2. 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 F = 3 × a + 1 F = 3a + 1 +5 P = 5g – 2 E = 2x + 3

3. Find a formula for the n th term of each of the following sequences and hence find the 20 th term of each sequence. (a)3, 7, 11, 15,........ (b)14, 26, 38, 50,....... (c) 2, 8, 14, 20,........ +4 n th term = 4n – 1 20 th term = 4 × 20 – 1 = 79 n th term = 12n + 2 20 th term = 12 × 20 + 2 = 242 n th term = 6n – 4 20 th term = 6 × 20 – 4 = 116 +12

4. Notation We write the terms of a sequence as u 1, u 2, u 3, …….., u n-1, u n, u n+1, ……... where u 1 is the 1 st term, u 2 is the 2 nd term etc…. u n and u n is the n th term ( n being any whole number.) Recurrence Relations

5. A formula for the n th term, u n in terms of n Or A REURRENCE RELATION calculating each term by using the previous term Consider 591317……. +4 u n = 4n + 1 u n+1 = u n + 4 u 1 = 4×1 + 1, u 2 = 4×2 + 1, u 3 = 4×3 +1, ………… u 1 = 5, u 2 = 5 + 4, u 3 = 9 + 4, u 4 = 13 + 4, ………

6. 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

7. 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 u n be the number of bacteria remaining after n hours. Removing 17% leaves behind 83% so the DECAY factor is 0∙83 and u n+1 = 0∙83 u n

8. (a) u 0 = 10000 u 1 = 0∙83u 0 = 0.83 × 10000 = 8300 u 2 = 0∙83u 1 = 0∙83 × 8300 = 6889 u 3 = 0∙83u 2 = 0∙83 × 6889 = 5718 So there are 5718 bacteria after 3 hours. (b) u 4 = 0∙83u 3 = 0∙83 × 5718 = 4746 u 5 = 0∙83u 4 = 0∙83 × 4746 = 3939 This is less than 4 000 so it takes 5 full hours to fall below 4000. u 0 = original value

9. Example 2The population of a town grows at a rate of 14% per annum. If P 0 is the initial population and P n is the population after n years. (a) Find a formula for P n in terms of P 0. (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 P n+1 = 1∙14 P n

10. P 1 = 1∙14 P 0 P 2 = 1∙14 P 1 = 1∙14 × 1∙14 P 0 = (1∙14) 2 P 0 P 3 = 1∙14 P 2 = 1∙14 × (1∙14) 2 P 0 = (1∙14) 3 P 0 So in general we have P n = (1∙14) n P 0 If the population trebles then we need to have P n > 3 P 0 or (1∙14) n P 0 > 3 P 0 Dividing by P 0 we get (1∙14) n > 3

11. 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.

12. 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 u n+1 = 1∙2u n u o = 150 150 (=) ENTER 1∙2 × ANS u 1 = 180u 2 = 216u 3 = 259u 4 = 311 259 after 3 days Doubles after 4 days

13. 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 u n+1 = 0∙95u n u o = 150 150 ENTER 0∙95 × ANS u 1 = 143u 2 = 135u 3 = 129u 7 = 104 Falls below 110 during 7 th year

14. 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 u n+1 = 0∙6u n u o = 25000 25000 ENTER 0∙6 × ANS u 1 = 15000u 2 = 9000u 3 = 5400 u 7 = 700 Falls below 1000 during 7 th week. u 11 = 91 Falls below 100 during 11 th week.

15. 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 u n+1 = 1∙03u n u o = 62 62 ENTER 1∙03 × ANS u 1 = 63∙8u 2 = 65∙8u 3 = 67∙7u 4 = 69∙8 2006 u 5 = 71∙9 2007200820092010 71∙9 million at the start of 2010

16. 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 u n+1 = 1∙06u n u o = 2∙5 2∙5 ENTER 1∙03 × ANS u 1 = 2∙575u 2 = 2∙65u 3 = 2∙73u 4 = 2∙81 £2∙81 million at start of 2007 20042005 20062007

17. 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 u n+1 = 0∙88u n u o = 5000 5000 ENTER 0∙88 × ANS u 1 = 4400u 2 = 3872u 3 = 3407 After 3 weeks 3407 healthy u 6 = 2322 Half the population falls ill during 6 th week

18. 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 v 0. Adding 15% gives us 115% or 1∙15 × previous amount Linear Recurrence Relations But 100 ml then escapes

19. v 1 = 1∙15v 0 – 100 v 2 = 1∙15 × 1625 – 100 v 3 = 1∙15 × 1768 – 100 In general v n+1 = 1∙15v n – 100 v 0 = 1 500 = 1625 = 1768 = 1934 Because 100 ml escapes

20. In general v n+1 = 1∙15v n – 100 Each press of = gives the next term v 0 = 1500 v 1 = 1625 v 2 = 1768 And so on v 5 = 2343 v 8 = 3216 Balloon bursts on 8 th puff (ii) We can now use this formula as follows Using ANS function1500 ENTER 1∙15 × ANS

21. 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 u n+1 = 0∙75u n + 20 u o = 160 160 ENTER 0∙75 × ANS + 20 u 1 = 140u 2 = 125u 3 = 113∙75u 4 = 105∙3 105 ml left in bloodstream

22. 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 u n+1 = 1∙075u n + 2000 u o = 24000 24000 ENTER 1∙075 × ANS + 2000 u 1 = 27800 u 2 = 31885And so on u 8 = £63 696∙21 Expected salary = £63 696∙21

23. 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 u n+1 = 0∙85u n + 2 u o = 50 50 ENTER 0∙85 × ANS + 2 u 1 = 44∙5 u 2 = 39∙8u 3 = 35∙9 Pressure after 3 weeks is 35∙9 psi u 5 = 29∙6 If you keep pressing (=) answer is 13∙33 ∙

24. Linear Recurrence Relations u n+1 = 0∙85u n + 2 These recurrence relations are called linear because they all fit a form that looks like : u n+1 = au n + b Which is the same as y = mx + c u n+1 = 1∙075u n + 2000 u n+1 = 0∙75u n + 20 v n+1 = 1∙15v n – 100

25. Linear Recurrence Relations Please note there are various ways to write the same recurrence relation. u n+1 = au n + b u n = au n-1 + b are the same since u n+1 is the term after u n and u n is the term after u n-1. Although we usually use u to represent the term in a sequence it is not necessary or essential. v n+1 = av n + b p n+1 = ap n + b x n+1 = ax n + b g n+1 = ag n + b

27. 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 W n+1 = 0∙25W n + 60 Enter 2000 0∙25 × ANS + 60 W 1 = 560kg End of 1 st August W 2 = 200kg W 7 = 80 kg

28. 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 N n+1 = 0∙7 N n + 25 Enter 300 0∙7 × ANS + 25 N 1 = 235 N 2 = 190 N 5 = 120 N 10 = 89

29. 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 A n+1 = 0∙2A n + 60 Enter 60 0∙2 × ANS + 60 A 1 = 72 kg A 2 = 74∙4 kg A 7 = 75 kg In the long term level of waste levels off at 75 kg, so safe. Statement is essential, always worth a mark!

30. 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 R n+1 = 1∙15 R n - 30 Enter 120 1∙15 × ANS - 30 R 1 = 108 R 6 = 15 R 10 = - 124 In the long term the rabbits will all be sold and the population disappears. Statement is essential, always worth a mark!

31. 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 D n+1 = 0∙7 D n + 20 Enter 90 0∙7 × ANS + 20 D 1 = 83 D 6 = 69 units 6 × 8hrs Note that in the long term number of units of the drug will level off at 66∙7 units.

32. 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 M n+1 = 1∙06M n – 6 000 Enter 80 000 1∙06 × ANS – 6 000 M 1 = £78 800 M 5 = £73 235 Mortgage paid off during year 27 M 10 = £64 183 M 28 = - £2 233

33. Divergence / Convergence/ Limits Consider the following linear recurrence relations (a) u n+1 = 2u n + 4 with u 0 = 3 u 0 = 3 u 1 = 10 u 2 = 24 u 3 = 52 u 10 = 7164 u 20 = 7340028 As n   u n   3 = 2 × ANS + 4 we say that the sequence DIVERGES. As n tends to infinity u n tends to infinity

34. (b) u n+1 = 0∙5u n + 4 with u 0 = 3 u 0 = 3 u 1 = 5∙5 u 2 = 6∙75 u 3 = 7∙375 u 10 = 7∙995 u 20 = 7∙999….. As n   u n  8 we say that the sequence CONVERGES to a limit of 8. 3 = 0∙5 × ANS + 4

35. (c) u n+1 = -2u n + 4 with u 0 = 3 u 0 = 3 u 1 = -2 u 2 = 8 u 3 = -12 u 10 = 1708 u 20 = 1747628 u 21 = -3495252 As n   u n  ±  and we say that the sequence DIVERGES. 3 = -2 × ANS + 4

36. (d) u n+1 = -0∙5u n + 4 with u 0 = 3 u 0 = 3 u 1 = 2∙5 u 2 = 2∙75 u 3 = 2∙625 u 10 = 2∙666 u 20 = 2∙666 As n   u n  2 2 / 3 we say that the sequence CONVERGES to a limit of 2 2 / 3 3 = -0∙5 × ANS + 4

37. As n   u n  ±  we say that the sequence DIVERGES As n   u n → 8 or u n  2 2 / 3 we say that the sequence CONVERGES to a limit For u n+1 = 0∙5u n + 4 and u n+1 = -0∙5u n + 4 For u n+1 = 2u n + 4 and u n+1 = – 2u n + 4

38. Conclusions The linear recurrence relation u n+1 = au n + b converges to a limit only if – 1 < a < 1 Otherwise the sequence diverges.

39. When the linear recurrence relation u n+1 = au n + b converges the answer remains the same no matter how many times you repeat the calculation. So u n = u n+1 = u n+2 + u n+3 …… If we let this limit be L, then u n+1 = au n + b becomes L = aL + b Limit formula

40. Other Factors (e) compare this with (b) u n+1 = 0∙5u n + 10 with u 0 = 3 u 0 = 3 u 1 = 11∙5 u 2 = 15∙75 u 3 = 17∙875 ….. u 10 = 19∙98... …… u 20 = 19∙99…. This is clearly heading to a limit of 20 Conclusion: if u n+1 = au n + b converges to a limit then changing b changes the limit.

41. (f) compare this with (b) u n+1 = 0∙5u n + 4 with u 0 = 200 u 0 = 200 u 1 = 104 u 2 = 56 u 3 = 32 u 10 = 8∙1875 u 20 = 8∙0001…. Again this is heading to a limit of 8 Conclusion: if u n+1 = au n + b converges to a limit then changing u 0 does not affect the limit.

42. A Formula for the Limit of a Converging Sequence u n+1 = au n + b, converges if – 1 < a < 1 The limit, L, is given by the formula The limit depends on a and b but not on u 0

43. 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 metre 3 each hour. The oxygen level in the tank should be between 5 and 7 parts per metre 3 for the survival of the fish. Initially the concentration of oxygen in the tank is 6 ppm 3 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 O n+1 = 0∙85 O n + 1 Limit exists since -1 < 0∙85 < 1 Fish survive since long term level is between 5 and 7 ppm 3

44. 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 F n+1 = 0∙38 F n + 20 In the long term the number of files levels off at 32 Enter 70 0∙38 × ANS + 20 F 1 = 46 F 5 = 32 Limit exists since -1 < 0∙38 < 1

45. 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 K n+1 = 0∙35 K n + 500 85% killed  15% left P n+1 = 0∙15 P n + 650 In the long term Pestkill is marginally better

46. 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 K n+1 = 0∙05 K n + 25 85% cleared  15% left P n+1 = 0∙15 P n + 20 In the long term Pickit Ltd is better

47. 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 C n+1 = 0∙4 C n + 45 In the long term level is constant at 75 units, so safe.

48. 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? u n+1 = 0∙3u n + 35 Burning off 70% leaves behind 30% or 0∙3  a = 0∙3, b = 35 In the long run level of drug in patients system levels out at 50 mg.

49. 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

50. (a) Problem 3.14% daily loss = ? Weekly loss. Decay factor = 0∙9686 Amount remaining after 1 week = (0.9686) 7 × A 0 = 0∙799854 × A 0 = 0∙80 × A 0 or 80% of A 0 The car is loses 20% of its brake fluid weekly So if A n is the fluid level after n weeks then we have A n+1 = 0∙8 A n + 50  7 day factor = (0∙9686) 7

51. (b) Using A n+1 = 0.8 A n + 50 with A 0 = 255 we get A 1 = 254ml 1 st week A 2 = 253∙2ml A 3 = 252∙6ml A 4 = 252∙0ml 4 th week NB : even before adding the 50ml the level is above 200ml Enter 225 0∙8 × ANS + 50

52. (c) considering A n+1 = 0∙8 A n + 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.

53. Given u 6 = 48, u 7 = 44 and u 8 = 42 find a & b. Finding a Formula A recurrence relation is defined by the formula u n+1 = au n + b u 8 = au 7 + b becomes 44a + b = 42 u 7 = au 6 + 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

54. The n th term in a sequence is given by the formula u n = an + b Given that u 10 = 25 and u 12 = 31 then find a & b. Hence find u 300, the 300 th term. u 10 = 10a + b becomes10a + b = 25 u 12 = 12a + b becomes 12a + b = 31 subtract up 2a = 6 a = 3 Now put a = 3 into 10a + b = 25 This gives us 30 + b = 25 So b = – 5 The actual formula is u n = 3n – 5 So u 300 = 3 × 300 – 5 = 895

55. eg8, 14, 20, 26, ……here d = u n+1 - u n = 6 u 1 = 8 = 8 + (0 × 6) Two Special Series In an arithmetic series there is a constant difference between consecutive terms. u 2 = 14 = 8 + (1 × 6) u 3 = 20 = 8 + (2 × 6) u 4 = 26 = 8 + (3 × 6) In general u n = u 1 + (n-1) × d So for the above u 100 = u 1 + 99d = 8 + (99 × 6)= 602

56. In a geometric series there is a constant ratio between consecutive terms. eg5, 10, 20, 40, …… here r = u n+1  u n = 2 u 1 = 5 = 5 × 2 0 u 2 = 10 = 5 × 2 1 u 3 = 20 = 5 × 2 2 u 4 = 40 = 5 × 2 3 In general u n = u 1 × r (n-1) So for the above u 100 = u 1 × r 99 = 5 × 2 99 = 3.17 × 10 30

57. The following slides show exam questions and their solutions. Remember for u n+1 = au n + b Limit only if – 1 < a < 1 4% decrease → 96% left → a = 0∙96 4% increase → 104% left → a = 1∙04 Given values for u n and u n+1 use simultaneous equations to find a and b

58. Put u 1 into recurrence relation Solve simultaneously: A recurrence relation is defined by u n+1 = pu n + q where -1 < p < -1 and u 0 = 12 a)If u 1 = 15 and u 2 = 16 find the values of p and q b)Find the limit of this recurrence relation as n   Put u 2 into recurrence relation State limit condition -1 < p < 1, so a limit L exists Use formula 15 = 12p + q …….. (1) 16 = 15p + q …….. (2) p = 1 / 3 q = 11

59. 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.

60. Construct a recurrence relation State limit condition -1 < 0∙8 < 1, so a limit L exists Use formula Where u n = height at the start of year n Use formula again with L = 2 a = 0∙75 means 75% left u n+1 = 0∙8u n + 0∙5 20% reduction → 80% (0∙8) left To find % reduction required make L = 2 metres Minimum prune = 25%

61. 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 u n and u n+1 and represent the amounts that he owes at the starts of two successive months. Write down a recurrence relation involving u n and u n+1 b) Find the date and amount of the final payment.

62. Construct a recurrence relation u 0 = 2500 Calculate each term in the recurrence relation 1 Maru 0 = 2500.00 1 Apru 1 = 2237.50 1 Mayu 2 = 1971.06 1 Junu 3 = 1700.62 1 Julu 4 = 1426.14 1 Augu 5 = 1147.53 1 Septu 6 = 864.74 1 Octu 7 = 577.71 1 Novu 8 = 286.38 1 DecFinal payment £290.68 u n+1 = 1∙015u n – 300 1∙5% interest → 101∙5% (1∙015) total You can use ANS function on your calculator. 2500 = then 1∙015 × ANS – 300

63. L 1 = L 2 so Sequence 1 Since limit exists a  1 Use formula for each sequence Limit = 25 Two sequences are generated by the recurrence relations u n+1 = au n + 10 and v n+1 = a 2 v n +16 The two sequences approach the same limit as n  . Determine the value of a and evaluate the limit. Sequence 2 Simplify

64. L 1 = L 2 so Cross multiply Sequence 1 Use formula for each sequence Sequence 2 Rearrange Two sequences are defined by the recurrence relations u n+1 = 0∙2u n + p, u 0 = 1 and v n+1 = 0∙6v n + q, v 0 = 1 If both sequences have the same limit, express p in terms of q.

65. Two sequences are defined by these recurrence relations u n+1 = 3u n - 0∙4, u 0 = 1 and v n+1 = 0∙3v n + 4, v 0 = 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 find i)the smallest value of n for which the n th term exceeds 1000, and ii)the value of that term.

66. Sequence 2 Requirement for a limit List terms of 1 st sequence Only the second sequence with a = 0∙3 has a limit u 0 = 1 u 1 = 2∙6 u 2 = 7 ∙ 4 u 3 = 21 ∙ 8 u 4 = 65 u 5 = 194 ∙ 6 u 6 = 583 ∙ 4 u 7 = 1749 ∙ 8 Smallest value of n is 8; value of 8 th term = 1749 ∙ 8 – 1 < a < 1 Exact value means leave as a fraction Use ANS function

67. 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 u n+1 =0∙35u n + 500 Pestkill u n+1 =0∙15u n + 650 In the long run Pestkill is slightly better with a limit 4 less than Killpest

69. A sequence is generated by the recurrence relation u n+1 = 0∙6u n + 5. Which of these is the limit of this sequence as n → ∞ A. 5 B. 2 / 5 C. 3 / 25 D. 25 / 2 × 10

70. A recurrence relation is defined by u n+1 = 0∙4u n – 24. The limit of this sequence is A. -40 B. -24 C. 0∙03 D. 50 × 10

71. A sequence is defined by the recurrence relation u n+1 = au n + b where u 0 = 5. Which of the following could be an expression for u 2, the second term of the sequence? A. 5a 2 + 2b B. 5a 2 + ab C. 5a + b D. 5a 2 + ab + b

72. A sequence is defined by the recurrence relation u n+1 = 3u n – 4, u 0 = –1. What is the value of u 2 ? A –25B –10C –4D –1 ENTER – 1, 3 × ANS – 4 u 1 = – 7 u 2 = – 25

73. Which of these recurrence relations has a limit? I u n+1 = 0∙4u n + 5 II u n+1 = 6 – 0∙5u n III u n+1 = 8 / 9 u n + 2 A.none of them.B. All of them C. I and II onlyD. I and III only All the multipliers of u n come between – 1 and 1

74. The exact limit of the sequence defined by the recurrence relation u n+1 = 0∙6u n + 1∙8 is A. 8 / 9 B. 3C. 9 / 2 D. 2 / 9 × 10

75. The two sequences defined by the recurrence relations u n+1 = 0∙2u n + 12 and v n+1 = 0∙3v n + k have the same limit. The value of k is A. 18B. 4∙5C. 12D. 10  5