Ks4 mathematics
Download
1 / 68

KS4 Mathematics - PowerPoint PPT Presentation


  • 129 Views
  • Uploaded on

KS4 Mathematics. A7 Sequences. A7 Sequences. Contents. A. A7.2 Linear sequences. A. A7.1 Generating sequences from rules. A7.3 Quadratic sequences. A. A7.4 Geometric sequences. A. A7.5 Other types of sequence. A. Sequences. 1 st term. 6 th term.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' KS4 Mathematics' - solomon-burch


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Ks4 mathematics

KS4 Mathematics

A7 Sequences


Contents

A7 Sequences

Contents

  • A

A7.2 Linear sequences

  • A

A7.1 Generating sequences from rules

A7.3 Quadratic sequences

  • A

A7.4 Geometric sequences

  • A

A7.5 Other types of sequence

  • A


Sequences
Sequences

1st term

6th term

2, 5, 8, 11, 14, 17, 20, 23, . . .

In mathematics, a sequence is a succession of numbers that follow a given rule.

Each number in a sequence is called a term.

For example,

If terms are next to each other they are referred to as consecutive terms.


Predicting terms in a sequence
Predicting terms in a sequence

For example,

What are the next two terms in the following sequence,

102, 95, 88, 81, 74 . . . ?

–7

–7

–7

–7

–7

–7

102

95

88

81

74

Sometimes, we can predict how a sequence will continue by looking for patterns.

Look at the difference between each consecutive term.

67

60

We can predict that this sequence continues by subtracting 7 each time. We can use this to find the next two terms.


Predicting terms in a sequence1
Predicting terms in a sequence

+2

+4

+6

+8

+10

+12

2

4

8

×2

×2

×2

×2

×2

×2

2

4

8

If we are not given the rule for a sequence, or if it is not generated from a practical context, we cannot be certain how it will continue.

For example, a sequence starts with the numbers 2, 4, 8, . . .

How could this sequence continue?

14

22

32

44

16

32

64

128


Defining sequences
Defining sequences

The first is to use a term-to-term rule.

The second is to use a position-to-term rule.

There are two ways to define a sequence.

To define a sequence using a term-to-term rule we need to know the first term in the sequence and what must be done to each term to give the value of the next term.

To define a sequence using a position-to-term rule we use a formula for the nth term of the sequence.

Term-to-term rules are usually easier to find for a given sequence.

Position-to-term rules are harder to find for a given sequence but are more useful for finding any term in a sequence.


Sequences from term to term rules
Sequences from term-to-term rules

Write the first five terms of each sequence given the first term and the term-to-term rule.

1st term

Term-to-term rule

First five terms

–7

Add 3

–7,

–4,

–1,

2,

5

100

Subtract 8

100,

92,

84,

76,

68

3

Double

3,

6,

12,

24,

48

Multiply by 10

0.04,

0.4,

4,

40,

400

0.04

1.3

Subtract 0.6

1.3,

0.7,

0.1,

–0.5,

–1.1

111

Divide by 5

111,

22.2,

4.44,

0.888,

0.1776


Sequences from position to term rules
Sequences from position-to-term rules

Position

1st

2nd

3rd

4th

5th

nth

Term

n2

When a sequence is defined by a position-to-term rule it can sometimes help to put the terms in a table. For example,

The nth term in a sequence is n2, where n is the term’s position in the sequence.

1

4

9

16

25

Each term can be found by squaring its position in the sequence.

What is the 20th term in this sequence?

202 = 400


Writing sequences from position to term rules
Writing sequences from position-to-term rules

The nth term, or the general term, of a sequence is often called un.

The 1st term is then called u1,

the 2nd term u2,

the 3rd term u3,

the 4th term u4,

the 5th term u5 and so on.

Any term in the sequence can then be found by substituting its position number into the formula for un.


Writing sequences from position to term rules1
Writing sequences from position-to-term rules

For example, suppose the nth term of a sequence is 4n – 5.

We can write this rule as:

un = 4n – 5

Find the first five terms in the sequence.

4 ×1 – 5 =

–1

u1 =

4 ×2 – 5 =

3

u2 =

4 ×3 – 5 =

7

u3 =

4 ×4 – 5 =

11

u4 =

4 ×5 – 5 =

15

u5 =

The first five terms in the sequence are: –1, 3, 7, 11 and 15.


Writing sequences from position to term rules2
Writing sequences from position-to-term rules

The nth term of a sequence is 2n2 + 3.

We can write this rule as:

un = 2n2 + 3

Find the first 4 terms in the sequence.

2 ×12 + 3 =

5

u1 =

2 ×22 + 3 =

11

u2 =

2 ×32 + 3 =

21

u3 =

2 ×42 + 3 =

35

u4 =

The first 4 terms in the sequence are: 5, 11, 21, and 35.

This sequence is a quadratic sequence.




Contents1

A7 Sequences

Contents

A7.1 Generating sequences from rules

  • A

  • A

A7.2 Linear sequences

A7.3 Quadratic sequences

  • A

A7.4 Geometric sequences

  • A

A7.5 Other types of sequence

  • A


Sequences that increase in equal steps
Sequences that increase in equal steps

+6

+6

+6

+6

+6

+6

+6

To work out a rule for a sequence it is often helpful to find the difference between consecutive terms.

For example, look at the difference between each term in this sequence:

2, 8, 14, 20, 26, 32, 38, 44, . . .

This sequence starts with 2 and increases by adding 6 each time.

Every term in this sequence is 4 less than a multiple of 6.


Sequences that decrease in equal steps
Sequences that decrease in equal steps

–3

–3

–3

–3

–3

–3

–3

Can you work out the next three terms in this sequence?

10, 7, 4, 1, –2,

–5,

–8,

–11, . . .

How did you work these out?

This sequence starts with 10 and decreases by subtracting 3 from each term to give the next term.

When the difference between each term in a sequence is always the same we have a linear or arithmetic sequence.


Generating linear sequences
Generating linear sequences

When we describe linear sequences we call the constant difference between consecutive terms the common difference, d.

We call the first term in a linear sequence a.

If we are given the values of a and d in a linear sequence then we can use them to generate the sequence.

For example, if a linear sequence has a = 5 and d = –3, we have the sequence:

5,

2,

–1,

–4,

–7,

–10, . . .

Linear sequences can also be generated given the rule for the nth term in the sequence.


The n th term of a linear sequence
The nth term of a linear sequence

× 3

× 3

× 3

× 3

+ 1

× 3

+ 1

+ 1

+ 1

+ 1

× 3

+ 1

Suppose we are given a linear sequence and asked to find the nth term, un, of the sequence. For example,

Find the nth term of the sequence 4, 7, 10, 13, 16, …

This sequence continues by adding 3 each time and so the common difference d is 3.

We compare the terms in the sequence to the multiples of 3.

2

3

4

5

n

Position

1

Multiples of 3

3

6

9

12

15

3n

Term

4

7

10

13

16

3n + 1

un = 3n + 1.


The n th term of a linear sequence1
The nth term of a linear sequence

× –2

× –2

× –2

× –2

+ 7

× –2

+ 7

+ 7

+ 7

+ 7

× –2

+ 7

Find the nth term of the sequence 5, 3, 1, –1, –3, …

This sequence continues by subtracting 2 each time and so the common difference d is –2.

We compare the terms in the sequence to the multiples of –2.

2

3

4

5

n

Position

1

Multiples of –2

–2

–4

–6

–8

–10

–2n

Term

5

3

1

–1

–3

7 – 2n

un = 7 – 2n.


The n th term of a linear sequence2
The nth term of a linear sequence

If a is the first term of a linear sequence and d is the common difference, we can find the general form of the nth term as follows:

u1 = a,

u2 = a + d,

u3 = a + 2d,

u4 = a + 3d,

un= a + (n – 1)d,

Multiplying out the bracket we have un= a + dn – d.

The nth term of a linear sequence with first term a and common difference d is:

(a – d) is the value of the 0th term.

un= dn + (a – d)


Plotting terms
Plotting terms

Value of term (un)

Position number (n)

We can also find the nth term of a linear sequence by plotting the value of each given term in the sequence against its position number. For example,

The sequence 3, 5, 7, 9, 11 … can be shown graphically:

The points lie on a straight line.

The equation of the line can be written in the form y = mx + c.

The value of the gradient m corresponds to the difference between the terms.

The value of the y-intercept c corresponds to the 0th term, to give

un=2n + 1


Finding the n th term of a linear sequence
Finding the nth term of a linear sequence


Tiling patterns
Tiling patterns

Pattern 1

Pattern 2

1 tile

Pattern 3

5 tiles

Pattern 4

9 tiles

13 tiles

The following patterns are made from tiles:

How many tiles will there be in the next pattern?

How many tiles will there be in the 20th pattern?


Tiling patterns1
Tiling patterns

+4

+4

+4

+4

To work out the number of tiles in the next pattern we can look at the difference between consecutive terms in the sequence:

1, 5, 9, 13,

17

The difference between each term is always equal to 4 and so we can use this to find the next term.

Using this pattern, we can predict that there will be 17 tiles in the next pattern.


Tiling patterns2
Tiling patterns

+4

+4

+4

+4

To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence.

Look at the pattern made by the differences:

1, 5, 9, 13,

17

This is a linear sequence and so the nth term will be of the form un = dn + c, where d is the constant difference between the terms.

For this sequence the constant difference d = 4.

How can we find the value of c?


Tiling patterns3
Tiling patterns

2

3

4

5

n

Position

1

Term

1

5

9

13

17

– 3

– 3

× 4

× 4

× 4

× 4

× 4

– 3

× 4

– 3

– 3

– 3

We can find the rule for the nth term by comparing the sequence to multiples of 4.

Multiples of 4

4

8

12

16

20

4n

4n – 3

un = 4n – 3.

We can use this formula to predict the number of tiles in the 20th pattern:

u20 = 4 × 20 – 3

= 77


Contents2

A7 Sequences

Contents

A7.1 Generating sequences from rules

  • A

A7.2 Linear sequences

  • A

A7.3 Quadratic sequences

  • A

A7.4 Geometric sequences

  • A

A7.5 Other types of sequence

  • A


Sequences that increase in increasing steps
Sequences that increase in increasing steps

+1

+2

+3

+4

+5

+6

+7

Some sequences increase or decrease in unequal steps.

For example, look at the differences between terms in this sequence:

4, 5, 7, 10, 14, 19, 25, 32, . . .

This sequence starts with 4 and increases by adding consecutive whole numbers to each term.

The differences between the terms form a linear sequence.


Sequences that decrease in decreasing steps
Sequences that decrease in decreasing steps

–0.1

–0.2

–0.3

–0.4

–0.5

–0.6

–0.7

Can you work out the next three terms in this sequence?

7, 6.9, 6.7, 6.4, 6,

5.5,

4.9,

4.2, . . .

How did you work these out?

This sequence starts with 7 and decreases by subtracting 0.1, 0.2, 0.3, 0.4, 0.5, …

With sequences of this type it is often helpful to find a second row of differences.


Using a second row of differences
Using a second row of differences

+20

+17

+11

+8

+5

+2

+3

+3

+3

+3

+3

+14

+3

Can you work out the next three terms in this sequence?

1, 3, 8, 16, 27,

41,

58,

78

First row of differences

Second row of differences

Look at the differences between terms.

A sequence is formed by the first row of differences so we look at the second row of differences.

This shows that the differences increase by 3 each time.


Quadratic sequences
Quadratic sequences

+3

+5

+7

+9

+2

+2

+2

When the second row of differences produces a constant number, the sequence is called a quadratic sequence.

This is because the rule for the nth term of the sequence is a quadratic expression of the form

un = an2 + bn + c

where a, b and c are constants and a≠ 0.

The simplest quadratic sequences is the sequence of square numbers.

1, 4, 9, 16, 25

The constant second difference is 2 and the nth term isn2.



The n th term of a quadratic sequence
The nth term of a quadratic sequence

When un = an2 + bn + c the value of a can be found by halving the value of the second difference, which is always equal to 2a.

We can prove this as follows,

un= an2 + bn + c

u1 = a × 12 + b × 1 + c =

a + b + c

u2 = a × 22 + b × 2 + c =

4a + 2b + c

u3 = a × 32 + b × 3 + c =

9a + 3b + c

u4 = a × 42 + b × 4 + c =

16a + 4b + c

u5 = a × 52 + b × 5 + c =

25a + 5b + c

This gives us the first five terms of a general quadratic sequence.


The n th term of a quadratic sequence1
The nth term of a quadratic sequence

Let’s find the first and second differences for these general terms.

a + b + c

4a + 2b + c

9a + 3b + c

16a + 4b + c

25a + 5b + c

3a + b

5a + b

7a + b

9a + b

2a

2a

2a

This shows that the second difference is always 2a when

un = an2 + bn + c.


The n th term of a quadratic sequence2
The nth term of a quadratic sequence

When un = an2 + bn + c the value of c can be found by calculating the value of the 0th term, u0.

We can prove this as follows,

un= an2 + bn + c

u0 = a × 02 + b × 0 + c

u0 = c

Of course, the 0th term is not really part of the sequence. We can easily work it out though by looking at the sequence formed by the first row of differences and counting back from the first term.


The n th term of a quadratic sequence3
The nth term of a quadratic sequence

When un = an2 + bn + c the value of b can be found by subtracting (a + c) from the value of the first term.

We can prove this as follows,

un= an2 + bn + c

u1 = a × 12 + b × 1 + c

u1 = a + b + c

Rearranging this formula to make b the subject gives,

b = u1– (a + c)

If we find the value of a and c first, we can then use them and the value of the first term, to find b.


The n th term of a quadratic sequence4
The nth term of a quadratic sequence

+5

+9

+13

+17

+4

+4

+4

Find the nth term of the sequence, 4, 9, 18, 31, 48, …

Let’s start by looking at the first and second differences.

4, 9, 18, 31, 48

The second differences are constant and so the nth term is in the form un = an2 + bn + c.

Let’s find a, b and c.

The second difference is 4, so we know 2a = 4

a = 2


The n th term of a quadratic sequence5
The nth term of a quadratic sequence

+4

+4

+17

+13

+4

+5

+9

+4

+1

Find the nth term of the sequence, 4, 9, 18, 31, 48, …

Let’s start by looking at the first and second differences.

3,

4, 9, 18, 31, 48

The value of c is the same as the value for the 0th term.

We can find this by continuing the pattern in the differences backwards from the first term.

The 0th term is 3,so:

c = 3


The n th term of a quadratic sequence6
The nth term of a quadratic sequence

Find the nth term of the sequence, 4, 9, 18, 31, 48, …

Putting a = 2 and c = 3 into un =an2 + bn + c gives us

un = 2n2 + bn + 3. We can use this to write an expression for the first term:

un = 2n2 + bn + 3

u1 = 2 × 12 + b × 1 + 3

u1 = 2 + b + 3

u1 = b + 5

The first term in the sequence is 4, so:

4 = b + 5

b = –1


The n th term of a quadratic sequence7
The nth term of a quadratic sequence

Once we have found the values of a, b, and c we can use them in un = an2 + bn + c to give the nthterm.

We have found that for the sequence 4, 9, 18, 31, 48, …

a = 2, b = –1 and c = 3,

un = 2n2 – n + 3

We can check this rule by substituting a chosen value for n into the formula and making sure that it corresponds to the required term in the sequence.

For example, when n = 5 we have,

u5 = 2 × 52 – 5 + 3

= 48

Check the rule for other terms in the sequence.


The n th term of a quadratic sequence8
The nth term of a quadratic sequence

+2

+3

+4

+5

+1

+1

+1

Find the nth term of the sequence 1, 3, 6, 10, 15, …

This is the sequence of triangular numbers.

1, 3, 6, 10, 15

The second differences are constant and so the nth term is in the form un =an2 + bn + c.

Let’s find a, b and c.

The second difference is 1, so 2a = 1

a = ½


The n th term of a quadratic sequence9
The nth term of a quadratic sequence

+1

+1

+5

+4

+1

+2

+3

+1

+1

Find the nth term of the sequence 1, 3, 6, 10, 15, …

This is the sequence of triangular numbers.

0,

1, 3, 6, 10, 15

The value of c is the same as the value for the 0th term.

We can find this by continuing the pattern in the differences backwards from the first term.

The 0th term is 0, so:

c = 0


The n th term of a quadratic sequence10
The nth term of a quadratic sequence

Find the nth term of the sequence 1, 3, 6, 10, 15, …

Putting a = ½ and c = 0 into un =an2 + bn + c gives us

un = ½n2 + bn. We can use this to write an expression for the first term:

un = ½n2 + bn

u1 = ½ × 12 + b× 1

u1 = ½ + b

The first term in the sequence is 1, so

1 = ½ + b

b = ½


The n th term of a quadratic sequence11
The nth term of a quadratic sequence

n2

n

un =

+

2

2

or

n2 + n

un =

2

Once we have found the values of a, b, and c we can use them in an2 + bn + c to give the nthterm.

We have found that for the sequence 1, 3, 6, 10, 15, …

a = ½, b = ½ and c = 0;

Checking, when n = 5, we have

u5 = ½(52 + 5)

= 15

Check the rule for other terms in the sequence.


Finding the n th term of a quadratic sequence
Finding the nth term of a quadratic sequence


Tiling patterns4
Tiling patterns

Pattern 1

Pattern 2

1 tile

Pattern 3

5 tiles

Pattern 4

13 tiles

25 tiles

The following patterns are made from tiles,

How many tiles will there be in the next pattern?

How many tiles will there be in the 20th pattern?


Tiling patterns5
Tiling patterns

+4

+8

+12

+16

+4

+4

+4

To work out the number of tiles in the next pattern we can look at the difference between consecutive terms in the sequence,

1, 5, 13, 25,

41

If the second difference is always equal to 4 then we can use this to find the next term.

Using this pattern we can predict that there will be 41 tiles in the next pattern.


Tiling patterns6
Tiling patterns

Look at the pattern made by the differences,

1, 5, 13, 25,

41

+4

+8

+12

+16

+4

+4

+4

To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence.

The second difference is constant and so the nth term will be a quadratic of the form un = an2 + bn + c.

The second difference is always equal to 2a:

2a = 4

a = 2


Tiling patterns7
Tiling patterns

+4

+4

+16

+12

+4

+4

+8

+4

+0

To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence.

Look at the pattern made by the differences,

1,

1, 5, 13, 25,

41

The value of c is equal to the 0th term.

The 0th term is equal to 1, so:

c = 1


Tiling patterns8
Tiling patterns

+4

+4

+16

+12

+4

+4

+8

To work out how many tiles there will be in the 20th pattern we can find the rule for the nth term in the sequence.

Look at the pattern made by the differences,

1, 5, 13, 25,

41

If a = 2 and c = 1 then un = 2n2 + bn + 1,

u1 = 2 + b + 1

u1 = b + 3

1 is the first term.

1 = b + 3

b = –2


Tiling patterns9
Tiling patterns

We can now substitute a = 2

b = –2

c = 1

into the general form for the nth term of the quadratic sequence, an2 + bn + c, to give us the following rule for the nth term:

un = 2n2– 2n + 1

We can now use this formula to find the value of the 20th term.

u20 = 2 × 202 – 2 × 20 + 1

= 800 – 40 + 1

= 761

We can predict that there will be 761 tiles in the 20th pattern.


Contents3

A7 Sequences

Contents

A7.1 Generating sequences from rules

  • A

A7.2 Linear sequences

  • A

A7.4 Geometric sequences

A7.3 Quadratic sequences

  • A

  • A

A7.5 Other types of sequence

  • A


Sequences that increase by multiplying
Sequences that increase by multiplying

×2

×2

×2

×2

×2

×2

×2

Some sequences increase or decrease by multiplying or dividing each term by a constant factor.

For example, look at this sequence:

2, 4, 8, 16, 32, 64, 128, 256, . . .

This sequence starts with 2 and increases by multiplying the previous term by 2.

All of the terms in this sequence are powers of 2.


Sequences that decrease by dividing
Sequences that decrease by dividing

,

, . . .

÷4

÷4

÷4

÷4

÷4

÷4

÷4

1

1

4

16

We could also continue this sequence by multiplying each term by to give the next term.

1

4

Can you work out the next three terms in this sequence?

1024, 256, 64, 16, 4,

1,

How did you work these out?

This sequence starts with 1024 and decreases by dividing by 4 each time.


Geometric sequences
Geometric sequences

r = 12 ÷ 8 =

1.5

Sequences that go from one term to the next by multiplying by a constant amount are called geometric sequences.

The amount that each term is multiplied by to get the next term is called the common ratio of the sequence.

The common ratio r of a geometric sequence can be found by dividing any term in the sequence by the one before it. For example,

Find the common ratio for the following geometric sequence:

8, 12, 18, 27, 40.5, …

The sequence continues by multiplying the previous term by 1.5.



The n th term of a geometric sequence
The nth term of a geometric sequence

Suppose we are given a geometric sequence and asked to find the nth term un of the sequence. For example:

Find the nth term of the sequence, 3, 6, 12, 24, 48, …

The first term of this sequence is 3 and it continues by multiplying the previous term by 2 each time. We can write this sequence as follows:

u1 = 3

u2 = 6 = 3 × 2

u3 = 12 = 3 × 2 × 2 = 3 × 22

u4 = 24 = 3 × 2 × 2 × 2 = 3 × 23

u5 = 24 = 3 × 2 × 2 × 2 × 2 = 3 × 24

un = 3 × 2(n – 1)


The n th term of a geometric sequence1
The nth term of a geometric sequence

In general, if the first term of a geometric sequence is a and the common ratio is r, then we can find the general form of the nth term as follows:

u1 = a,

u2 = a × r,

u3 = a × r2,

u4 = a × r3,

u5 = a × r4,

un= a × rn – 1,

The nth term of a geometric sequence with first term a and common ratio r is

un= arn – 1


Geometric sequences in real life
Geometric sequences in real-life

×1.05

×1.05

×1.05

Geometric sequences often occur in real-life situations where there is a repeated percentage change.

For example, £800 is invested in an account with an annual interest rate of 5%.

Write a formula for the value of the investment at the beginning of the nth year.

Every year the amount in the account is multiplied by 1.05.

The amount in the account at the beginning of each year forms a geometric sequence with £800 as the first term and 1.05 as the common ratio.

£800,

£840,

£882,

£926.10, …


Geometric sequences in real life1
Geometric sequences in real-life

Geometric sequences often occur in real-life situations where there is a repeated percentage change.

For example, £800 is invested in an account with an annual interest rate of 5%.

Write a formula for the value of the investment at the beginning of the nth year.

We can write the sequence as,

u1 = £800

u2 = £800 × 1.05

u3 = £800 × 1.052

u4 = £800 × 1.053

un = £800 × 1.05n – 1


Contents4

A7 Sequences

Contents

A7.1 Generating sequences from rules

  • A

A7.2 Linear sequences

  • A

A7.5 Other types of sequence

A7.3 Quadratic sequences

  • A

A7.4 Geometric sequences

  • A

  • A


Fraction sequences
Fraction sequences

2

3

4

5

6

,

,

,

,

, …

3

6

9

12

15

+3

+3

+3

+3

+3

+3

+1

+1

+1

+1

+1

+1

2

3

4

5

6

7

8

,

,

,

,

,

,

, …

The terms in the following sequence are all fractions:

Find the next two terms in the sequence.

Look at the sequence formed by the numerators and the sequence formed by the denominators separately.

3

6

9

12

15

18

21


Fraction sequences1
Fraction sequences

2

3

4

5

6

,

,

,

,

, …

3

6

9

12

15

The terms in the following sequence are all fractions,

Find the formula for the nth term of the sequence.

The sequence formed by the numerators is:

The sequence formed by the numerators is:

u1 = 2

u1 = 3

u2 = 3

u2 = 6

u3 = 4

u3 = 9

un =

n + 1

un =

3n


Fraction sequences2
Fraction sequences

2

3

4

5

6

,

,

,

,

, …

3

6

9

12

15

n + 1

un =

3n

The terms in the following sequence are all fractions:

Find the formula for the nth term of the sequence.

Given that the nth term for the numerators is n + 1 and the nth term for the denominators is 3n, we can write the formula for the nth term of the above sequence as:


The fibonacci sequence
The Fibonacci Sequence

21,

34,

55, . . .

13+21

5+8

8+13

1+1

1+2

2+3

3+5

21+34

Can you work out the next three terms in this sequence?

1, 1, 2, 3, 5, 8, 13,

How did you work these out?

This sequence starts 1, 1 and each term is found by adding together the two previous terms.

This sequence is called the Fibonacci Sequence after the Italian mathematician who first wrote about it.


The n th term of the fibonacci sequence
The nth term of the Fibonacci Sequence

The Fibonacci Sequence is an example of a sequence for which the general term cannot be written in terms of its position in the sequence.

To find any given term we have to know the value of the previous two terms.

We can write:

u1 = 1

u2 = 1

u3 = u2 + u1

u4 = u3 + u2

u5 = u4 + u3

The nth term can then be written as:

un = un – 1 + un – 2


The n th term of the fibonacci sequence1
The nth term of the Fibonacci sequence

The formula is called an iterative formula.

un = un – 1 + un – 2

The sequence cannot be defined by this formula alone. We have to be given the value of the first two terms.

Can you use this formula to find the 100th term in the sequence?

Using this formula would give us,

u100 = u99 + u98

All this tells us is that the 100th term is equal to the sum of the 99th term and the 98th term.

The only way to find the 100th term is to use a computer or spend some time doing a lot of arithmetic!


Using iterative formulae to generate sequences
Using iterative formulae to generate sequences

Write down the first five terms of the sequence generated by the iterative formula

un = 2un – 1 + 1

when u1 = 3.

u1 = 3

u2 = 6 + 1 = 7

u3 = 14 + 1 = 15

u4 = 30 + 1 = 31

u5 = 62 + 1 = 63

The first five terms of the sequence are 3, 7, 15, 31, 63, …

Can you find the formula for the nth term of this sequence?