P1 Chapter P1 Chapter 88

CIE Centre A-level Pure CIE Centre A-level Pure MathsMaths

© Adam Gibson

Sequences

Look at the following lists of numbers – what is next?

3 10 17 24 …

-1 2 11 26 47 …

2 4 8 16 77 605154 1111

We cannot always deduce the sequence rule from the first few terms

A “sequence”

“terms”

!

Defining a sequence or series – deductive and inductiveThe second sequence on the previous slide can be expressed with a simple formula:

23 1, 0, 1, 2 ...ru r r

This formula allows you to deduce the value of rufor any r.

However, sometimes this is not convenient and we should define the sequence inductively.

Example: write the first 5 terms of the sequence1

1

10| 8 | 3r r

uu u

Try different startingvalues; look for patterns

The prolific geniusGauss was born in Brunswick (Ger. Braunschweig), in the Duchy of Brunswick-Lüneburg as the only son of uneducated lower-class parents. According to legend, his gifts became apparent at the age of three when he corrected, in his head, an error his father had made on paper while calculating finances.

Another story has it that in elementary school his teacher tried to occupy pupils by making them add up the integers from 1 to 100. The young Gauss produced the correct answer within seconds by a flash of mathematical insight, to the astonishment of all. Gauss had realized that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050.

Important sequencesThe triangle sequence is the one discovered by theyoung Gauss:

12

1

( 1)r

ri

u r r i

The factorial sequence is as follows:

0 1 2 30 1 3 6 ...u u u u

Can we give an inductive definition of this sequence?

1 1r ru u r

2 3 42 2.3 6 2.3.4 24 ...u u u and is marked with ‘!’Give an inductive definition. Then calculate : 5!, 6!, 1!, 0!, (32!/30!)

Pascal sequences are a key part of mathematics. They take this form:

0

1

1

1r r

pa rp pr

a is any integer

We can write the terms of Pascal sequences as: ar

Important sequences .. Contd.

Write the whole Pascal sequence for a = 6. What is 6

4

You want to save money at the bank.

At the beginning of 2006 you have $150.00 in the account.Your salary is $2,000 per month.Your salary increases by $1,000 each year (not each month!)At the end of each year, you will add 5% of your totalannual salary to your bank savings account.Ignoring interest payment, how much will be in the account:

•After 32 years?•After 41 years?

Arithmetic sequences

Arithmetic sequences Arithmetic sequences have a common difference:

5 9 13 17 21 …. has a common difference 4

5 2 -1 -4 -7 …. has a common difference -3

We write the common difference as d, the startingValue a, and the last value l

Remember Gauss’ problem?

1+2+3+4+………………………….100

1, 100, 1a l d So:

Arithmetic sequences

1 2 .......... 99 1002

100 99 ........... 2 1101 101 101 .................

S

General formula is: (see page 122 of textbook)

12

( 1)( )

l a n dS n a l

The savings problem

•a = $1200 (not $150!)

•d = $50 (why? It’s 5% of the $1,000 salary increase)

•n = 32

•l = $1,200+31x50=$2,750

•S = 0.5x32x(1200+2750)=$63,200

•The balance in the account = S+$150 = $63,350

Pascal’s sequence – what does it mean?

•Pascal’s sequence tells us the number of ways we can choose a objects from a set of r objects

0

41

0p

by definition; all Pascal sequences startwith 1 because there is one way you can choose zero objects from 4.

1 0

4 4 0 41 0 1

p p

because there are 4 ways you can choose 1 object from 4.

Look:

Pascal’s sequence – what does it mean?

2 1

4 4 1 62 1 1

p p

because there are 6 ways you canchoose 2 objects from 4

Let’s choose 2

6 ways – assuming order doesn’t matter

Think

Pascal’s sequence and algebra Consider the expression na b n

3 2 2 33 3 3n a a b ab b

4 3 2 2 3 44 4 6 4n a a b a b ab b

It can be shown that the coefficients are the same as those in Pascal’s triangle.

Can we use this to find the coefficient of 10 3a bin the expression 13a b ?

Why?

Pascal’s sequence and factorials We can! But we first need to know that:

13 13! 13.12.11 2863 (13 3)!3! 3.2.1

See page 132 of your textbook.

So the coefficient of 10 3a b in 13a b is 286.

!

! !n nr n r r

Because

The Binomial Theorem States that …

1 2 2 3 3 ....0 1 2 3

n n n n n nn n n n na b a a b a b a b b

n

where a and b are real numbers and n is a natural number.

and where nr

can be calculated using the formula:

!( )! !

n nr n r r

YOU MUST COMMIT THIS TO MEMORY!

More succinctly, 0

( )n

n n r r

r

na b a b

r

Tasks

•Calculate these: 8 8 7 5 24, , , and

2 0 3 5 1

Give the complete expansion of 12x y

Give the complete expansion of 6

3 7xx

Calculate 101.005 correct to 5 decimal places.

An aside: the Fibonacci sequence

1,1,2,3,5,8,13,21,34,55,89,144,233......

0

20

40

60

80

100

120

140

0 20 40 60 80 100

F(n-1)

Fn

Gradient = ?

Why is the gradient Φ? The key concept is self-

similarityFractals

This one is based on the“FibonacciSpiral”

Why is the gradient Φ? … Contd. Fibonacci numbers very often occur in nature:

02nF 1nF nF

ab

c

c bb a

By definition, c a b In the limit,

a b bb a

Let a xb

Solve the equation!

Examples – how to use the Binomial Theorem Give the value of 101.005 to 5 decimal places

Consider the expression 1 nx A special case of the Binomial Theorem

0 1 1 2 2 0(1 ) 1 1 1 ...... 10 1 2

n n n n nn n n nx x x x x

n

Let’s expand it out:

Simplify:2(1 ) 1 ......

2n nn

x nx x x

What can we say about the terms if x is small …?

Examples – how to use the Binomial Theorem If x is small, the terms in the sequence

get rapidly (exponentially) smaller. 10 105 5

1000 1000(1 0.005) (1 ) , i.e. we let x

1st term:2nd term:3rd term:

4th term:

5th term:

110.5 0.051000

210 5 5.9.5.5 0.0011252 1000 (5.2.5.2.5.2).1000

310 5 0.0000153 1000

410 5 0.00000014 1000

What is the lastterm?

Examples – how to use the Binomial Theorem There is no need to calculate the rest

of the expansion; the terms are very small. So, to 5 decimal places:

101.005 1 0.05 0.001125 0.000015 1.05114

To extend this idea to a slightly more difficult case, tryQ21 p. 136