numbers.

Square

Use

the

–

the

investigations

to

complete

this

table.

1st

2nd

3rd

4th

5th

6th

1

3

6

10

15

21

1

4

9

16

25

1

5

12

1

6

15

7th

8th

nth

numbers

numbers

numbers

table

to

make

a

conjecture

about

the

nth

term

of

any

polygonal

number .

Chapter

1

9

.

The

Arithmetic

picture

arches

To

on

find

level

you

This

gives

Each

➔

will

If

5

common

you

write

count

arches

20

is

ari thmetic

If

to

of

a

series

building

with

5,

the

10,

more

will

arches

be

on

on

the

each

four th

level.

15.

than

between

constant

first

two

then

progression .

and

dierence

the

there

the

previous

one,

so

the

next

arches.

difference

sequence

front

arches

sequence

has

have

the

many

need

the

the

and

level.

how

first

floor

floor

represents

each

out

sequences

term

u

consecutive

this

We

is

an

call

denote

then

ari thmetic

this

it

the

numbers

constant

by

in

a

sequence

difference

or

an

the

d.

sequence

of

arches

is

1

This

u

=

5

=

5

is

recursive

1

u

+

5

=

u

2

+

d

Y ou

1

=

u

10

+

5

=

u

3

+

d

any

can

term

is

found

from

the

previous

one

by

adding

the

common

equation

know

difference

u

=

u

n

+

n

Look

–

again

=

5

=

5

d

1

at

the

terms:

1

+

5

=

u

2

u

=

10

+

5

=

u

=

15

+

5

=

u

d

+

d

=

u

+

d

=

u

2

4

This

+

1

3

u

3

leads

to

the

+

2d

+

3d

1

1

general

term

u

=

u

n

➔

An

arithmetic

sequence

+

(n

–

1)d

1

with

the

first

term u

and

common

1

difference

d

is

u

,

1

The

general

term

u

+

d,

u

1

is

Example

Three

Their

Find

sum

2d,

. . .

,

u

=

u

u

+

(n

–

1)d.

1

+

(n

–

1)d

in

an

1



numbers

the

+

1

n

is

are

48

three

consecutive

and

their

terms

product

is

arithmetic

sequence.

2800.

numbers.

Answer

Let

the

u

–

d,

u

–

d

three

u,

+

u

u

+

+

numbers

be

d

u

+

d

=

48

3u

=

48

u

=

16

Write

the

sum

{

10

out

using

or

only

can

if

this

you

generate

d.

the

u

equation.

work

term

a

2

Each

u

called

Mathematics

as

the

science

of

patterns

of

the

three

Continued

numbers.

on

next

page

previous

term.

(u

–

d) u(u

+

d)

2

=

Write

2800

u (u

–

d

)

2

=

2800

=

175

–

d

by =

±

d

=

±

The

7,

256

numbers

Find

the

a

50,

b

a,

47,

16

and

divide

two

values

of

d

give

two

possible

sequences:

7,

16,

25

or

25,

16,

7



number

3a,

are

25.

Example

=

16.

The

and

u

175

9

three

16

product.

Substitute

2

16

d

the

2

44,

5a,

of

...,

...,

terms

in

these

arithmetic

sequences:

14

21a

Answers

u

a

=

50,

=

53

d

=

−3

Using

u

1

=

u

n

u

−

+

(n

−

1)d

1

3n

n

53

−

3n

=

14

3n

=

39

n

=

13

Using

u

=

14

and

solving

for

n

n

u

b

=

a,

=

(2n

d

=

2a

1

u

−

1)a

n

(2n

−

1)a

=

21a

2an

=

22a

n

=

11

Example

The



second

Find

the

term

of

common

an

arithmetic

difference

and

sequence

the

first

is

15

term

and

of

the

the

fifth

term

is

21.

sequence.

Answer

u

=

2

u

+

d

=

15

Using

u

1

=

5

u

= n

+

4d

=

u

+

(n

−

1)d

1

21

1

3d

=

d

u

u

=

=

6

Solving

simultaneously

2

15

−

2

=

13

1

Exercise

1

2

Find

a

5,

c

a,

Find

a

2,

1B

the

11,

a

+

the

11,

nth

17,

2,

−1,

20,

c

3,

7,

+

of

these

sequences.

…

4,

a

10,

b

+

6,

3,

– 4,

–11

…

indicated

in

each

...

15th

term

...

12th

term

of

these

arithmetic

sequences:

3

, 4

23,

terms

1 b

a

term

, 2

11,

...

(n

+

1)th

term

Chapter

1

11

The

3

four th

common

The

4

four th

first

A

would

initial

per

is

arithmetic

–5.

an

40.

Find

sequence

the

arithmetic

Find

the

first

is

term

sequence

common

18

and

and

is

0

the

the nth

and

difference

term.

the

and

the

to

of

sum

be

hold

of

Here

S

=

S

=

an

was

Gauss

5050.

rising

salar y

have

Gauss

the

year

states

by

after

this

that

the

annual

15

carries

increments

years?

position

job

How

for

a

a

of

many

50%

salar y

€500.

years

salar y

of

How

much

would

increase

a

on

the

salar y?

sum

surprise

as

of

term

the

person

find

term

is

adver tisement

€48000

When

an

term.

job

The

of

difference

four teenth

5

term

1

11

the

years

the

how

+

+

old

numbers

gave

is

100

arithmetic

he

his

did

teacher

from

correct

series

1

to

challenged

100.

answer

2

+

3

+

4

+

...

99

+

+

98

+

97

+

...

2

+

101

+

101

+

...

101

+

=

101

+

101

2S

=

101

×

100

+

the

to

teacher’s

immediately

it:

99

2S

T o

almost

him

100

1

101

10 100 * 2

Carl

Friedrich

(1777–1855)

German

mathematician

The

numbers

1

to

100

are

an

arithmetic

series

with

1

and

method

common

for

calculating

Generalizing

term

u

difference

and

this

the

method

common

1.

sum

for

a

Gauss

of

a

finite

series

difference

d

had

found

is

queen

of

all

sciences”

a

arithmetic

containing n

“Mathematics

first the

term

Gauss

series.

terms,

with

first

gives

1

S

=

u

n

S

=

u

n

2S

u

+

(n

=

–

1)d

+

u

+

+

(n

–

1)d

+

+

(n

+

(n

–

–

2)d

+

(n

–

1)d

1)d]

n

[2u

n

+

(n

–

1)d]

1

2

12

+

Mathematics

as

the

science

u

+

of

+

2u 1

1

=

S

+

2d

+

…

+

1

1

n[2u

u

u

+

patterns

(n

–

2)d

+

1

(n

–

3)d

+

…

+

u

+

(n

+

d

+

(n

–

1)d

+

…

+

2u 1

1)d

u

1

+

–

1

u

1

2u

1

=

d

1

2u

n

+ 1

1

n

2S

+

1

+

(n

1

–

1)d

+

2u 1

+

(n

–

1)d

This

formula

can

also

be

written

as

n

S

=

[u

n

+

u

1

+

(n

–

1)d ]

1

u

2

+

(n

–

1)d

=

u

1

n

n

=

[u

+

u

1

]

n

2

➔

The

sum

of

a

finite

arithmetic

series

is

n

S

=

[

2u

n

+

(n

–

1)d ]

1

2

n

(u

=

+ u

1

)

n

2

where

n

is

the

number

of

terms

in

the

series, u

is

the

first

1

term,

is

u

the

last

term

and

d

is

the

common

difference.

n

Example

The

first

series

Find



term

has

the

9

of

an

arithmetic

series

is

2

and

the

last

term

is

26.

The

terms.

sum

of

the

series.

Answer

u

=

2,

u

1

=

26,

n

=

9

9

n

9

S

=

(2

+

26)

=

Using

126

S

=

(u

n

9

Example

The

first

The

sum

Find

+

u

1

) n

2

2



term

the

of

of

the

an

arithmetic

series

number

of

is

series

is

25

and

the

four th

term

is

13.

–119.

terms

in

the

series.

Answer

u

=

25

+

3d

1

u u

=

=

u

4

13

+

3d

1

1

Find 3d

=

13

d

=

–4

–

d.

25

n

S

=

[2u

n

+

(n

–

1)d]

1

2

n

–119

=

[50

–

4(n

–

1)]

Solve

for

n.

2

–238

=

n(54

–

4n)

2

4n

–

54n

–

238

=

0

2n

–

27n

–

119

=

0

(2n

+

=

0

Divide

by

2. The

natural

numbers



2

7)(n

–

17)

is

the

set

{0,

Factorize.

1,

2,

...}

It

+

differs

from



because

+ +

Since

n

∈



,

n

=

17



is

the

set

of

positive

integers. it

includes

zero.

Chapter

1

13

Example



10

Calculate

∑ r

(5r

− 7)

=1

Answer

u

=

2

1

d

=

u

5

=

43

10

10

S

=

(

2

+

43)

=

Use

205

the

for mula

for

the

sum

of

a

10

2

finite

arithmetic

series

or

Using

a

GDC:

10

∑ r

(5r

− 7)

Exercise

1

2

=

207

=1

1C

Evaluate

a

6

b

52

+

c

–78

these

19

+

+

41

–

32

+

82

series.

+

30

–

…

+

110

+

…

–

25

–

90

–

…

86

–

142

Calculate:

10 15

a

 r

(5r

 7)

b

∑ r

3

Find

first

the

sum

term

is

EXAM-STYLE

4

The

four th

The

sum

Find the

5

The

(5 − 3r )

1

sum

of

60

an

and

arithmetic

the

10th

series

term

is

with

16

terms,

given

that

–3.

QUESTIONS

term

the

first

of

of

=1

of

first

five

an

an

arithmetic

five

terms

is

series

is

8.

25.

terms.

arithmetic

series

is

given

by S

=

n(2n

+

3).

Find

n

the

14

common

Mathematics

difference

as

the

and

science

of

the

first

patterns

four

terms

of

the

series.

the

.

The

Geometric

diagram

named

shows

after

described

the

it

in

sequences

the

Polish

rst

three

steps

mathematician

in

and

series

constructing

Waclaw

Sierpinski’ s

Sierpinski

who

triangle,

Claudia

Zaslavsky

(1917–2006)

rst

Ethnomathematician

1915.

Researched

expressions

of

mathematics

African

1

If

you

2

count

sequence

the

the

1,

number

3,

of

white

9,

27

white

3

triangles

Figure

5

triangles

in

each

would

is

of

the

have

three

gures

81

times

white

the

1

to

4

you

triangles.

number

in

get

At

the

words

the

each

stage

previous

the

sides

of

the

original

triangle

are

1

unit

long ,

then

the

length

of

stage.

each

number

and

reckoning

games,

Africa If

culture,

including

4

and

green

triangle

will

time,

patterns.

Counts

published

the

signs,

of

was

side in

1973.

1

1

of

in

be

unit;

each

side

of

the

orange

triangles

will

be 4

2 1

unit

and

each

side

of

the

blue

triangles

is

unit. 8

2

If

the

area

of

the

rst

triangle

is

1

unit

,

then

the

area

of

the

green

triangle

1 2

is

unit

,

since

four

of

the

green

triangles

make

up

the

original

triangle.

The

4 1 2

area

of

each

orange

triangle

will

be

unit

and

of

each

blue

triangle

will

be

16 1 2

unit 64

Hidden

in

Sier pinski’s

triangle

are

the

sequences: The

1,

3,

9,

27,

...

 To

get

next

term

multiply

previous

term

by

ratio

of

3 consecutive

1

1,

1 ,

2

,

4

multiply

previous

term

To

get

next

term

multiply

previous

term

4

ratio

this

call

the

is

of

a

this

two

consecutive

geometric

ratio

sequence

the

1,

3,

terms

9,

or

sequence

common

27,

ratio

=

…u

in

1,

a

a

geometric

and

r

sequence

=

denote

3

and

it

the

is

constant

progression .

by

r.

recursive A

For

u

equation

–

1

sequence:

=

1

=

1

one

in

which

the

r

n

this

u

recursive

is: is

=

n

each

by

1

u

for

by

1 , ...

equation

constant

sequences.

the

We

for

term

64

then

So

next

1 ,

If

get

8

16

➔

To

...

a

2

1 ,

1,

is

,

4

1

terms

1

1

next

term

as

function

a

is

dened

of

ear lier

terms.

1

u

×

3

=

u

2

×

r

×

r

1

2

u

=

3

×

3

=

u

3

=

u

2

×

r

×

r

1

3

u

=

9

×

3

=

u

4

×

r

=

u

3

1

n

This

leads

to

the

general

term

u

=

n

➔

In

a

geometric

sequence

u

×

–

1

r

1

with

first

term u

and

common

ratio

r

Why

have

these

values

1

n–1

the

general

term

is

given

by

u n

=

u

×

r

of

,

r

≠

–1,

0,

r

been

omitted?

1.

1

Chapter

1

15

Example

Find

two

a

the



common

terms

10,

of

25,

1

of

these

sequences

and

write

the

next

sequence.

...

18

3

2a

each

...

6

a,

of

1

,

2

c

each

62.5,

1

,

b

ratio

5

,

4a

,

...

Answers

25 a

r

=

= 2.5 10

The

next

two

1 b

r

=

terms

1

−

156.25

and

=

−

2

3

1

The

next

390.625

1

÷

6

are

two

terms

1

are

and

54

162

3

2a 2

c

=

r

=

2a

a

7

The

next

Example

Find

the

two

2,

4,

b

5,

10,

are

8a

9

and

16a



number

a

terms

8,

...,

of

terms

in

each

of

these

geometric

sequences.

256

k

20,

...,

5

×

2

=

2

×

128

=

5

×

2

Answers

a

u

=

2,

=

2

r

=

2

1

n−1

u

×

2

7

=

2

×

2

n

n

b

=

u

8

=

5,

r

=

2

1

n−1

u

=

5

×

2

k

n

n

=

k

+

Exercise

1

Write

1

1D

down

the

6th

term

and

the

nth

term

3

a

2

1,

Find

2,

4,

the

…

b

common

9,

ratio

3,

1

and

...

the

c

terms

sequences.

a

48,

24,

16

,

b

3

16

12,

8

−

...

10th

,

5th

term

4

, 9

Mathematics

...

term

27

as

the

science

of

patterns

x

of

each

sequence.

2

,

x

,

x,

…

indicated

in

each

of

these

Find

3

the

number

of

terms

in

each

of

these

sequences.

1

0.03,

a

0.06,

0.12,

...,

1.92

81,

b

27,

9,

...., 81

The

4

third

term

Find two

term

5

6

in

first

is

What

term

is

of

of

geometric

The

sum

How

could

values

a

the

numbers

common

geometrical

of

the

sequence

common

is

2

and

ratio

the

and

fifth

the

is

18.

second

case.

The

a

a

possible

each

The

9.

of

a

geometric

value

−

4,

a

of

+

the

2

sequence

seventh

and

progression.

3a

+

Find

1

is

and

the

fifth

term

term?

are

the

16

three

two

consecutive

possible

values

terms

of

the

ratio.

of

a

you

geometric

find

the

sum

of

series

the

first

10

terms

of

the

series

n–1

1

+

3

+

9

+

…

+

+

3

…

?

Write

2

S

=

1

+

3

+

3

3

+

3

9

+

…

+

3

10

Multiply

the

whole

series

by

3

and

subtract

from S:

2

S

=

1

+

3

+

3

3

+

3

3

+

3

+

3

9

+

…

+

3

+

…

+

3

10

2

3S

=

3

9

10

+

3

–

3

10

10

(1

–

3)S

=

1

10

This

gives

10

1

3

1 10

S

=

=

(3

− 1)

10

1

To

by

3

find

2

the

sum

multiplying

of

by

the

the

first

n

terms

common

of

a

ratio r

geometric

and

series

2

S

=

+

u

n

1

r

+

use

the

same

process

subtracting:

3

+

r

r

n

+

…

+

1

u

3

rS

=

+

u

n

–

1

r

1

1

u

1

n

+

…

1

+

u

r

–

1

n

+

1

u

r

1

n

(1

–

r)S

=

➔

The

–

u

n

1

sum

of

a

u

r

1

geometric

series

is

given

by

n

u

(1

r

)

1

S

=

, r

≠ 1

n

1

where

n

r

is

the

number

of

terms,

u

is

the

first

term

and

r

is

the

1

common

ratio.

Chapter

1

17

Investigation

In

the

diagram,

The

string

half

is

AB

–

infinite

represents

a

sums

piece

of

string

11.28

cm

long.

is

cut

in

half

and

one

half,

CD,

is

placed

underneath.

again

cut

in

half

and

one

half,

DE,

is

placed

next

repeated

twice

more

and

the

total

length

of

the

pieces

A

to

CD.

placed

The

The

side

process

by

AB

=

11.28 cm

CD

=

5.64 cm

CD

+

D

E

DE

DE

=

2.82 cm

C

D

E

CD

CD

C

D

E

F

If

this

BUT

process

can

is

never

continued

exceed

indenitely,

11.28

cm,

the

+

DE

=

+

8.46 cm

the

=

EF

=

9.87 cm

+

+

FG

1.41 cm

DE

+

EF

=

total

original

length

length.

=

0.705 cm

will

continue

u

=

=

5.64

DE

u

=

1

=

5.64

×

2

2

3

2

⎛ 1 ⎞

⎛ 1 ⎞ EF

u

=

=

5.64

×

3

FG ⎜

u

=

5.64

× ⎜

4

⎟ 2

⎝

=

⎝

⎠

n

⎟ 2

⎠

1

⎛ 1 ⎞

u

=

5.64

× ⎜

n

⎟ 2

⎝

⎠

1

This

is

a

geometric

sequence,

rst

term

5.64

and

common

ratio

2

The

sum

of

n

terms

of

this

series

will

therefore

be

n

⎛

5.64

(1

r

⎞

⎛ 1 ⎞ 1

⎜

⎟

⎟

⎜

n

u

⎜

⎟ ⎝ 2 ⎠

)

⎝

⎠

1

S

=

=

n

1

r

1 ⎞

⎛ 1

⎜

⎟ 2

⎝

Enter

as

n

Note

ver y

this

on

gets

that

the

when

close

geometric

18

as

the

string,

to

⎠

see

what

happens

bigger .

to

you

and

series

is

number

the

series

geometric

Mathematics

sum

11.28,

Convergent

value

GDC

series

as

the

the

15

terms

original

of

terms

does

not

science

of

to

is

series

when

gets

converges

result

length.

divergent

convergent

the

the

bigger.

a

sum S

converge

patterns

it

sum

In

=

the

tends

to

a

finite

investigation

11.28

increasing

Mathematically:

1

CD

10.575 cm

G FG

a

noted.

F EF

If

is

is

D

C

the

side

B

C

A

remaining

cm.

is divergent

with

Investigation

In

at

order

the

to

–

understand

formula

for

the

convergent

the

condition

sum

of

n

for

series

convergence

now

look

terms:

n

u

(1

r

)

1

S

= n

(1

r )

n

Use

your

GDC

to

calculate

these

values

+

r

of

for

n

∈



1

,

≤

n

≤

10: What

happens

as

n

n

3

⎛ n

⎞

n

3

a

increases?

2

b

c

⎜

⎟ 2

n

n

1

⎛ d

⎜

e

⎟

values

your

f

⎞

⎜

⎟ 4

⎠

r

of

results

3

⎛

⎟ 5

⎝

other

Use

n

⎞

⎜

5

T ry

1

⎛

⎞

to

justify

these

statements:

n

●

r

>

⇒ r

1

increases

n

as

gets

larger .

The

larger

the

value

of

r,

the

n

faster

the

value

of

r

increases.

n

●

0

r




2)

7k

+

1

then

2

2k

∴

–

6

7(k

>

+

0

since

1)

+

1

k

+

>

+

(2k

–

6)

k

8

(2k

7k

–

6)

>

7(k

+

1)

+

+

+

2k

1

+

+

1

2k

>

+

1

1

2

It

∴

follows

P (k

Since

+

1)

P (8)

proved

is

the

by

(k

+

1)

>

7(k

+

1)

+

1

tr ue.

was

that

follows

that

that

if

the

shown

P (k)

is

to

tr ue,

principle

statement

is

be

of

true

tr ue,

P (k

+

and

1)

is

it

was

also

mathematical

for

all

positive

also

tr ue,

it

induction

integers

n

≥

8.

Chapter

1

27

Example

Use

the



principle

of

mathematical

induction

to

prove

that

n

2

∑

r ( 2r

+ 6)

=

n (n

+ 1)( n

+ 5)

3 r =1

Answer

n

2

P

=

r ( 2r

∑

n

+ 6)

=

n (n

+ 1)( n

+ 5)

3 r =1

P

:

1

LHS

=

1

×

8

=

8

2

RHS =

× 1 ×

2

×

6

=

8

3

LHS

=

RHS

⇒

P

is

tr ue.

1

Assume

P

is

tr ue,

i.e.,

Add

on

the

next

number

in

k

k

the

2

sequence

+

∑

r ( 2r

+ 6)

=

k (k

+ 1)( k

+ 5) ,

k

∈



3

(k

r =1

Show

P

is

tr ue,

+

1)(2(k

+

1)

+

6)

i.e.,

k+1

k +1

2

∑

r (r

+ 1)

=

(k

+ 1)(( k

+ 1) + 1)(( k

+ 1) + 5)

Using

the

assumption

from

3 r =1

step

2

=

(k (k

+ 1)( k

+ 2 )( k

2.

+

1)

is

a

common

factor.

+ 6)

3

LHS:

k +1

k

⎛

∑

r ( 2r

+ 6)

⎞

=

∑

⎜ ⎝

r =1

r

r ( 2r

+ 6)

Expand

+ u ⎟

and

simplify.

k +1

⎠

=1

2

=

k (k

+ 1)( k

+ 5) + ( k

+ 1)( 2( k

+ 1) + 6 )

3

using

the

assumption

2

=

k (k

+ 1)( k

+ 5) + ( k

+ 1)( 2k

+ 8)

This

is

the

expression

3

obtained ⎛ 2

= (k

k (k

+ 5) + ( 2k

+ 8)

⎜

substituted

⎟ 3

⎠

2

=

(k

+ 1)( k

+ 2 )( k

+ 6)

3

=

RHS:

Since

was

P (k

P

(1)

also

+

1)

the

28

was

shown

proved

is

principle

n

=

⎞

+ 1)

⎝

when

also

of

that

true,

if

it

to

is

Mathematics

true

as

for

the

tr ue,

is

follows

mathematical

statement

be

P (k)

all

and

by

the

induction

values

science

it

tr ue,

of

of

that

n

≥

1.

patterns

in

P(n).

k

+

1

is

Exercise

Use

1

1H

mathematical

series

with

first

induction

term

u

and

to

prove

common

that

for

ratio

r,

a

geometric

the

sum

of

the

1

first

n

terms

S

is

given

by

n

n

u

(1

r

)

1

S

 n

1

Use

2

r

mathematical

induction

to

prove

these

statements.

n

n 2

a

r

∑

=

( n + 1)( 2 n + 1) 6

r =1

n

r

b

1

n

2

∑

=

2

−1

r =1

2

n 3

3

1

c

+

3

2

+

3

3

+

. . .

+

n

2

=

(n

+

1)

4

n

n d

r (r

∑

+ 2)

=

( n + 1)( 2n + 7 ) 6

r =1

Example



2n

2n

Use

by

mathematical

8

for

all

n

∈

induction

to

prove

that

3

+

7

is

This

divisible

is



a

means

that

multiple

of

3

+

7

8.

Answer

2n

P (n):

Step

3

+

7

=

8A

1

0

When

so

n

P (0)

Step

=

is

0,

LHS

=

+

7

=

8

tr ue.

Recall

2

value

that

include

Assume

k

that

≥

0,

P (k)

k

∈

is

tr ue

for

some

n

=

0

the

so

we

natural

must

numbers

star t

+

+

Step

7

=

8A,

A

∈



3

Prove

that

P (k

+

1)

is

2(k+1)

3

tr ue.

+

+

7

=

8B,

B

∈



Proof:

Using

2(k+1)

assumption

2k

3

+

7

=

9

=

9(8A

×

3

=

72A

=

8(9A

=

8B

+

–

–

7)

63

7

+

+

7

7

+

Since

tr ue,

P (k)

it

with

0.



2k

3

3

P (0)

and

is

it

was

statement

numbers

is

to

proved

+

the

mathematical

9A

B

P (k

by

7)

shown

was

tr ue,

follows

–

1)

is

also

principle

for

all

a

∈



,

positive

+

since

A

∈



integer.

be

that

induction

tr ue

is

7

if

tr ue,

of

that

the

natural

n

Chapter

1

29

Example



+

Prove

that

for

all

values

of

n

∈



3

,

n

+

5n

is

a

multiple

of

6.

Answer

3

Let

P (n)

=

n

+

5n

×

1

3

P (1)

=

1

+

Statement

Assume

5

is

tr ue

P (k)

is

+

k

∈

6

for

tr ue

n

=

for

1.

k

≥

1,

3



Use

=

then

k

+

assumption

5k

to

=

6A,

show

A

that

∈

P (k

3

(k

+



+

1)

+

1)

+

5(k

+

1)

=

6B,

B

∈



LHS

3

2

=

k

+

=

6A

3k

+

3k

+

1

+

5k

+

5

+

5k

3

2

–

5k

+

3k

+

3k

+

1

+

Using

5

k

=

6A

–

5k

from

step

2.

2

=

6A

+

3(k

=

6A

+

3[k (k

=

6(A

=

6B

=

RHS

if

+

2)

1)

+

2]

k(k

C)

it

was

it

+

1)

is

was

also

tr ue,

integers

is

to

proved

P (k

follows

mathematical

statement

shown

by

+

is

for

is

=

of

2C,

any

even

numbers

C

and

is

∈

two

the

also



since

consecutive

sum

of

two

even.

also

principle

induction

tr ue

even

tr ue,

2

that

1)

the

be

+

product

integers

P (1)

P (k)

tr ue,

k

+

the

Since

and

+

+

all

that

of

the

positive

n

Exercise

1I

n

1

Use

mathematical

induction

to

prove

that

7

–

n

∈

1

is

divisible

+

by

2

6

for

Prove

all

by

n

∈



mathematical

induction

that

2

1

+

3

+

5

+

7

+

…

+

(2n

–

1)

=

+

n

for



n

3

Prove

by

mathematical

induction

that

9

induction

that n

−

1

is

a

multiple

–

n

is

a

multiple

+

of

8

for

n

∈



3

4

Prove

by

mathematical

of

+

for

n

∈



n

1

5

Show

using

mathematical

induction

that

r =1

30

Mathematics

as

the

science

of

patterns

n =

 r

r

+ 1

n +1

6

Prove

6

by

mathematical

n+2

values

Find

7

of

n,

2

the

first

2u

1

induction

that

for

all

positive

integer

2n+1

+

five

3

is

terms

exactly

of

the

divisible

sequence

by

7.

given

by u

=

1,

1

r

u

=

r+1

3

n

⎛

Prove

using

mathematical

induction

that

u

=

− ⎜

n

⎝

.

Counting

Some

a

number

of

Look

⎧

at

the

u

problems

involve

large

counting

Factorial

1

⎟ 3

⎠

methods

mathematical

combinations

2 ⎞

3

about

numbers

arrangements

so

you

will

and

need

to

develop

techniques.

notation

first

four

terms

of

this

sequence.

= 1

0

:

u n

⎨ =

u ⎩

u

×

n

u

n

=

1

=

1

1

n

0

u

×

u

1

=

1

=

2

×

1

=

3

×

2

0

u

=

2

×

u

=

3

×

u

2

1

u 3

×

1

2

The

u

=

general

n

×

(n

term

–

1)

of

×

(n

this

–

2)

sequence

×

…

×

3

is:

×

2

×

1

n

This

A

simpler

way

to

denote

this

sequence

is

to

use

factorial

is

read

as

notation u

equals

n

factorial’.

n

where

u

=

n!

n

It

follows

that

u

=

0!

=

1

0

Working

with

Here

the

are

large

first

numbers

few

is

factorial

easier

using

factorial

notation.

numbers.

Christian

0!

=

1

1!

=

1

Kramp

(1760–1826),

=

1

×

French

2!

=

2

×

1

=

3!

=

3

×

2

×

a

0!

2

×

mathematician,

1!

introduced

1

=

3

×

factorial

2! notation.

4!

=

➔

4

n!

×

3

=

×

n

2

×

×

(n

1

–

=

1)

4

×

×

3!

(n

–

2)

×

…

×

3

×

2

×

1

=

n

×

(n

–

1)!

8!

Y ou

can

use

this

patter n

to

calculate

expressions

such

as 6!

8!

8 × 7 ×

6 × 5 × 4 × 3 × 2 ×1

= 8

=

6!

×

7

= 56

6 × 5 × 4 × 3 × 2 ×1

Chapter

1

31

Example



10 !

×

5!

Evaluate 7!

×

6!

Answer

10 !

×

5!

10

×

9 ×

8 ×

7! ×

5!

=

7!

×

6!

7!

10 ×

×

9 ×

6

×

5!

8

=

6

=

120

Example



Simplify

(n

+ 1) !

(n

a

+ 1) !

+

n !

b

(n

1) !

(n

1) !

Answers

(n

+ 1) !

(n

+ 1) × n

× (n

− 1) !

= a (n

1) !

(n

= n (n

(n

+ 1) !+

1) !

+ 1)

n !

Rewrite

(n

+

1)!

b (n

1) !

as

(n

+ 1) n ( n

− 1) ! +

n (n

(n

and

(n

+ 1) n

+

1)

×

n

×

(n

–

1)!

− 1) !

=

(n

+

n!

=

n

×

(n

–

1)!

1) !

n

= n (n

=

+ 2) )

1

Exercise

1

Copy

8!

and

–

10!

complete

–

4!

95!

–

94!

1)!

–

simplifying

the

expressions.

n!

Evaluate:

4 !

5!

a

×

8!

3!

b

×

6!

c

6!

3

table

9!

–

+

this

7!

5!

(n

2

1J

5!

6!

Simplify:

2

n !

+

( n − 1) !

n !

a

−

(n

2) !

QUESTION

2

( 2n + 2 ) !( n !) 4

Show

2( 2n

+ 1)

=

that 2

[ (n

32

Mathematics

as

+ 1) !]

the

!)

1!

c

( n + 1) !

EXAM-STYLE

(n

( n − 1) !

b

(n

( 2n ) !

science

of

+ 1)

patterns

n !

+ 1

Arrangements

Alma

lays

the

eggcup,

a

arrange

them

each

glass

person,

three

breakfast

Having

from,

Here

from

made

which

are

a

cup

she

the

left

her

different

person.

decides

that

is

can

her

Christmas

each

choice

leaves

on

she

Here

she

first

for

so

thinks

objects.

then

the

a

row ,

since

different

Starting

and

in

table

there

her

with

She

for

people

cannot

arrange

are

6

six

decide

them

ways

with

how

differently

of

an

to

for

arranging

reasoning:

choose

she

to

Day

is

the

left

one

eggcup,

with

way

two

to

the

glass

objects

choose

or

to

the

the

cup.

choose

third

object.

arrangements:

E

EGC

CEG

ECG

So

CGE

there

are

3

represents

eggcup

GEC

GCE

×

2

×

1

=

6

ways

of

arranging

three

distinct

objects

in

G

represents

glass

C

represents

cup

a

row .

Similarly

with

choosing

the

of

2

choosing

ways

object,

This

in

n

of

the

–

objects

object

a

n

1)

total

can

distinct

×

(n

the

–

The

number

row

is

of

be

×

of

arrange

for

each

Having

third

4

×

3

×

…

×

ways

can

3

of

×

2

of

1

=

a

row

these

the

and

×

to

be

2

in

chosen

object

extended

objects

2)

to

and

second.

choosing

reasoning

(n

➔

first

giving

which

×

four

deduce

1

=

two

way

that

in

of

the

a

are

4

there

ways

are

objects

3

of

ways

there

choosing

different

arranged

×

ways

first

one

24

there

the

are

last

arrangements.

number

row

of

ways

is

n!

arranging

n

distinct

objects

in

The

a

different

which

objects

arranges

angels,

ways.

a

some

snowman

She

reasons,

indistinguishable

this.

Here

AASB

AABS

can

is

a

list

AASB

Christmas

and

a

however,

the

of

all

of

possible

ASBA

decorations

She

that

number

ABSA

AABS

bell.

can

since

in

a

arrange

the

different

two

line:

them

two

in

angels

4!

identical

=

are

permutations

24

is

less

than

arrangements:

ABSA

ASBA

BSAA

BSAA

SABA

A

SABA

ABAS

BAAS

BAAS

SAAB

SAAB

ASAB

ASAB

BASA

BASA

SBAA

SBAA

–

angel

Fan

Rong

B

–

bell

Chung

S

–

snowman

(1949–)

K

Graham

Taiwanese

mathematician

many

The

arrangements

arrangements

differently .

in

on

Because

the

the

the

left-hand

right

angels

column

because

are

the

are

two

identical

only

different

angels

there

are

are

only

from

shown

12

ways

problems

combinatorics

easily

could

arranging

the

four

“...

from

were

explained,

get

quickly

of

be

called

are

arrangements

ABAS

the

in

n!

arranged

Alma

ways

into

but

you

them

getting

objects.

out

was

hand

often

ver y

... ”

Chapter

1

33

The

number

of

ways

of

arranging

4

objects,

two

of

which

are

the 2!

4 !

same,

is

is

ways

 12



2!

No

The

of

number

of

arranging

the

2 1 two

➔

the

4  3  2 1

number

of

permutations

of

n

objects,

k

of

which

objects.

questions

will

be

are

asked

n!

identical

identical

in

your

exam

is: about

permutations

of

k !

identical

Alma

three

decides

arrange

or naments

choose

ways

to

her

so

Alma

first

there

to

choose

or nament

are

changes

only

3

her

×

2

=

mind

two

from

in

6

3

(snowman,

ways

and

different

again

or naments.

dierent

and

bell

her

and

second

She

angel).

has

She

or nament

can

in

2

arrangements.

decides

to

arrange

two

of

the The

objects:

snowman,

bell,

angel

and

candle.

Now

she

has

to

choose

from.

She

can

choose

her

first

candle

is

one

four distinct of

objects

objects.

or nament

in

the

four

distinct

four objects.

4 !

ways

and

her

second

in

three

4

giving

×

3 =

= (4

12

different

2) !

arrangements.

Using

and

the

same

wanted

to

reasoning

arrange

4

if

she

had n

or naments

different

in

a

line,

or naments

she

could

do

this

in

n !

n

×

(n

−

1)

×

(n

−

2)

×

(n

−

3)

=

ways.

n

➔

The

number

of

4

permutations

!

of

r

objects

out

of

n

distinct

n ! n

objects

P

is



r

n

The

number

of

ways

r

of

!

arranging

n

objects

in

a

row

is

simply

the We

numbers

of

permutations

n !

of

n

objects

out

of

n.

The

formula

is

dene

0!

as

1.

then

n !

n

P



 n !



n

n

Alma

n

!

0 !

chooses

6

different

napkins

out

of

the

10

patterns

that

she

has.

⎛ n ⎞ n

If

she

wanted

to

arrange

6

napkins

out

of

10

in

a

line

she

could

do

it

⎜

⎟

and

C

are

two

r

⎝

r

⎠

10 !

in

ways,

10

6

but

this

includes

the

6!

ways

of

arranging

the

6

different

notations

combinations.

napkins

in

a

line.

Now

the

order

is

not

impor tant,

so

she

equivalent

arrangements

by

dividing

by

Both

excludes are

the

for

!

equally

correct

6! ⎛ n ⎞

but

10 !

So

Alma

has

ways

(10

6

of

choosing

the

six

napkins.

⎜

⎝

⎟

r

is

)! 6! throughout

When

the

order

combinations

34

Mathematics

of

and

as

used

⎠

arrangements

you

the

can

science

is

not

generalize

of

patterns

relevant

the

result.

they

are

called

this

book.

➔

The

number

of

ways

of

choosing

(order

is

not

impor tant)

n

r

objects

from

n

is

p n

C

r

= r

⎛ n ⎞

r

n !

p

n

⎜ ⎝

Y ou

⎟

r

is

a

r

= r

(n

⎠

can

letters

the

= C

also

from

same

think

A,

as

B,

BA.

different

r

)! r

of

C,

!

combinations

you

can

However,

arrangement

have

you

as

a

three

could

selection.

different

arrange

(permutation)

to

To

select

selections

them

in

six

two

as

AB

ways

as

is

AB

BA.

Note

Having

chosen

her

number

of

six

different

napkins

Alma

would

now

like

to

are

out

the

ways

of

arranging

them

round

the

table.

that

we

find interested

arrangements

realizes

that

the

number

of

ways

is

no

longer

6!

because

they

are

arranged

in

a

they

two

are

arrangements

in

However,

a

straight

arranged

arrangement

is

relative

are

round

just

different

other .

because

line.

a

rotated

table

by

the

one

same

Another

place.

way

reasoning

to

x

Then

of

the

fact

you

without

could

rotate

repeating

an

arrangement

positions.

So

the

six

number

times

of

round

distinct

the

you

ways

in

of

nishing

6!

arranging

6

distinct

objects

in

a

circle

will

be

of

can

the

number

of

ways

of

arranging n

is

(n

–

Example

a

In

b

In

at

objects

around

1)!

5

on

think

the

napkins

with

rst

each

either

end

side

napkin.

number

of

ways

a arranging

objects

in

a

5

distinct

line

is

5!



how

many

ways

can

the

letters

of

the

how

many

ways

can

the

letters

be

arranged

a

line

the

The

distinct

of

circle

a

be

napkin.

 5! 6

Hence

would

arranging

table

of

rst

remaining

In

to

circle. each

These

of

to napkins

be

in

She

word

special

taking

be

arranged?

them

two

time?

Answer

a

There

can

be

are

7

different

arranged

in

7!

letters

in

the

word

special

which

ways.

b

Chapter

1

35

Example

How



many

four-digit

a

1,

2,

3

and

4

b

0,

1,

2

and

3?

Digits

may

be

used

numbers

more

than

can

be

made

using

the

digits

once.

Answers

There

a

are

choosing

giving

a

four

ways

each

total

of

of

the

Since

four

digits

we

distinct

of

more

are

only

concer ned

numbers,

than

digits

can

with

be

used

once.

4

4

=

256

There

b

are

choosing

there

are

3

a

of

digit

ways

each

giving

ways

first

four

choosing

digits

only

the

of

four-digit

with

number

cannot

star t

zero.

of

the

total

A

but

other

of

3

3

×

4

=

Example

In

8

a



Model

boys.

from

In

the

a

there

b

the

c

at

192

United

how

15

many

students

are

no

ways

three

is

of

(MUN)

can

a

club

at

school

delegation

of

10

there

are

students

7

be

girls

and

chosen

if:

gender

delegation

least

Nations

to

restrictions

be

each

made

gender

up

of

are

5

boys

included

and

in

5

girls

the

delegation?

Answers

⎛ 15 ⎞ a

⎜

⎝

⎟

10

Calculate

15!

=

=

⎠

combination,

the

order

of

impor tant.

Now ⎛ 7 ⎞ b

⎜

⎝

⎠

⎜

⎝

⎟ 5

=

The

out

number

⎛ 7 ⎞

⎜

It

is

2

⎠

⎝

1

⎝

36

only

of

having

two

girls

a

is:

choose

5

girls

of

8

a

7

and

5

boys

out

of

8.

GDC.

Cannot

0,

1

or

have

2

a

girls

other

delegation

OR

possible

0,

1,

2

with

boys.

delegations

21 allowed.

⎠

to

have

no

girls

or

are

only

7

girls

combinations

three

will

all

include

boys.

Subtract

the

unwanted

combinations

⎞

⎟ 10

to

GDC.

girl.

least

15

ways

All

=

⎟

there

possible

⎜

⎜

impossible

Since

⎛

of

with

⎛ 8 ⎞ ×

⎟

only

at

need

a

not

⎠

delegation

⎝

Use

is

1176

Use

c

we

choosing

⎛ 8 ⎞ ×

⎟ 5

as

3003

10!× 5!

21

=

3003

−

21

=

number ⎠

Mathematics

as

the

science

from

2982

of

patterns

of

ways.

the

total

are

Exercise

1

2

To

1K

open

your

three

distinct

How

many

Three

three

In

a

In

b

many

how

many

same

par t

times

r un.

a

of

A

In

If

of

a

ways

be

are

on

in

a

the

code

consisting

of

lock.

there?

books,

placed

ways

how

can

on

the

are

students

There

many

he

two

a

geography

books

and

bookshelf.

books

be

be

arranged

on

arranged

so

that

books

of

together?

to

ways

can

up

is

are

to

8

to

can

a

his

team

team

be

routes

next

12

be

least

a

10

along

to

r un

a

km

route

which

he

different

four

can

set

of

race.

Mark’s

selected

at

r uns

manage

boys and

include

a

just

before

be

Mark

different

will

there

ways

many

for

next

a

girls

race?

mathematics

to

choose

from.

chosen?

one

girl

and

one

boy ,

in

how

selected?

QUESTION

many

2,

books

training

eight

leading

is

1,

the

grouped

are

that

team

How

can

are

There

the

0,

to

punch

letters

codes

cross-countr y

week

4

EXAM-STYLE

5

26

science

different

weeks

competition.

b

must

different.

calculates

each

team

a

are

locker

are

you

QUESTION

his

How many

4

There

four

subject

week.

He

routes

all

locker,

shelf ?

EXAM-STYLE

As

books

are

how

the

3

books,

histor y

the

letters.

different

maths

The books

school

3,

4,

four-digit

5

and

even

numbers

can

be

made

using

the

digits

6?

b

How

many

of

these

four-digit

even

numbers

are

divisible

c

How

many

of

these

four-digit

even

numbers

have

no

by

5?

repeated

digits?

6

In

the

UK

contained

How

between

three

many

1932

letters

different

of

and

the

number

1945

car

alphabet

plates

in

registration

followed

this

by

format

numbers

three

were

digits.

possible?

Chapter

1

37

.

The

Repeated

binomial

algebraic

theorem

multiplication

gives These

are

called

0

(1

+

x)

=

1

binomial

expansions

=

1

+

x

because

the

=

1

+

2x

1

(1

+

x)

(1

+

x)

2

2

+

(1

+

x)

(1

+

x)

(1

+

x)

two

x

3

2

3

=

1

+

3x

+

3x

=

1

+

4x

+

6x

=

1

+

5x

+

10x

4

●

these

The

●

expressions

of

common

The

x

+

4x

3

highest

x

for

form

an

difference

index

of

x

4

10x

powers

+

●

of

(1

+

+ x)

arithmetic

you

can

sequence

see

of

with

first

term

0

the

same

as

the

power

to

which

(1

+ x) triangular

first

and

last

term

are

always

1.

is

of

called

T riangle

If

you

write

all

the

coefficients

like

this

you

will

recognize

a

numbers

Pascal’ s

after

the

number French

of

a

1.

is

the

as

that:

pattern

coefficients

terms

known

x

raised.

The

is

5

5x

This

is

x)

x

3

+

algebraic

+

of

binomial.

4

+

2

indices

and

+

2

5

In

(1

sum

mathematician

patter ns. Blaise

Pascal

(1623–62).

actually

row

0:

was

1

as

early

11th

row

It

recorded

2:

1

2

as

the

centur y

by

the

1

Persians

and

the

Chinese.

row

4:

There

1

is

a

obtained

can

4

ver tical

by

extend

Looking

adding

this

at

line

the

6

of

the

patter n

symmetr y .

two

by

numbers,

4

Each

numbers

counting

the

two

above

or

1’s

as

combinations:

in

⎝

In

fact

the

first

and

last

0

be

written

⎝

number,

which

which

(1

+

x)

is

is

0

⎟

⎜

⎠

⎝

the

at

the

= ⎜

⎠

⎝

same

⎟ n

the

second

side.

is

Y ou

row

can

be

of

1

⎟ 1

all

⎠

the

rows

as

where

n

is

the

row

⎠

the

power

to

raised.

second

row

2

 2 



 1!(2

so

either

row

technology .

first

and

2!

Looking

a

⎛ n ⎞ and

as ⎜

1

⎟

coefficients

⎛ n ⎞ can

the

to

in

⎛ 1⎞ =

⎜

it

using

⎛ 1 ⎞ written

number

1

row

is

1) !







1



[

actually

This

illustration

book

⎛ 2 ⎞

⎛ 2 ⎞

⎝

0

⎟

⎜

⎠

⎝

1

⎟

⎜

⎠

⎝

written

is

from

a

Shih-Chieh.

China

in

It

1303,

more

than

300

years

before

⎠

in

Mathematics

in

⎟

2

Pascal.

38

Chu

⎛ 2 ⎞ was

⎜

by

as

the

science

of

patterns

It

shows

Chinese

the

triangle

numerals.

As

each

number

immediately

⎛ 2 ⎞

3 =

⎜

⎝

obtained

above

it,

then

by

in

adding

the

the

third

two

numbers

row

⎛ 2 ⎞

+

⎟

0

is

⎜

⎠

⎝

⎟

1

⎠

2!

2!

=

+ 0 !( 2 − 0 ) !

1!( 2 − 1) !

2!

2!

=

Remember ,

we

dene

+ 0 ! 2

×

( 2 − 1) !

2!

1

⎡ 1

×

0 !( 2 − 1) !

0!

as

1

1⎤

=

+ ⎢ 0 !( 2

1) !

3

⎥ 2

⎣

1

⎦

2!

×

= 1

0 ! 2

×

( 2 − 1) !

×

⎛ 2⎞ Show

that

3!

⎝

= 1!(3

that

⎜

⎝

To

can

⎠

 2 





 2 

0









 3 

1









 3 



2



⎝

0









⎠

the

be

third

row

written

of

Pascal’ s

triangle

as

⎛ 3 ⎞

⎛ 3 ⎞

⎛ 3 ⎞

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎠

⎝

⎠

⎝

⎠

⎝

0

1

2

⎟ 3

⎠



 3 

x



1





leads

⎛ n ⎞

x )





to

a



2

3  3

x











3

x



conjecture

⎛ n ⎞

⎝

0

⎛ n ⎞ x

+

= ⎜

for

the

binomial

theorem

which

⎟

⎜

⎠

⎝

1

⎛ n ⎞

2

x

+

⎟

⎜

⎠

⎝

2

+

x

⎟

⎜

⎠

⎝

r

Coefcient

⎛ n ⎞

r

+

...

+

...

+

formal

proof

of

the

binomial

⎟

⎜

⎠

⎝

theorem

is

of

general

n

x ⎛ n ⎞

⎟ n

term

⎠

is

⎜

⎝

A

and

⎟ 2

that

n

(1 +



patter n

states

⎜

⎠

x

2





This

2

 2 

x



3

x )

⎝

2





(1 

⎠

⎟

⎛ 3 ⎞

⎝

2

x )

1

⎜

⎟

1

summarise

(1 

⎛ 3 ⎞ =

⎟

1) !

⎛ 3⎞

=

⎛ 2⎞ +

⎜

beyond

the

scope

of

⎟

r

⎠

this

course.

n

(1

+

x)

When

=

(1

you

degree

n.

+

x)

×

(1

multiply

Y ou

considering

can

+

x)

these

work

different

×

n

(1

+

x)

factors

out

how

powers

of

. . .

you

to

(1

+

get

obtain

x)

a

polynomial

the

of

expansion

by

x:

Number

of

ways

of

⎛ n ⎞ n

0

x

:

multiplying

all

the

1’s

⇒

C

= 0

choosing

⎜

0

x ’s

out

of

⎟ 0 n

factors.

1

x

:

choose

one

x

from

the

n factors

(1

+

x)

and

multiply

it

by

the

1’s

⎛ n ⎞ n

in

all

the

other

factors

⇒ C

x 1

=

⎜

⎟

x

1

Chapter

1

39

2

x

:

choose

two

x’s

from

the

n

factors

and

multiply

them

together

⎛ n ⎞ n

and

by

the

1’s

in

all

the

factors ⇒ C

other

2

2

=

x 2

⎜ ⎝

⎟ 2

x

⎠

r

:

x

choose

r

of

and

the

the

x’s

from

the

n

factors,

multiply

them

together

⎛ n ⎞ n

by

1’s

in

all

the

other

factors

r

⇒ C

r

x

=

⎜

r

⎟

x

r

n

For

the

expansion

of

(a

+

x)

you

n

⎛

x )

=

x

⎛

n

(a +

a

⎜

a

⎝

x

⎛

n

=

⎟ ⎟

⎜

⎝

deduce

a

⎛ n ⎞

like

this:

⎞

⎜

⎠ ⎠

⎟ a

⎝

⎛ n ⎞

x

result

1 +

⎠

r

2

⎛ ⎛ n ⎞

the

n

⎞ ⎞

1 +

can

⎛ n ⎞

x

n

n

⎛ n ⎞

x

a

Distributing

⎞

x

over

the

n

=

a

⎜ ⎜

⎝ ⎝

+

⎟

0

⎜

⎠

⎝

+

⎟

1

⎜

a

⎠

⎝

⎟

2

...

+

+

⎜

2

a

⎠

⎝

+

⎟

r

...

+

⎜

r

a

⎠

⎝

⎟

n

n

⎟ expansion.

a

⎠

⎠

x

⎛ n ⎞

⎛ n ⎞

⎛ n ⎞

n

=

⎜

⎝

⎟

0

➔

a

+

⎠

⎜

⎝

The

⎟

1

a

n −2

x

+

⎜

⎠

⎝

binomial

2

 n  



 0

...

+

+

⎜ ⎝

states

 n 

⎟

r









 1

a

r

x

+

...

⎝





⎟

n

x

in

x the

expansion

above.

⎠

 n  n2

x

⎜

⎠

 n  a

+

for a

n

that:

n 1

a





n −r

⎠

n

x )

2

x

theorem

n

(a

a

⎟

Substituting

⎛ n ⎞

⎛ n ⎞

n −1

 

 2

a

 n 

2

x

n r

 ... 



 

 r

a

r

x

n

 ... 



 

 n

x



n

⎛ n ⎞ n

=

∑⎜

⎟

r

a

r

x

r r =0

Example

Find

the



values

of

a,

b

+

ax

and

8

a

(1

+

2x)

c

in

these

2

≡

1

+

bx

3

+

cx

Use

identities.

8

+

a

…

+

x

⎛

⎞

= 1 + bx 2

⎝

+ cx

theorem

256x

–

⎜

⎠

⎝

ax)

to

write

down

b

–

64x

+

Then

⎞

⎟ 2

simplify

cx

and

compare

⎠

coefcients.

2

=

expansion.

+ ... +

⎟

6

(2

c

x

⎛

2

1 + ⎜

binomial

10

the

b

the

+

…

Answers

⎛ 8 ⎞

⎛ 8 ⎞

⎛ 8 ⎞

⎛ 8 ⎞

8

a

(1 + 2 x )

⎛ 8⎞

2

=

+

(2 x ) +

(2 x )

3

+

(2 x )

8

+ ... +

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎟

⎜

⎠

⎝

⎠

⎝

⎠

⎝

⎠

⎝

⎝

0

1

8!

2

8!

3

×

2x

1! 7 !

+

×

4x

+ 112 x

(2 x )

⎟ ⎠

3

+

2! 6!

8

× 8x

+ ... + 256 x

3! 5!

2

1 + 16 x

⎟

8

8! 2

1 +

⎜

3

+ 44 8 x

8

+ ... + 256 x

∴

∴ a

= 16,

b

= 112

and

c

=

448

{

40

Mathematics

as

the

science

of

patterns

Continued

on

next

page

means

‘therefore’.

a

x

⎛ b

10

⎞

= 1 + bx 2

⎝

+ cx

⎜

⎠

⎝

a

=

⎞

+ ... +

⎟

⇒

x

⎛

2

1 + ⎜

⎟ 2

⎠

10

10

x

⎛

2

⎛ 10 ⎞

⎞

1 +

= 1 +

⎜

⎟ 2

⎝

⎜

⎟

⎜

⎠

⎟

1

⎝

⎠

x

⎛

⎛ 10 ⎞

⎞

+ ⎜

⎜

⎟ 2

⎝

⎟

⎜

⎠

⎟

2

⎝

⎠

x

⎛

10

x

⎛

⎞

⎞

+ ... . + ⎜ 2

⎝

⎟

⎜

⎠

⎝

⎟ 2

⎠

2

10 !

x

⎛

10 !

⎞

= 1 +

⎛

x

10

⎞

⎛

1! 9 !

⎟ 2

⎝

x

⎞

+ ... +

+ ⎜

⎜ 2 ! 8!

⎠

⎝

⎟ 2

⎜

⎠

⎝

⎟ 2

⎠

10

45

⎛

2

= 1 + 5x

+

x

x

⎞

+ ... + ⎜

4

⎝

⎟ 2

⎠

45

∴

a

=

10,

b

=

5

and

c

= 4

n

⎛ 6 ⎞ 6

c

(2

−

⎛ 6 ⎞

6

ax )

=

5

2

+

⎜ ⎝

2

⎟ 1

− ax

(

+

)

⎜

⎠

⎟ 2

⎝

2

(a

2

4

×

×

(

− ax

+

)

−

b)

n

=

(a

+

(−

b))

...

⎠

2

=

64

+

6

×

32

×

(–ax)

+

2

=

∴

b

=

192a

64

–

192ax

+

240a

15

×

16

(–

ax)

+

…

2

x

+

…

64

= 64

1

⇒ a

= 3

240

80

2

c

= 240 a

=

=

9

Example

3



5

Use

the

binomial

theorem

to

expand

(a

+

3x)

.

5

Hence

find

the

value

of

(1.03)

correct

to

5

decimal

places.

Answer

5

(a

+

3x)

5

=

a

4

+

5a

3

(3x)

2

+

10a

a

10a

+

5a(3x)

4

+

15a

3

x

+

90a

4

+

405ax

+

(1

+

2

2

x

+

270a

3(10

1

+

15

×

5

(10

+

270

×

(10

+

243

×

(10

+

≈

+

0.15

+

(5

90

×

(10

2

)

Substituting

–2

+

405

×

(10

a

4

)

=

1,

2

x

=

0.01

=

10

5

)

0.009

0.000 004 05

1.159 27

+

3

)

–2

1

–2

)

–2

=

3

x

))

–2

=

(3x)

243x

–2

=

5

+

5

5

(1.03)

2

(3x)

4

(3x)

5

=

+

3

dp)

+

+

0.000 27

…

Only

ter ms

consider

to

cor rect

give

to

5

the

the

first

4

answer

decimal

places.

Chapter

1

41

Example

Find

the



term

that

is

independent

of

x

in

the

expansion

9

1

⎛

⎞

2

3x

of

⎜

⎟ 2x

⎝

⎠

Answer

The

general

term,

T

of

the

r

expansion

is

given

by:

r

⎛ 9 ⎞

9

r

2

T

=

r

⎜

(3 x

⎟ r

⎝

)

1

⎛

⎞

⎜

⎟ 2x

⎝

⎠

⎠

r

⎛ 9 ⎞ =

⎜

⎝

⎟ r

9 −r

×

x

=

3r

r

⎠

0

x

–

x

⎟ 2

⎝

r

18

× ⎜

×

−r

×

⎠

18–2r

x

1 ⎞

⎛

18 − 2 r

3

x

=

0

=

6

For

ter m

must

be

independent

of

x,

index

of

x

0.

6

⎛ 9 ⎞ 3

T

=

6

⎜

⎝

×

⎟ 6

3



2

2

n

 



Write

Give

answer

as

an

exact

fraction.

16

⎠

that

 r

567

1L

 n 



⎟

⎝

Show

a

1 ⎞

⎜

=

⎠

Exercise

1

⎛

×



⎛ n





b

 n

r

down

the

+ 1⎞

⎜

⎛ n ⎞ =

⎟

⎜

r



first

four

terms

in

⎟ r

binomial

⎞

⎜

r

the

n

⎛ +

⎟

− 1

expansion

of:

9

x

⎛ 11

a

(1

+

7

2x)

(1

b

–

(2

c

+

5x)

2 −

d

⎜ ⎝

3

Write

down

the

required

term

in

each

of

these

binomial

expansions.

20

x

⎛

⎞

7

a

4th

term

of

(1

–

4x)

3rd

b

term

of

1− ⎜

⎟ 2

⎝

⎠

8

c

4th

term

EXAM-STYLE

of

(2a

–

b)

QUESTION

12

1

⎛ 4

Find

the

term

independent

of

x

in

the

expansion

of

2x

⎞

+

⎜

2

⎟

x

5

x

⎛ 5

Use

the

binomial

theorem

to

expand

⎞

.

2 + ⎜

Hence

find

the

⎟ 5

⎝

⎠

5

value

of

(2.01)

correct

to

5

decimal

places.

4

6

a

Express

(

2

3

)

in

the

form

a +b

where

6

a,

b

∈

3

1

⎛ b

Express

2

⎞

+

⎜

in

⎟ 5

⎝

the

42

Express

Mathematics

1+

(

as

7

the

)

form

a

2 +b

5

where

a,

b

∈

⎠

5

c

5

−

1 −

(

science

of

7

)

⎞

5

3x)

in

the

patterns

form

a

7,

where

a

∈

⎟ 3

⎠

7

Let

a

=

x

+

y

and

2

Write

a

2

b

=

x

−

y

2

a

−

b

in

terms

of

x

and

y

and

hence

b

=

(a

−

b)

(a

+

x

the

and

3

y

binomial

and

use

theorem

your

3

−

a

to

results

2

b

=

(a

−

b)

write

to

3

a

and

show

b

in

the

(a

binomial

y

and

d

Use

your

e

Prove

use

of

that

+

ab

+

b

)

expansion

to

4

write a

and

b

4

and

terms

2

4

Use

c

that

b)

3

Use

b

show

2

−

a

your

results

to

factorize

a

in

terms

of

x

4

−

b

n

results

your

Review

to

make

conjecture

a

conjecture

using

for

the

mathematical

factors

of

a

n

−

b

induction.

exercise

✗ 1

2

Show

term

is

Find

the

1

3

that

+

3

16

+

Three

there

and

sum

4

+

6

are

the

of

+

The numbers

+

a,

c,

sum

the

7

numbers

two

a

b

geometric

of

the

first

sequences

three

such

terms

is

that

the

second

84.

series.

9

+

10

and

and

b

c

+

12

form

form

+

an

a

…

+

46

arithmetic

geometric

sequence.

sequence.

9

If the

sum

of

the

numbers

is

,

find

the

three

numbers.

2

4

Write

down

⎧u ⎪ 1

the

first

six

terms

of

the

sequence

given

by:

= 1

⎨ +

u ⎪ n +1 ⎩

=

2u

+ 1 ,

n ∈ 

n

n

Use

mathematical

induction

to

prove

that u

=

2

–

1.

n

2n

5

Prove

by

multiple

6

Write

mathematical

of

in

induction

that

3

+

–

8n

–

1,

n

∈

,

is

a

64.

factorial

notation:

4

a

the

coefficient

of

x

b

the

coefficient

of

x

n+1

in

the

expansion

of

(1

+

x)

in

the

expansion

of

(1

+

x)

2

n–1

4

Find

c

n

given

that

the

coefficient

of

x

in

n+1

(1

+

the

expansion

of

2

x)

is

six

times

the

coefficient

of

x

in

the

expansion

n–1

of

7

(1

+

Evaluate

x)

these

by

choosing

an

appropriate

value

n

for

x

in

the

⎛ n ⎞

a

⎜

⎝

⎟

0

⎠

expansion

⎛ n ⎞ +

⎜

⎝

⎟

1

⎛ n ⎞

⎜

⎟

⎜

⎠

⎝

(1

⎛ n ⎞ +

⎠

⎛ n ⎞

of

⎜

⎝

⎟

2

+

x)

⎛ n ⎞ +

...

+

⎠

⎜

⎝

⎟

r

⎛ n ⎞ +

...

+

⎠

⎜

⎝

⎛ n ⎞

⎟

n

⎠

⎛ n ⎞

n

n

b

⎝

0

⎟

1

⎠

+

⎜

⎝

⎟

2

⎠

−

...

+

( −1)

⎜

⎝

⎟

r

⎠

+

...

+

(

−1)

⎛ n ⎞

⎜

⎝

⎟

n

⎠

Chapter

1

43

Review

exercise

EXAM-STYLE

1

The

QUESTION

diagram

shows

largest

square,

square

of

a

If

the

the

spiral

lines

b

in

of

is

Use

of

midpoints

the

the

This

largest

second,

formed

the

sequence

sequence.

sides

lengths

A

the

a

by

of

are

squares.

joined

process

square

third

joining

and

to

can

have

form

be

with

the

1,

the

second

continued

length

four th

segments

Star ting

infinitely .

calculate

the

squares.

shown

as

red

diagram.

your

answers

to

par t

to

a

find

the

length

of

the

spiral

shown.

c

What

happens

process

A

to

the

length

of

the

spiral

if

we

continue

the

infinitely?

different

spiral

is

formed

by

shading

triangles

as

shown

in

the

diagram.

d

Find

e

What

of

the

is

total

the

forming

area

total

of

the

area

squares

of

and

shaded

the

triangles.

spiral

shading

formed

triangles

if

is

the

process

continued

infinitely?

2

a

In

how

many

characteristic

b

How

5

c

many

can

Four

be

person

In

how

from

and

formed

is

next

many

six

one

men

using

ways

to

ways

and

female

bigger

couples

different

ways

can

the

letters

of

the

word

arranged?

numbers

married

many

3

different

be

his

can

four

and

the

are

can

or

a

than

digits

to

be

they

her

20 000

0,

in

3,

divisible

5,

and

group

in

a

7

9?

photo.

line

so

by

In

that

how

each

spouse?

women,

are

a

stand

committee

there

1,

and

so

of

that

more

five

people

there

women

is

at

than

be

selected

least

men

one

on

male

the

committee?

4

Write

down

and

simplify

the

term

independent

of

x

in

the

8

3



expansion

3

of

 x



EXAM-STYLE



x 



QUESTION

r–1

5

Given

that

the

coefficients

of

x

r

,

x

r+1

,

x

in

the

expansion

n

of

(1

+

x)

2

n

are

in

arithmetic

sequence,

show

that

2

+

4r

–

Hence find

2

–

n(4r

three

+

1)

=

0

consecutive

coefficients

of

the

14

(1

44

+

x)

which

form

an

arithmetic

sequence.

expansion

of

CHAPTER

1

SUMMARY

Sequences

●

An

and

arithmetic

series

sequence,

or

arithmetic

progression,

with

the

first

term u 1

and

common

,

u

u

1

and

+

d,

difference

u

1

the

+

2d,

d

…

is

,

u

1

+

general

term

is

=

u

u

n

●

The

sum

of

n

=

terms

[ 2u

of

a

+ (n

− 1) d

1

n

1)d

+

(n

–

1)d

finite

arithmetic

series

is

n

n

where

–

1

n

S

(n

1

is

the

number

=

]

[u

+

u

1

of

]

n

terms

in

the

series, u

is

the

first

with

first

1

term

●

A

and

d

is

geometric

the

common

sequence

or

difference

geometric

progression

term u 1

and

common

ratio

2

,

u

u

1

r,

u

1

r

r

is

3

,

u

1

r

4

,

u

1

r

,

...

1

n–1

and

the

general

term

is

=

u

u

n

×

r

,

r

≠

–1,

0,

1

1 n

(1 − r

u

)

1

●

The

sum

of

n

terms

of

a

geometric

series

is

S

,

=

r

≠

1

n

(1

r ) u 1

●

When

–1




0 there

are

ii

Ιf

Δ

=

0

there

is

iii

Ιf

Δ




0

and

x 4

2

y

=

e

+ 1.

Find

the

area

of

the

shaded

region.

2

8 6

Consider

the

cur ves

y

x

=

and

y

=

.

2

4

a

Find

the

b

Write

down

enclosed

c

points

by

Calculate

the

the

the

of

+

4

x

intersection.

integral

that

represents

the

area

of

the

region

cur ves.

area

of

the

region.

Chapter

9

485

The

volume

V

of

a

solid

formed

by

a

cur ve

y

=

f

(x),

between

Y ou

x

=

a

and

x

=

b

rotated

2

through

radians

about

the

in

isgiven

met

this

formula

x-axis Section

7.

by

b

2

V



=

y

dx

a

Example

Find

the

y

sin



volume

of

a

solid

obtained

by

rotating

the

cur ve

y



x ,

0



 

x

through

2π

radians

about

the

x-axis. 2

1

Answer

b

π

2

sin x

π

=

Use

dx

the

for mula

V



=

y

x

0

2

V

1

dx

2

3

4

–1

a

0

–2

π



π

=

sin x dx

 



cos x

 0

Simplify. 0

3

π (– cos π

=

The

volume

–

cos 0)

of

a

=

solid

2π

units

formed

Evaluate.

by

rotating

a

cur ve, y

=

f (x),

b

2

2π

through

radians

about

the

y-axis

is

given

by

V

=

π

x

dy.

a

1

To

use

this

Example

Find

y

=

the

formula

you

must

first

find x

=

f

(y).



volume

arccos

directly

x,

0

of

≤

x

a

≤

solid

1

that

through

is

obtained

2π

radians

by

rotating

about

the

the

cur ve

y-axis.

Answer

y

=

y

arccos x,

0

≤

x

≤

0

≤

y

≤

1

Express

x

in

ter ms



find

domain

of

y

r

and

2

⇒

x

=

cos y,

the

of

y

values.

2 

1 2

V

=

π

(cos y)

dy

Use

the

volume

Use

the

half

for mula.

0

π

1

=

cos 2 y

π

dy

2

0

p

⎡ y

sin 2 y ⎤

= p ⎢

⎥

⎣ 2

4

⎦

0

2

⎛ p

= p

486

The

sin 2p

⎜ ⎝

x

0

2

⎞ ⎟

2

power

4

of

⎠

⎜

p

sin 0 ⎞

⎛

−

0 −

⎝

calculus

⎟ 4

⎠

3

=

units 2

Evaluate.

angle

for mula.

1

Y ou

can

subtract

Example

Find

the

volumes

of

two

of

that

different

cur ves.



volume

a

solid

is

obtained

by

rotating

the

finite

region

x

enclosed

by

the

cur ves

y

=

ln x

+

1

and

y



2π

through

tan

radians

2

about

the

x-axis.

Answer

y

Sketch

the

graph

to

identify

the

finite

3 x y

=

region

and

the

points

of

intersection.

tan 2

2

x y

=lnx

+1

The

1

equation

ln

x

=

tan

cannot

be

2

solved

algebraically

so

use

a

GDC.

x

0 1

2

3

x

Solve

the

equation

ln

x

=

tan

and 2

store

b.

the

Apply

rotating

solutions

the

as

for mula

region

the

for

between

variables

the

two

a

volume

and

of

cur ves.

3

Vol.

=

3.58

Exercise

1

Find

by

9X

the

the

units

volume

given

of

a

cur ves

solid

generated

through

2π

by

radians

rotating

about

the

the

region

bounded

x-axis.

 a

y



cos x ,

x



0,

x



,

y



y



0

,

y

0

2

 b

y



sec x ,

x



0,

x



, 4

 c

y



cos x ,

x

5



,

x



6

 d

y



sin x ,

x



Find

the

volume

by

the

given

a

y

=

arcsin x,

x



,

b

y

=

arcsin x,

of

=

y



0

3

cur ves

x

0

2

, 3

2



6

a

solid

generated

through

0,

x

=

0,



0,

x

y

x

=

=

2π

by

radians

rotating

about

the

the

region

bounded

y-axis.

1

1,

y

=

0

 c

y



tan x ,

x



,

y



0

4

d

y

=

tan x,

x

=

0,

y

=

1

Chapter

9

487

3

Find

by

the

the

volume

of

a

solid

generated

rotating

the

region

enclosed

cur ves



y

a

by



sin x ,

y



5

cos x ,



x

through



4

2π

radians

about

the

4

x-axis.

y

b

=

sin 2x,

y

=

sin x,

0

≤

x

≤

π

2π

through

radians

about

the

x-axis.

x

3

y

c

= e

− 1

and

y

through

= arctan x

2π

radians

about

the

x-axis.

See

the x

3

y

d

= e

− 1

and

y

Investigation

F ind

by

the

volume

rotating

a

of

a

circle

through

= arctan x

–

volume

tor us

with

that

the

is

of

2π

a

radians

about

the

torus

y-axis.

y

obtained

centre

at

(h,

k) r

and

a

r,

radius

k,

r

>

0

Look

and

back

chapter

k

at

>

r,

about

x

the

(h

axis.

k)

the

introduction x 0

on

Exercise

1

Use

page

441.

9Y

volume

of

revolution

to

find

the

volume

2

by

rotating

about

2

Use

the

the

circle

(x

−

4)

(y

+

3)

volume

rotating

about

3

Use

the

of

the

revolution

to

find

the

4

Find

the

2

5

Find

circle

of

(x

−

4)

revolution

6

rotating

=

the

36

+

−

The

the

of

a

+

to

(y

find

circle

solid

through

volume

9y

=

the

36

of

a

2π

solid

x)

power

through

volume

2

488

obtained

radians

volume

of

a

tor us

obtained

+

3)

=

4

through

2π

radians

the

x

volume

of

a

sphere

2

+

y

=

9

through

2π

radians

obtained

by

rotating

the

ellipse

radians

about

obtained

by

the

x-axis.

rotating

the

ellipse

2

Find

(5

2π

x-axis.

volume

9y

2

4x

tor us

2

+

4x

by

the

a

through

2

2

obtained

4

y-axis.

volume

about

=

x-axis.

2

by

of

2

+

of

a

2π

solid

radians

about

obtained

by

the

y-axis.

rotating

the

ellipse

2

+

of

9y

=

36

calculus

through

2π

radians

about

the

x-axis.

further

CD.

questions

on

Review

exercise

✗ 1

Differentiate

a

f

(x)

=

with

(2x

+

3)

respect

to

x:

sinx

x

b

g(x)

c

h

=

e

cos3x

tan x

(x )

= 2

2x

2x

2

Find

the

equation

3

Find

the

value

of

of

m

a

tangent

that

to

satisfies

the

this

cur ve

sin y

+

e

=1

at

the

origin.

equation

m



⎛

2

sec

x dx

=

2

⎜



cos

⎞

− sin

⎟

p

6

⎝

6

⎠

4

4

Use

the

method

of

integration

by

par ts

to

solve:

2x

a

(2x

–

5) e

b

(x

c

e

The

diagonal

dx;

2

–

5x) cosx

dx;

x

cos 3x

dx

–1

5

Find

has

a

the

of

rate

length

of

of

a

square

change

5

is

of

increasing

the

area

at

of

a

the

rate

of

square

0.2

cm

when

s

the

.

side

cm.

2x–1

6

The

a

cur ve

Find

y

the

through

b

Find

the

c

=

e

the

rotating

of

the

tangent

to

the

cur ve

that

passes

origin.

area,

tangent

Find

given.

equation

the

the

is

in

and

the

volume

the

terms

e,

of

the

region

bounded

by

the

cur ve,

y-axis.

of

region

of

the

in

revolution,

par t

b

about

in

terms

the

of

,

obtained

by

x-axis.

2

7

Use

the

8

The

region

x

=

1

substitution

and

a

Sketch

b

Find

this

x

bounded

the

x-axis

the

the

by

is

region

exact

=

3

cos θ

the

cur ve

rotated

in

value

the

of

to

y

=

ln

volume

(2x),

2π

through

coordinate

the

9 −

find

x

dx

the

ver tical

radians

about

line

the

y-axis.

system.

of

revolution

obtained

by

rotation.

2t

3

9

The

velocity ,

v,

of

an

object,

at

a

time

t,

is

given

by

v

= 5e

k

>

,

–1

where

t

is

a

Find

b

What

in

the

is

seconds

distance

the

total

and

v

is

in

travelled

distance

m

in

.

s

the

first k

travelled

by

seconds,

the

object?

2

10

Find

point

the

(1,

equation

of

the

normal

to

the

0.

cur ve x

3

y

=

cos(x)

at

the

–1).

Chapter

9

489

Review

exercise

2

1

Find

the

2

Given

points

of

inflection

of

the

cur ve y

=

find

the

equation

x

sin 2x,

–1

≤

x

≤

1.

3

point

the

cur ve

where

x

=

y

=

cos x,

of

Find

the

value

of

a,

0




5)

Exercise

1

The

2

the

2

b

in

0.5

is

a

a

0.868

ml

of

the

of

modeled

in

25

≤

P(Y

P(Y

+

≤

≤

P(Y

4 )

0 )

=

+

3 )

P(Y

+

=

P(Y

1 )

=

+

P(Y

=

2 )

4 )

4)

of

is

by

a

two

per

millilitre

of

a

given

be

than

liquid

is

3.5

that

there

liquid

of

0.01.

there

flaws

If

Poisson

will

per

flaws

will

100

be

square

occur

variable

chosen

fewer

least

metre

the

2

of

randomly

find

square

at

7

bacteria

bacteria.

fabric

and

probability

metres

of

produced

their

fabric

number

that

there

will

be

flaws

square

one

bacteria

liquid

the

randomly

least

=

(3 sf)

number

machine

a

P(Y

number

ml

exactly

b

–

Use

is

Po(7.5)

=

mean

in

~

1

probability

in

on

Y

the

10H

a

The

by

arriving

=

mean

Find

inter val,

cars

metres

of

chosen

randomly

fabric

there

will

be

at

flaw .

Chapter

10

519

The

3

number

and

If

9:00

the

mean

the

a

the

X

~

number

8:15

in

and

Po

calls

weekday

of

a

given

8:10

a

of

by

modeled

calls

phone

a

by

per

school

a

hour

calls

between

Poisson

is

12,

received

8:00

distribution.

calculate

between

8:00

day

that

in

received

is

phone

number

probability

8:00

4

phone

any

expected

and

b

of

on

more

given

than

five

calls

are

received

between

day .

(3.5)

Calculate

a

P(X

i

Write

b

=

3)

P(X

ii

down

the

values

>

of

3)

P(X

iii

E(X)

and




3)

Var(X)

2

Hence

c

find

EXAM-STYLE

The

5

the

value

of

E(X

).

QUESTION

random

variable

X

is

Poisson

distributed

with

mean

m

and

satisfies

P(X

Find

b

Hence

Let

6

=

a

X

P(X

>

Find

0)

the

be

3)

The

=

Let

p

λ

be

>

1)

m

P(2

–

P(X

correct

≤

X

≤

variable

=

4)

to

=

four

0

decimal

places

4).

with

a

Poisson

distribution,

such

that

3)

QUESTION

variable

P

has

a

Poisson

distribution

with

0.

the

probability

a

Write

down

an

b

Show

that

=

c

Sketch

p

the

that

P

expression

p(λ)

graph

concavities

.

=

0.555




for

0,

u

some

and

scalar

v

have

k

the

same

Geometrically,

direction;

points

if

k