Find a formula connecting the variables in each table below :

g

1 2 3 4 5

P 3 8 13 18 23

a

1 2 3 4 5

F 4 7 10 13 16

x

5 6 7 8 9

E 13 15 17 19 21

+3 +3 +3 +3

F = 3 × a + 1

F = 3a + 1

+5 +5 +5 +5

P = 5g – 2

E = 2x + 3

Find a formula for the nth term of each of the following sequences and hence find the 20th term of each sequence.

(a)3 , 7 , 11 , 15 , ........

(b)14 , 26 , 38 , 50 , .......

(c) 2 , 8 , 14 , 20 , ........

+4 +4nth term = 4n – 1

20th term = 4 × 20 – 1 = 79

nth term = 12n + 2

20th term = 12 × 20 + 2 = 242

nth term = 6n – 4

20th term = 6 × 20 – 4 = 116

+12 +12

Notation We write the terms of a

sequence asu1 , u2 , u3 , …….., un-1 , un , un+1 ,

……...where u1 is the 1st term, u2 is the 2nd

term etc….and uunn is the nth term

( n being any whole number.)

Recurrence Relations

A formula for the nth term, un in terms of n

Or

A REURRENCE RELATION calculating each term by

using the previous term

Consider 5 9 13 17……. +4 +4 +4

un = 4n + 1

un+1 = un + 4

u1 = 4×1 + 1, u2 = 4×2 + 1, u3 = 4×3 +1, …………

u1 = 5, u2 = 5 + 4, u3 = 9 + 4, u4 = 13 + 4, ………

Growth & Decay

Removing 15% leaves behind 85% or 0∙85 which is called the DECAY factor or multiplier

Adding on 21% gives us 121% or 1∙21 and this is called the GROWTH factor or multiplier

Growth and decay factors allow us a quick method of tackling repeated % changes

Example 1

An oven contains 10 000 bacteria which are being killed off at a rate of 17% per hour by a particular disinfectant.

(a) How many bacteria are left after 3 hours?

(b) How many full hours are needed so that there are fewer than 4 000 bacteria?

Let un be the number of bacteria remaining after n hours.

Removing 17% leaves behind 83%

so the DECAY factor is 0∙83 and un+1 = 0∙83 un

(a) u0 = 10000

u1 = 0∙83u0 = 0.83 × 10000 = 8300

u2 = 0∙83u1 = 0∙83 × 8300 = 6889

u3 = 0∙83u2 = 0∙83 × 6889 = 5718

So there are 5718 bacteria after 3 hours.

(b) u4 = 0∙83u3 = 0∙83 × 5718 = 4746

u5 = 0∙83u4 = 0∙83 × 4746 = 3939

This is less than 4 000 so it takes 5 full hours to fall below 4000.

u0 = original value

Example 2 The population of a town grows at a rate of 14% per annum.

If P0 is the initial population and Pn is the population after n years.

(a) Find a formula for Pn in terms of P0.

(b) Find roughly how long it takes the population to treble.

Adding on 14% gives us 114%

so the GROWTH factor is 1∙14 and Pn+1 = 1∙14 Pn

P1 = 1∙14 P0

P2 = 1∙14 P1 = 1∙14 × 1∙14 P0 = (1∙14)2 P0

P3 = 1∙14 P2 = 1∙14 × (1∙14)2 P0 = (1∙14)3 P0

So in general we have Pn = (1∙14)n P0

If the population trebles then we need to have

Pn > 3 P0

or (1∙14)n P0 > 3 P0 Dividing by P0

we get (1∙14)n > 3

We can use the calculator answer (ANS) button

Insert 1∙14 by 1∙14 (=)ENTER

1∙14 × ANS

Now each time you press the = button we multiply by another 1∙14

Count the number of times you press = to get a number greater than 3

From the above we can say it takes just over 8 years for the population to treble.

Examples to try

1. A population of bacteria grows at a rate of 20% per day. In an experiment the initial number of bacteria was 150.

a) Find a recurrence relation to model this experiment.

b) How many bacteria are there after 3 days?

c) How long will it take the population to double?

20% increase 120% left un+1 = 1∙2un

uo = 150 150 (=)ENTER 1∙2 × ANS

u1 = 180 u2= 216 u3= 259 u4= 311

259 after 3 days Doubles after 4 days

2. A health report states that the level of harmful gasses in the atmosphere should be no more than 110 units. The current level is 150.

Environmental Health officials introduce a plan to reduce these gasses by 5% per annum.

How many years will it take for a safe level to be attained?

5% decrease 95% left un+1 = 0∙95un

uo = 150 150 ENTER 0∙95 × ANS

u1 = 143 u2= 135 u3= 129 u7= 104

Falls below 110 during 7th year

3. When an oil tanker runs aground it spills 25 000 tonnes of oil. The natural action of the waves will disperse the oil at a rate of 40% per week.

a) How long will it take to reduce the amount of oil to 1 000 tonnes?

b) After how many weeks will it be less than 100 tonnes.

40% decrease 60% left un+1 = 0∙6un

uo = 25000 25000 ENTER 0∙6 × ANS

u1 = 15000 u2= 9000 u3= 5400 u7= 700

Falls below 1000 during 7th week.

u11= 91 Falls below 100 during 11th week.

4. Ethiopia’s population at the start of 2005 was 62 million. The rate of increase is 3% per annum. What is the population likely to be at the start of 2010?

3% increase 103% left un+1 = 1∙03un

uo = 62 62 ENTER 1∙03 × ANS

u1 = 63∙8 u2= 65∙8 u3= 67∙7 u4= 69∙8

2006

u5= 71∙9

2007 2008 2009 2010

71∙9 million at the start of 2010

5. An art dealer bought a painting for £2∙5 million at the beginning of 2003. She expects the value to increase by 6% per annum.

a) Find a recurrence relation for the value of the painting.

b) What is its expected value at the start of 2007?

6% increase 106% left un+1 = 1∙06un

uo = 2∙5 2∙5 ENTER 1∙03 × ANS

u1 = 2∙575 u2= 2∙65 u3= 2∙73 u4= 2∙81

£2∙81 million at start of 2007

2004 2005 2006 2007

6. A virulent strain of ‘flu will affect 12% of the population per week if no preventative measures are taken.

a) What percentage remains healthy?

b) For a village of 5 000 people, how many will be healthy after 3 weeks.

c) How long will it take for half the population to be infected.

12% ill 88% healthy un+1 = 0∙88un

uo = 5000 5000 ENTER 0∙88 × ANS

u1 = 4400 u2= 3872 u3= 3407

After 3 weeks 3407 healthy u6= 2322

Half the population falls ill during 6th week

A balloon contains 1 500ml of air and is being inflated by mouth. Each puff inflates

the balloon by 15% but at the same time 100 ml of air escapes.

(i) Find a linear recurrence relation to describe this.(ii) How much air is in the balloon after 5 puffs? (iii) If the volume reaches 3 litres then the balloon will burst. How many puffs will this take?(i) Suppose the starting volume is v0.Adding 15% gives us 115% or 1∙15 ×

previous amount

Linear Recurrence Relations

But 100 ml then escapes

v1 = 1∙15v0 – 100

v2 = 1∙15 × 1625 – 100

v3 = 1∙15 × 1768 – 100

In general vn+1 = 1∙15vn – 100

v0 = 1 500

= 1625

= 1768

= 1934

Because 100 ml Because 100 ml escapesescapes

In general vn+1 = 1∙15vn – 100

Each press of = gives the next termv0 = 1500

v1 = 1625

v2 = 1768

And so on

v5 = 2343

v8 = 3216

Balloon bursts on 8th puff

(ii) We can now use this formula as follows

Using ANS function 1500 ENTER 1∙15 × ANS

A patient is given 160 ml of a drug. Every 6 hours 25% of the drug passes out of her bloodstream. To compensate she is given a further 20 ml dose every 6 hours.

a) Write down a recurrence relation for the amount of drug in her bloodstream.

b) How much drug will be in her bloodstream after 24 hours?

25% removed 75% left un+1 = 0∙75un + 20

uo = 160 160 ENTER 0∙75 × ANS + 20

u1 = 140 u2= 125 u3= 113∙75 u4= 105∙3

105 ml left in bloodstream

A job is advertised a starting salary of £24 000, with annual percentage increase of 7∙5%

and an annual increment of £2 000.

a) Find a recurrence relation for the total annual salary.

b) Calculate the expected salary after 8 years.

7∙5% increase 107∙5% left un+1 = 1∙075un + 2000

uo = 24000 24000 ENTER 1∙075 × ANS + 2000

u1 = 27800 u2= 31885 And so on

u8= £63 696∙21

Expected salary = £63 696∙21

The air pressure in a tractor tyre is 50 psi. Each week it loses 15% and 2 psi are pumped back in to compensate. The

manufacturer states that it is dangerous to run the tyre at under 30 psi.

a) Calculate the pressure after 3 weeks.

b) Does the pressure ever drop below 30 psi?

15% loss 85% left un+1 = 0∙85un + 2

uo = 50 50 ENTER 0∙85 × ANS + 2

u1 = 44∙5 u2= 39∙8 u3 = 35∙9

Pressure after 3 weeks is 35∙9 psi

u5 = 29∙6If you keep pressing (=) answer is 13∙33∙

Linear Recurrence Relations

un+1 = 0∙85un + 2

These recurrence relations are called linear because they all fit a form that looks like :

un+1 = aun + b

Which is the same as y = mx + c

un+1 = 1∙075un + 2000

un+1 = 0∙75un + 20 vn+1 = 1∙15vn – 100

Linear Recurrence Relations

Please note there are various ways to write the same recurrence relation.

un+1 = aun + b un = aun-1 + b

are the same since un+1 is the term after un and un is the term after un-1.

Although we usually use u to represent the term in a sequence it is not necessary or essential.

vn+1 = avn + b pn+1 = apn + b

xn+1 = axn + b gn+1 = agn + b

A farmer grows a variety of plum tree which ripens during the months of July and August. On the last day in July there was 2000 kg of ripe fruit ready to be picked. At the beginning of August the farmer hires some fruit pickers who manage to pick 75% of the ripe fruit each day. Also,

each day 60 kg more of the plums become ripe.

a) Find a recurrence for the weight of ripe plums left in the orchard.

b) What is the estimated weight of ripe plums left in the orchard at the end of the day on the 7th August ?

75% picked 25% left

Wn+1 = 0∙25Wn + 60

Enter 2000

0∙25 × ANS + 60

W1 = 560kg End of 1st August

W2 = 200kg

W7 = 80 kg

A large shoal of 300 fish are observed and it is noticed that every minute 30% of the fish leave the shoal and 25 join the shoal.

a) Find a recurrence relation for the size of the shoal of fish.

b) How many fish are in the shoal after 5 minutes and after 10 minutes?

30% leave 70% left

Nn+1 = 0∙7 Nn + 25

Enter 300 0∙7 × ANS + 25

N1 = 235

N2 = 190

N5 = 120

N10 = 89

A pharmaceutical company is given permission to discharge a maximum of 60 kg of chemical waste into a section of river each day. The natural

tidal nature of the river disperse 80% of the chemical each day.

a) Find a recurrence relation for the amount of chemical in the river.

b) How many kg of chemical waste are there in the river after 2 days and after 7 days?

c) Due to environmental concerns the chemical waste in the section of river must not exceed 80 kg. Is it safe for the pharmaceutical company

to continue discharging waste at this rate? Give a reason.

80% removed 20% left

An+1 = 0∙2An + 60

Enter 60

0∙2 × ANS + 60

A1 = 72 kg

A2= 74∙4 kg

A7 = 75 kg

In the long term level of waste levels off at 75 kg, so safe.

Statement is essential,

always worth a mark!

120 rabbits are bred for sale. Each month the rabbit population increases by 15% and each month 30 rabbits are sold to customers.

a) Write down a recurrence relation for the population of rabbits.

b) Calculate how many rabbits there are after 6 months.

c) What will happen to the rabbit population if it continues in this way?

15% increase 115% left

Rn+1 = 1∙15 Rn - 30

Enter 120 1∙15 × ANS - 30

R1 = 108

R6 = 15

R10 = - 124

In the long term the rabbits will all be sold and the population disappears.

Statement is essential,

always worth a mark!

A patient is injected with 90 units of a drug in hospital. Every eight hours 30% of the drug passes out of the bloodstream. The patient is

therefore given a further dose of 20 units of the drug at 8 hr intervals.

a) Find a recurrence relation for the amount of the drug in the bloodstream.

b) Calculate how many units of the drug you would expect there to be in the patient’s bloodstream after 48 hours.

30% loss 70% left

Dn+1 = 0∙7 Dn + 20

Enter 90

0∙7 × ANS + 20

D1 = 83

D6 = 69 units 6 × 8hrs

Note that in the long term number of units of the drug will level off at 66∙7 units.

A family have taken out a £80000 mortgage to buy a cottage. The building society charge interest at 6% per annum. The family pay

back £6000 each year.

a) Find a recurrence relation for the amount of money owed to the building society.

b) How much money do they owe after 5 years, after 10 years?

c) During what year will the mortgage be paid off?

6% interest 106% left

Mn+1 = 1∙06Mn – 6 000

Enter 80 000

1∙06 × ANS – 6 000

M1 = £78 800

M5 = £73 235

Mortgage paid off during year 27

M10 = £64 183

M28 = - £2 233

Divergence / Convergence/ Limits

Consider the following linear recurrence relations

(a) un+1 = 2un + 4 with u0 = 3

u0 = 3

u1 = 10

u2 = 24

u3 = 52

u10 = 7164

u20 = 7340028

As n un

3 =

2 × ANS + 4

we say that the sequence DIVERGES.

As n tends to infinity un tends to infinity

(b) un+1 = 0∙5un + 4 with u0 = 3

u0 = 3

u1 = 5∙5

u2 = 6∙75

u3 = 7∙375

u10 = 7∙995

u20 = 7∙999…..

As n un 8

we say that the sequence CONVERGES to a limit of 8.

3 =

0∙5 × ANS + 4

(c) un+1 = -2un + 4 with u0 = 3

u0 = 3

u1 = -2

u2 = 8

u3 = -12

u10 = 1708

u20 = 1747628

u21 = -3495252

As n un ±

and we say that the sequence DIVERGES.

3 =

-2 × ANS + 4

(d) un+1 = -0∙5un + 4 with u0 = 3

u0 = 3

u1 = 2∙5

u2 = 2∙75

u3 = 2∙625

u10 = 2∙666

u20 = 2∙666

As n un 22/3

we say that the sequence CONVERGES to a limit of 22/3

3 =

-0∙5 × ANS + 4

As n un ±

we say that the sequence DIVERGES

As n un → 8 or un 22/3

we say that the sequence CONVERGES to a limit

For un+1 = 0∙5un + 4 and un+1 = -0∙5un + 4

For un+1 = 2un + 4 and un+1 = – 2un + 4

Conclusions

The linear recurrence relation

un+1 = aun + b

converges to a limit only if

– 1 < a < 1

Otherwise the sequence diverges.

When the linear recurrence relation

un+1 = aun + b

converges the answer remains the same no matter how many times you repeat the calculation.

So un = un+1 = un+2 + un+3 ……

If we let this limit be L, then

un+1 = aun + b becomes L = aL + b

baLL So baL )1(

ab

L

1

Limit formul

a

Other Factors

(e) compare this with (b)

un+1 = 0∙5un + 10 with u0 = 3

u0 = 3

u1 = 11∙5

u2 = 15∙75

u3 = 17∙875

…..

u10 = 19∙98...

……

u20 = 19∙99….

This is clearly heading to a limit of 20

Conclusion:

if un+1 = aun + b converges to a limit

then changing b changes the limit.

(f) compare this with (b)

un+1 = 0∙5un + 4 with u0 = 200

u0 = 200

u1 = 104

u2 = 56

u3 = 32

u10 = 8∙1875

u20 = 8∙0001…. Again this is heading to a limit of 8

Conclusion:

if un+1 = aun + b converges to a limit then changing u0 does not affect the

limit.

A Formula for the Limit of a Converging Sequence

un+1 = aun + b, converges if – 1 < a < 1

The limit, L, is given by the formula

a1

bL

The limit depends on a and b but not on u0

Fish, like all animals need oxygen to survive. The fish in a certain tank use up 15% of the oxygen in the water each hour. However, due to the action of a pump, oxygen is added to the water at a rate of 1 part per

metre3 each hour. The oxygen level in the tank should be between 5 and 7 parts per metre3 for the survival of the fish. Initially the concentration

of oxygen in the tank is 6 ppm3

a) Write down a recurrence relation to describe the oxygen level . b) Say whether or not a limit exists, giving a reason.

c) Determine, in the long term, whether the fish will survive.

15% used 85% left

On+1 = 0∙85 On + 1

Limit exists since -1 < 0∙85 < 1

ab

L

1 8501

1

676

Fish survive since long term level is between 5 and 7 ppm3

An office worker has 70 folders on his desk ready to be filed. Each hour he manages to file 62% of the folders. However, another 20 are also

added to his pile each hour.

a) Letting Fn represent the number of folders on his desk after n hours, write down a recurrence relation to model this situation.

b) How many folders are ready to be filed after 5 hours?

c) In the long term how many folders should he expect on his desk ?

62% filed 38% left

Fn+1 = 0∙38 Fn + 20

In the long term the number of files levels off at 32

Enter 70 0∙38 × ANS + 20

F1 = 46

F5 = 32

Limit exists since -1 < 0∙38 < 1

Trees are sprayed weekly with the pesticide, ’Killpest’, whose manufacturers claim it will destroy 65% of all pests. Between the

weekly sprayings, it is estimated that 500 new pests invade the trees. A new pesticide, ’Pestkill’, comes onto the market. The manufacturers claim it will destroy 85% of existing pests but it is estimated that 650

new pests per week will invade the trees.

Which pesticide will be more effective in the long term?

65% killed 35% left

Kn+1 = 0∙35 Kn + 500

ab

L

1 3501

500

769

85% killed 15% left

Pn+1 = 0∙15 Pn + 650

ab

L

1 1501

650

764

In the long term Pestkill is marginally better

A Local Authority puts its Park cleaning out to tender. Kleenall claims it will remove 95% of all rubbish dropped each week. It is known that the

public drop 25kg of rubbish each week. Pickit Ltd claim to remove 85% of all the litter dropped each week. They also say they will install litter bins which will reduce the amount of litter dropped each week to 20kg per week. If the two firms charge the same amount, which one

should the council employ?

95% cleared 5% left

Kn+1 = 0∙05 Kn + 25

ab

L

1 0501

25

kg326

85% cleared 15% left

Pn+1 = 0∙15 Pn + 20

ab

L

1 1501

20

kg523

In the long term Pickit Ltd is better

A chemical firm has applied to release 45 units of waste per week into the sea. It is estimated that the natural tidal action will disperse 60% of the

waste each week. If a safe level is 80 units, is it safe to accept the application.

60% cleared 40% left

Cn+1 = 0∙4 Cn + 45

ab

L

1 401

45

units75

In the long term level is constant at 75 units, so safe.

A hospital patient is put on medication which is taken once per day. The dose is 35mg and each day the patient’s metabolism burns off 70% of the drug

in her system. It is known that if the level of the drug in the patients system reaches 54mg then the

consequences could be fatal. Is it safe for the patient to take the medication indefinitely?

un+1 = 0∙3un + 35

Burning off 70% leaves behind 30% or 0∙3

a = 0∙3, b = 35

a

bL

1 301

35

50

In the long run level of drug in patients system levels out at 50 mg.

The brake fluid reservoir in a car is leaky. Each day it loses 3∙14% of its contents. The driver “tops up” once per

week with 50ml of fluid. For safety reasons the level of fluid in the reservoir should always be between 200ml &

260ml. Initially it has 255ml.

3∙14% loss leaves 96∙86%

(a) Determine the fluid levels after 1 week and 4 weeks.

(b) Is the process effective in the long run?

decay factor 0∙9686

(a) Problem 3.14% daily loss = ? Weekly loss.

Decay factor = 0∙9686

Amount remaining after 1 week = (0.9686)7 × A0

= 0∙799854 × A0

= 0∙80 × A0

or 80% of A0

The car is loses 20% of its brake fluid weekly

So if An is the fluid level after n weeks then we have

An+1 = 0∙8 An + 50

7 day factor = (0∙9686)7

(b) Using An+1 = 0.8 An + 50 with A0 = 255 we get

A1 = 254ml 1st weekA2 = 253∙2mlA3 = 252∙6mlA4 = 252∙0ml 4th week

NB : even before adding the 50ml

the level is above 200ml

Enter 225 0∙8 × ANS + 50

(c) considering An+1 = 0∙8 An + 50

Since – 1 < 0.8 < 1 then a limit must exist

where a = 0∙8, b = 50

In the long run the weekly level will be 250ml and won’t fall below 200ml so the driver should be

OK with this routine.

a

bL

1

801

50

L 250

Given u6 = 48 , u7 = 44 and u8 = 42 find a & b .

Finding a Formula

A recurrence relation is defined by the formula

un+1 = aun + b

u8 = au7 + b becomes 44a + b = 42u7 = au6 + b becomes 48a + b = 44

Subtract up 4a = 2 so a = 0∙5

Now put a = 0∙5 into 44a + b = 42

to get 22 + b = 42 so b = 20

The nth term in a sequence is given by the formula

un = an + b

Given that u10 = 25 and u12 = 31 then find a & b.

Hence find u300 , the 300th term.

u10 = 10a + b becomes 10a + b = 25

u12 = 12a + b becomes 12a + b = 31

subtract up 2a = 6 a = 3

Now put a = 3 into 10a + b = 25Now put a = 3 into 10a + b = 25

This gives us 30 + b = 25 So b = – 5

The actual formula is un = 3n – 5

So u300 = 3 × 300 – 5 = 895

eg 8, 14, 20, 26, ……here d = un+1 - un = 6

u1 = 8 = 8 + (0 × 6)

Two Special Series

In an arithmetic series there is a constant difference between consecutive terms.

u2 = 14 = 8 + (1 × 6)

u3 = 20 = 8 + (2 × 6)

u4 = 26 = 8 + (3 × 6)

In general un = u1 + (n-1) × d

So for the above u100 = u1 + 99d = 8 + (99 × 6) = 602

In a geometric series there is a constant ratio between consecutive terms.

eg 5, 10, 20, 40, ……here r = un+1 un = 2

u1 = 5 = 5 × 20

u2 = 10 = 5 × 21

u3 = 20 = 5 × 22

u4 = 40 = 5 × 23

In general un = u1 × r(n-1)

So for the above u100 = u1 × r 99 = 5 × 299 = 3.17 × 1030

The following slides show exam The following slides show exam questions and their solutions.questions and their solutions.

Remember for uRemember for un+1n+1 = = aauunn + + bb

Limit only if Limit only if – 1 < – 1 < aa < 1 < 1

4% decrease 4% decrease → 96% left → → 96% left → aa = 0∙96 = 0∙96

4% increase 4% increase → 104% left → → 104% left → aa = 1∙04 = 1∙04

Given values for Given values for uunn and and uun+1n+1 use use simultaneoussimultaneous

equationsequations to find to find aa and and bb

Put u1 into recurrence relation

Solve simultaneously:

A recurrence relation is defined by un+1 = pun + q

where -1 < p < -1 and u0 = 12

a) If u1 = 15 and u2 = 16 find the values of p and q

b)Find the limit of this recurrence relation as n

Put u2 into recurrence relation

State limit condition -1 < p < 1, so a limit L exists

Use formula

15 = 12p + q …….. (1)

16 = 15p + q …….. (2)

pp = = 11//33 qq = 11 = 11

a

bL

13

11

11

L

2

116 L

A man decides to plant a number of fast-growing trees as a boundary between his property and

the property of his neighbour. He has been warned however by the local garden centre, that

during any year, the trees are expected to increase in height by 0.5 metres.

In response to this warning, he decides to trim 20% off the height of the trees at the start of any

year.

(a)If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the

long run.

(b)His neighbour is concerned that the trees are growing at an alarming rate and wants

assurance

that the trees will grow no taller than 2 metres. What is the minimum percentage that the trees

will need to be trimmed each year so as to meet this condition.

Construct a recurrence relation

State limit condition -1 < 0∙8 < 1, so a limit L exists

Use formula

Where un = height at the start of year n

Use formula again with L = 2

a = 0∙75 means 75% left

un+1 = 0∙8un + 0∙5

20% reduction → 80% (0∙8) left

a

bL

1 801

50

L metresL 25

To find % reduction required make L = 2 metres

a

1

502 5022 a 512 a 750 a

Minimum prune = 25%

On the first day of March, a bank loans a man £2500 at a fixed rate of interest of 1.5% per month.

This interest is added on the last day of each month and is calculated on the amount due on the first day of the month.

He agrees to make repayments on the first day of each subsequent month. Each repayment is £300 except for the

smaller final amount which will pay off the loan.

 a) The amount that he owes at the start of each month is taken to be the amount still owing just after the monthly

repayment has been made.

Let un and un+1 and represent the amounts that he owes at the starts of two successive months.

Write down a recurrence relation involving un and un+1

b) Find the date and amount of the final payment.

Construct a recurrence relation

u0 = 2500

Calculate each term in the recurrence relation

1 Mar u0 = 2500.00

1 Apr u1 = 2237.50

1 May u2 = 1971.06

1 Jun u3 = 1700.62

1 Jul u4 = 1426.14

1 Aug u5 = 1147.53

1 Sept u6 = 864.74

1 Oct u7 = 577.71

1 Nov u8 = 286.38

1 Dec Final payment £290.68

un+1 = 1∙015un – 300

1∙5% interest → 101∙5% (1∙015) total

You can use ANS function on your calculator.

2500 = then 1∙015 ×ANS – 300

L1 = L2 so

Sequence 1

Since limit exists a 1

Use formula for each sequence

Limit = 25

Two sequences are generated by the recurrence relations un+1 = aun + 10 and vn+1 = a2vn +16

The two sequences approach the same limit as n .

Determine the value of a and evaluate the limit.

Sequence 2

Simplify

a

bL

1

aL

1

101 22 1

16

aL

21

16

1

10

aa

aa 16161010 2

061610 2 aa

0385 2 aa0)1)(35( aa

15

3 aora

5

3 a

L1 = L2 so Cross multiply

Sequence 1

Use formula for each sequence

Sequence 2

Rearrange

Two sequences are defined by the recurrence relations

un+1 = 0∙2un + p, u0 = 1 and vn+1 = 0∙6vn + q, v0 = 1

If both sequences have the same limit, express p in terms of q.

a

bL

1

2011

pL

6012

qL

4080

qp

qp 8040

40

80

q

p qp 2

Two sequences are defined by these recurrence relations

un+1 = 3un - 0∙4, u0 = 1 and vn+1 = 0∙3vn + 4, v0 = 1

a) Explain why only one of these sequences approaches a limit as n

b) Find algebraically the exact value of the limit.

c) For the other sequence findi) the smallest value of n for which the nth term

exceeds 1000, andii)the value of that term.

Sequence 2

Requirement for a limit

List terms of 1st sequence

Only the second sequence with a = 0∙3 has a limit

u0 = 1

u1 = 2∙6

u2 = 7∙4

u3 = 21∙8

u4 = 65

u5 = 194∙6

u6 = 583∙4

u7 = 1749∙8

Smallest value of n is 8; value of 8th term = 1749∙8

– – 1 < 1 < aa < 1 < 1

a

bL

1 301

4

L

70

4

7

40

7

55

Exact value means leave as a fractionExact value means leave as a fraction

Use ANS functionUse ANS function

Trees are sprayed weekly with the pesticide “Killpest”, whose manufacturers claim it will destroy 65% of all pests. Between sprayings, it is estimated that 500 new pests invade the trees.

A new pesticide “Pestkill”, comes on the market. It is claimed to destroy 85% of existing pests but it is estimated that 650 new pests

will invade the trees.

Which pesticide will be more effective in the long run?

Killpest un+1 =0∙35un + 500

11 since1

aa

bL

3501

500

L

769L

Pestkill un+1 =0∙15un + 650

11 since1

aa

bL

1501

650

L

765L

In the long run Pestkill is slightly better with a limit 4 less than Killpest

A sequence is generated by the recurrence relation un+1 = 0∙6un + 5.

Which of these is the limit of this sequence as n → ∞

A. 5 B. 2/5 C. 3/25 D. 25/2

ab

L

1 601

5

L

405

L × 10× 10

× 10× 10

450

L2

25 L

A recurrence relation is defined by un+1 = 0∙4un – 24. The limit of this sequence is

A. -40 B. -24 C. 0∙03 D. 50

ab

L

1 401

24

L

6024

L × 10× 10

× 10× 10

6240

L 40 L

A sequence is defined by the recurrence relation un+1 = aun + b where u0 = 5. Which of the following could be

an expression for u2, the second term of the sequence?

A. 5a2 + 2b B. 5a2 + ab C. 5a + b D. 5a2 + ab + b

50 u bau 51 ba 5

buau 12 bbaa )5(

baba 25

A sequence is defined by the recurrence relation

un+1 = 3un – 4, u0 = –1.

What is the value of u2?

A –25 B –10 C –4 D –1

ENTER – 1, 3 ENTER – 1, 3 × ANS – 4 × ANS – 4

uu11 = – 7 = – 7 uu22 = – 25 = – 25

Which of these recurrence relations has a limit?

I un+1 = 0∙4un + 5

II un+1 = 6 – 0∙5un

III un+1 = 8/9un + 2

A. none of them. B. All of them

C. I and II only D. I and III only

All the multipliers of un come between – 1 and 1

The exact limit of the sequence defined by the recurrence relation un+1 = 0∙6un + 1∙8 is

A. 8/9 B. 3 C. 9/2 D. 2/9

ab

L

1 601

81

4081

× 10× 10

× 10× 10

418

29

The two sequences defined by the recurrence relations un+1 = 0∙2un + 12 and vn+1 = 0∙3vn + k have the same limit.

The value of k is

A. 18 B. 4∙5 C. 12 D. 105

ab

L

1

20112

15

ab

L

1

301

k

70

k

70

k15

1570 k

510 k