Sequences and Series

A sequence is an ordered list of numbers where each term is obtained according to a fixed rule. 2 1nU n

A series, or progression, is a sum. The terms of which form a sequence.

1

2 1n

r

r

The nth term of a sequence is often denoted Un, so that, for example, U is the first term.

A sequence can be defined by a recurrence relation where Un+1 is given as a function of lower, earlier terms.

A first – order recurrence relation is where Un+1=rUn + d, where r and d are constants. This relation is linear.

A sequence can be defined by a formula for Un, given as a function.Un = f(n)

Being given the first few terms of a sequence is not enough to identify the sequence.

For example. Identify the next term in the sequence1, 2, 3, …, …,

Possible answers include: 2 316 9 6

2nU n n n

2 316 8 6

3nU n n n

3 216 13 6

2nU n n n

nU n

If however we also know that the sequence is generated by a first order linear recurrence relation, then we know

1 1 2 3, 1, 2, 3.n nU rU d U U U

1 1 2 3, 1, 2, 3.n nU rU d U U U

2 r d and 3 2r d

1 1r d

1 1n nU U

Example 1

Find the first order linear recurrence relation when:U3 = 7, U4 = 15 and U5 = 31.

1n nU rU d 15 7

31 15

r d

r d

16 8 2 hence 1r r d 1 2 1n nU U

Note

Given a relation 1 3 4n nU U

If 3, the sequence would proceed 3, 5, 11, 29, .....nU

If 2, the sequence would proceed 2, 2, 2, 2, .....nU

When this repetition happens, Un is referred to as a fixed point.

In this case, for any other value of Un, the relation generates values that move away or diverge from the value of 2.

Un =2 is an unstable fixed point.

Given the relation , then if for some value of n,

Un = 4, the sequence would proceed 4, 4, 4, 4, ……

1

12

2n nU U

If any other value of Un is used apart from 4, the relation generates terms whose value moves towards or converges on 4.

Un = 4 is a stable fixed point, often referred to as the limit of the recurrence relation.

In general, for the relation , we have a fixed point when

1n nU rU d

1n nU U

Solving simutaneously we get

, 11

only when 1 is the fixed point stable

n

dU r

rr

Arithmetic Sequences

If a sequence is generated so that, for all n,

2 1 1n n n nU U U U d

then the sequence is known as an arithmetic sequence. The constant d is referred to as the common difference.

1 1n n n nU U d U U d

This is a first order linear recurrence relation. 1, 0r d

Traditionally, U1 is represented by the letter a: U1 = a.

( 1)nU a n d

This can be proved by induction – LATER !!!

a) Find the nth termb) The 10th term of the arithmetic sequence 6, 11, 16, ……..

6 by inspectiona

a) Using ( 1)nU a n d

11 6 5d

( 1)nU a n d

6 5( 1)nU n 1 5nU n

10) 1 5 10 51b U

b) Find the arithmetic sequence for which U3 = 9 and U7 = 17.

3 9 (3 1) 9 2 9U a d a d

7 17 (7 1) 17 6 17U a d a d

Subtracting gives: 4 8 2d d

Substituting gives: 4 9 5a a

5 1 2 2 3nU n n

c) Given the arithmetic sequence 2, 8, 14, 20, …. For what value of n is Un = 62?

2, 6 2 1 6 6 4na d U n n

62 6 4n

6 66

11

n

n

Page 117 Exercise 2A Questions 1 a, c, d2 a to e3, 4, 6.

TJ Exercise 1 Questions 1 to 3

The Sum to n Terms on an Arithmetic Sequences

12 1

2nS n a n d

Proof

( ) ( 2 ) ( 3 ) ......... ( ( 2) ) ( ( 1) )nS a a d a d a d a n d a n d

( ( 1) ) ( ( 2) ) .... ( 3 ) ( 2 ) ( )nS a n d a n d a d a d a d a

Adding

2 (2 ( 1) ) (2 ( 1) ) ........(2 ( 1) ) (2 ( 1) )nS a n d a n d a n d a n d

(2 ( 1) )n a n d

12 1

2nS n a n d

Find the sum of the first 15 terms of the arithmetic sequence which starts 3, 8, 13, 18, ……….

By inspection 3, 5, 15a d n

15

156 14 5

2S

570

When does the sum of the arithmetic sequence which starts 2, 10, 18, 26,…. First exceed 300?

By inspection 2, 8a d

4 1 82n

nS n

2 4( 1)n n (2 4 4)n n

24 2n n

We require 24 2 300n n 24 2 300 0n n

Solving 24 2 300 0n n we get 8.4, 8.9n and n

y

x

8

8

– 8

– 8

100

100

200

200

300

300

400

400

– 100

– 100

– 200

– 200

– 300

– 300

– 400

– 400

Given that , we see that 9n n

The sum of the first four terms of an arithmetic sequence is 26. The sum of the first twelve terms is 222. What is the sum of the first 20 terms?

4 2 2 3 26 2 3 13S a d a d

12 6 2 11 222 2 11 37S a d a d 8 24

3

d

d

When 3, 2 9 13 2 4d a a

4 3 12n

nS n

20 10 4 3 19S

610

NOTE Given and knowing ( 1) ,n nU U a n d

2n n

nS a U

Page 120 Exercise 3A Questions 1, 3, 4, 5 and 8

TJ Exercise 1 Questions 4 to 9

Geometric SequencesIf a sequence is generated so that for all n

1 2

1

n n

n n

U Ur

U U

then the sequence is known as a geometric sequence. The constant r is referred to as the common ratio.

1Since n

n

Ur

U

1 a first order linear recurrence relation. ( 0, 0)n nU rU d r

1 1Again is traditionally represented by the letter .U a U a

1nnU ar The nth term:

a) Find the nth term and the 10th term of the geometric sequence3, 12, 48, …….

12By inspection 3, 4

3a r

1 13 4n nnU ar

910 3 4 786,432U

b) Find the geometric sequence whose 3rd term is 18 and whose 8th term is 4374

23

78

18

4374

U ar

U ar

7

2

4374

18

ar

ar 5 243 3r r

Substituting gives 9 18 2a a 12 3n

nU

c) Given the geometric sequence 5, 10, 20, 40,…… find the value of n for which 20480nU

10By inspection 5, 2

5a r

120480 5 2n 12 4096n

1 ln 2 ln 4096n

ln 40961 12

ln 2n

13n

Page 123 Exercise 4A Questions 1 a - e, 2, 3, 5 and 7

TJ Exercise 2A Questions 1 to 4

The Sum to n Terms of a Geometric Sequence

1

1

n

n

a rS

r

PROOF2 3 2 1...... n n

nS a ar ar ar ar ar 2 3 2 1....... n n n

nrS ar ar ar ar ar ar Multiplying by r:

Subtracting: nn nS rS a ar

(1 ) (1 )nnS r a r

11

1

n

n

a rS r

r

a) Find the sum to 6 terms of the geometric sequence whose first term is 6 and whose common ratio is 1.5.

6, 1.5a r 6

6

6(1 1.5 )

1 1.5S

83.125

b) A geometric sequence starts 12, 15, 18.75,……What is the smallest value of n for which Sn>100?

15 512,

12 4a r

512 1

4We require 100

51

4

n

512 1

4100

51

4

n

548 1 100

4

n

548 1 100

4

n

51 2.083

4

n

53.083

4

n

ln3.083

ln1.25n

5.046n

Since , 6n n

c) A geometric series is such that S3 = 14 and S6 = 126. Identify the series.

3114

1

a r

r

61

1261

a r

r

1r

Dividing we get:6

3

19

1

r

r

3 69 1 1r r 3 69 9 1 0r r

6 39 8 0r r 3 31 8 0r r 3 31 8r or r

Since 1, 2r r

1 8Substituting Gives: 14 2

1

aa

1 2 32, 4, 8U U U

12 2nnU

The series is 2, 4, 8, .....

Page 127 Exercise 5a Questions 1, 2, 3, 4,

TJ Exercise 2A Questions 5 to 7

Infinite Series, Partial Sums, Sum to infinity.

An Infinite series is a series which has an infinite number of terms.

When we have an infinite series then Sn is defined as the sum to n terms of that series. Such a sum is referred to as a partial sum of the series.

If the partial sum, Sn, tends towards a limit as n tends to infinity, then the limit is called the sum to infinity of the series.

Arithmetic Series

2 ( 1)2n

nS a n d

2 2Rearranging:

2n

dadS n

n

2, 0da

As nn

2

So becomes a good approximation for the sum. 2

n d

But , depending on n S d

The sum to infinity for an arithmetic series is undefined.

Geometric Series

11

1

n

n

a rS r

r

When 1,r as nr n nS

So is undefined when 1.nS r

When 1,r 0 as nr n

1

aS

r

a) Find the sum to infinity of the geometric series 24 + 12 + 6 + …….. If it exists.

1By inspection: 24, .

2a r Since 1, exists. r S

2411 2

S

48

b) Express the recurring decimal 0.121212…… as a vulgar fraction.

0.121212...... 0.12 0.0012 0.000012 .......

0.0012 12 10.12,

0.12 1200 100a r Since 1, exists. r S

0.1211 100

S

0.12

0.99

12

99 4

33

c) Given that 12 and 3 are two adjacent terms of an infinite geometric progression with find the first term. 64S

1

4r 64

34

a

464

3

a

48a

Hence the first term is 48.

Page 131 Exercise 6A Questions 1 to 4 and 7

T.J. Exercise 2B

Expanding (1-x)-1 and Related Functions

Remember: 0

nn n r r

r

nx y x y

r

!

!( )!

n n

r r n r

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

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

n

1 2 21.........

1 1 2n n n nn nn

x x y x y y

Knowing ,1

aS

r

1consider

1 r

1 2 21( ) .........

1 1 2n n n n nn nn

x y x x y x y y

1 2 3 2 4 31 ( 1)( 2) ( 1)( 2)( 3)(1 ( )) 1 (1) ( ) (1) ( ) (1) ( ) .......

1 1 2 1 2 3r r r r

2 31 ..............r r r

This is a geometric series with first term 1 and common ration r.

2 311 ..............

1x x x

x

0

1

1r

r

xx

Now Consider1

a b

can be re written as 1b

a b aa

11 1 1 1

Hence 11

ba b a b a a a

2 31 ( 2)1 1 ( 1)( 2)( 3)1 (1) (1) (1) .........

1 1 2 1 2 3b b b

a a a a

2 3 4

11 .........

b b b ba a a a a

This is a geometric series with common ratio ba

a) Expand in ascending powers of x giving the first four terms.

11 2x

2 31From 1 ............

1r r r

r

2 311 (2 ) (2 ) (2 ) ..........

1 2x x x

x

2 31 2 4 8 ..............x x x

b) Expand giving the first four terms. 1

1 3x

2 31 11 ( 3 ) ( 3 ) ( 3 ) ............

1 3 1 ( 3 )x x x

x x

2 31 3 9 27 ..............x x x

c) Evaluate to 4 decimal places.1

0.9

2 3 41 11 0.1 0.1 0.1 0.1 .....

0.9 1 0.1

1 0.1 0.01 0.001 0.0001 .....

1.1111 correct to 4 decimal places

d) Expand in ascending powers of x giving the first four terms 12 3x

1 1 1 1 13 32 3 2 21 12 2

x x x

2 3

1 3 3 31 .....

2 2 2 2x x x

2 31 3 9 271 ..............

2 2 4 8x x x

2 31 3 9 27............

2 4 8 16x x x

Page 134 Exercise 7A Questions 2, 4.Page 134 Exercise 7B Questions 2, 5.

TJ Exercise 3

The Sequence and Limit of 11

n

n

1 2 2( 1)..........

1 1 2n n n nn n n

x y x x y x y

1Replace with 1 and with , then:-x y

n

2 31 1 ( 1) 1 ( 1)( 2) 1

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

nn n n n n

nn n n n

2 3

1 ( 1) 1 ( 1)( 2) 1 ( 1)( 2)( 3)2 ....

2! 3! 4!n n n n n n

n n n

1 1 1 1 2 1 1 2 32 1 1 1 1 1 1 ......

2! 3! 4!n n n n n n

1 1 1 1 2 1 1 2 32 1 1 1 1 1 1 ......

2! 3! 4!n n n n n n

1 1 1 11 2 .............

2! 3! 4!nLim

n

2.718.....

e

Summation of a Series

The Sigma Notation The sigma notation is used as a more concise way of writing a series.

e.g. 12 + 22 + 32 + 42 + 52 +…………+n2 can be written more concisely as

2

1

n

k

k (i.e. the sum of all k2 for k = 1 to k = n)

In general is the series with the first term f(1), second term

f(2), third term f(3) and last term f(n)

1

( )n

k

f k

a) Write the following series in full.10

5

( 1)k

k k

For 5, ( 1) 5(6) 30k k k

For 6, ( 1) 6(7) 42k k k

and so on to k = 1010

5

Thus ( 1) 30 42 56 72 90 110k

k k

b) Write the following series in full. 4

2

1

2 1k

k

2For 1, (2 1) 1k k 2For 2, (2 1) 7k k

and so on to k = 44

2

1

Thus (2 1) 1 7 17 31k

k

c) Express the following in notation. 1+4+7+10+…….+298

Here, ( ) is the expression for the term of the series. thf k k

This is an arithmetic series where 1 3.a and d

The term = ( 1) 1 ( 1)3 3 2thk a k d k k

The final term is 298 so 3 2 298k 100k 100

1

Thus 1 4 7 10 13 .... 298 (3 2)k

k

Summation of a Series

The sums of certain finite series can be found by a number of methods

1

( 1)1 2 3 4 ....

2

n

k

n nk n

Proof: (2 ( 1) )2n

nS a n d When 1, 1a and d

( 1)2

n n

We can use this to help evaluate many summation series.

10

1

a) Evaluate ( 2)k

k

10 10 10

1 1 1

( 2) 2k k k

k k

10(10 1)(10 2)

2

75

20

1

b) Evaluate (4 5)k

k

20 20 20

1 1 1

(4 5) 4 5k k k

k k

20(20 1)4 (20 5)

2

940

Page 137 Exercise 8 Questions 1 (a) to (d) 2 (a) to (b) and 4

TJ Exercise 4 and 5