Higher Applications www mathsrevision com Higher nth term

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Higher Applications www. mathsrevision. com Higher nth term of a sequence nth term of

Higher Applications www. mathsrevision. com Higher nth term of a sequence nth term of a Recurrence Relation Problem Solving Recurrence Relation Find a formula Divergence / Convergence Limit Applications Exam Type Questions www. mathsrevision. com

Recurrence Relations Higher www. mathsrevision. com Sequences A B C D E 5 3

Recurrence Relations Higher www. mathsrevision. com Sequences A B C D E 5 3 2 17 2 9 6 3 23 3 13 12 5 41 5 17 24 8 77 7 ……. 13 137 11 ……. . ……… In the above sequences some have obvious patterns while others don’t. However this does not mean that a pattern doesn’t exist.

Recurrence Relations Higher www. mathsrevision. com Notation Suppose we write the term of a

Recurrence Relations Higher www. mathsrevision. com Notation Suppose we write the term of a sequence as u 1 , u 2 , u 3 , ……. . , un-1 , un+1 , ……. . . where u 1 is the 1 st term, u 2 is the 2 nd term etc…. and un is the nth term ( n being any whole number. ) The terms of a sequence can then be defined in two ways

Recurrence Relations Higher www. mathsrevision. com Either Using a formula for the nth term,

Recurrence Relations Higher www. mathsrevision. com Either Using a formula for the nth term, un in terms of the value n Or By expressing each term using the previous term(s) in the sequence. This is called a Recurrence Relation Now reconsider the sequences at the start

Recurrence Relations www. mathsrevision. com Higher A 5 Formula: un = 4 n +

Recurrence Relations www. mathsrevision. com Higher A 5 Formula: un = 4 n + 1 So 9 13 17 ……. u 100 = 4 X 100 + 1 = 401 Recurrence Relation: u 2 = u 1 + 4 = 5 + 4 = 9 un+1 = un + 4 with u 1 = 5 u 3 = u 2 + 4 = 9 + 4 = 13

Recurrence Relations Higher www. mathsrevision. com B 3 Formula: So 6 12 24 ……

Recurrence Relations Higher www. mathsrevision. com B 3 Formula: So 6 12 24 …… un = 3 2 n-1 X u 10 = 3 29 = 3 512 = 1536 X Recurrence Relation: u 2 = 2 u 1 = 2 3 = 6, X X un+1 = 2 un with u 1 = 3. u 3 = 2 u 2 = 2 6 = 12, etc X

Recurrence Relations Higher www. mathsrevision. com C 2 3 5 8 13 ……. .

Recurrence Relations Higher www. mathsrevision. com C 2 3 5 8 13 ……. . No formula this time but we have a special type of recurrence relation called a FIBONACCI SEQUENCE. Here u 1 = 2 , u 2 = 3 then we have u 3 = u 2 + u 1 = 3 + 2 = 5 , u 4 = u 3 + u 2 = 5 + 3 = 8 In general un+2 = un+1 + un ie apart from 1 st two, each term is the sum of the two previous terms.

Recurrence Relations Higher www. mathsrevision. com D 17 23 41 77 137 ……… This

Recurrence Relations Higher www. mathsrevision. com D 17 23 41 77 137 ……… This sequence doesn’t have a recurrence relation but the terms can be found using the formula un = n 3 - n + 17 Quite a tricky formula but it does work. . . u 1 = 13 - 1 + 17 = 17 u 2 = 23 - 2 + 17 = 8 - 2 + 17 = 23 u 10 = 103 - 10 + 17 = 1000 - 10 + 17 = 1007

Recurrence Relations www. mathsrevision. com Higher E 2 3 This sequence is the 5

Recurrence Relations www. mathsrevision. com Higher E 2 3 This sequence is the 5 7 11 ……… PRIME NUMBERS (NB: Primes have exactly two factors !!) There is neither a formula nor a recurrence relation which will give us all the primes.

Linear Sequence www. mathsrevision. com Higher We are going to consider Linear Sequences. Example

Linear Sequence www. mathsrevision. com Higher We are going to consider Linear Sequences. Example : Find a formula for the nth term of the sequence below and the value of the 20 th term. 4 , 7, 10 , 13. .

Linear Sequence www. mathsrevision. com Higher Example : 4 , 7, 10 , 13.

Linear Sequence www. mathsrevision. com Higher Example : 4 , 7, 10 , 13. . Un = 3 n + b Sim. Equations U 1 = 3(1) + b = 4 b=1 U 2 = 3(2) + b = 7 Un = 3 n + 1 U 1 = 3 + b = 4 U 2 = 6 + b = 7 U 20 = 3(20) + 1 = 61

Linear Sequence www. mathsrevision. com Higher HG Ex 2. 1 Page 20

Linear Sequence www. mathsrevision. com Higher HG Ex 2. 1 Page 20

Linear Recurrence Relation www. mathsrevision. com Higher Example : Find the value of U

Linear Recurrence Relation www. mathsrevision. com Higher Example : Find the value of U 4 for the recurrence relation Un+1 = 2 Un + 5 U 0 =10 U 1 = 2(10) + 5 = 25 U 2 = 2(25) + 5 = 55 U 3 = 2(55) + 5 = 115 U 4 = 2(115) + 5 = 235

Linear Recurrence Relation www. mathsrevision. com Higher Example : Find the smallest value of

Linear Recurrence Relation www. mathsrevision. com Higher Example : Find the smallest value of n such that Un < 10. Un+1 = 0. 5 Un - 1 U 0 = 62 U 1 = 0. 5(62) - 1 = 30 U 2 = 0. 5(30) - 1 = 14 U 3 = 0. 5(14) - 1 = 6 So smallest value of n is 3

Linear Sequence www. mathsrevision. com Higher HG Ex 2. 2 Page 21

Linear Sequence www. mathsrevision. com Higher HG Ex 2. 2 Page 21

Linear Recurrence Relations www. mathsrevision. com Higher Many recurrence relations take the form un+1

Linear Recurrence Relations www. mathsrevision. com Higher Many recurrence relations take the form un+1 = aun + b where a & b are real nos. If we think about un+1 like y and un like x then we get y = ax + b and this is basically the same as y = mx + c which is the equation of a straight line Hence the expression “Linear Recurrence Relations” Many day to day scenarios can be modelled by this.

Linear Recurrence Relations www. mathsrevision. com Higher Example A balloon contains 1500 ml of

Linear Recurrence Relations www. mathsrevision. com Higher Example A balloon contains 1500 ml 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 situation. (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? (NB: 3 litres = 3000 ml) (i) Suppose the starting volume is V 0. Adding 15% gives us 115% or 1. 15 X previous amount,

Linear Recurrence Relations Higher www. mathsrevision. com however we also lose 100 ml so

Linear Recurrence Relations Higher www. mathsrevision. com however we also lose 100 ml so we have… V 1 = 1. 15 V 0 - 100 similarly V 2 = 1. 15 V 1 - 100 and V 3 = 1. 15 V 2 - 100 In general Vn+1 = 1. 15 Vn - 100 (ii) We can now use this formula as follows V 0 = 1500 V 1 = 1. 15 X 1500 - 100 = 1625

Linear Recurrence Relations www. mathsrevision. com Higher V 2 V 3 V 4 V

Linear Recurrence Relations www. mathsrevision. com Higher V 2 V 3 V 4 V 5 = = 1. 15 X 1625 - 100 = X 1769 - 100 = X 1934 - 100= X 2124 - 100= 1769 1934 2124 2343 So after 5 puffs the balloon contains 2343 ml of air. (iii) continuing the above V 6 = 1. 15 X 2343 - 100 = 2594 V 7 = 1. 15 X 2594 - 100= 2883 V 8 = 1. 15 X 2883 - 100 = 3216 The balloon bursts on the 8 th puff. BANG!!!

Linear Recurrence Relations www. mathsrevision. com Higher Example A factory wishes to dump 150

Linear Recurrence Relations www. mathsrevision. com Higher Example A factory wishes to dump 150 kg of a particular waste product into a local steam once per week. The flow of the water removes 60% of this material from the stream bed each week. However it has been calculated that if the level of deposit on the stream bed reaches 265 kg then there will be a serious risk to the aquatic life. Should the factory be allowed to dump this waste indefinitely?

Linear Recurrence Relations Higher www. mathsrevision. com Let An be the amount of waste

Linear Recurrence Relations Higher www. mathsrevision. com Let An be the amount of waste deposited after n weeks. So A 0 = 150 Removing 60% leaves behind 40% or 0. 4. This means that A 1 = 0. 4 A 0 + 150 Similarly A 2 = 0. 4 A 1 + 150 In general we get the recurrence relation An+1 = 0. 4 An + 150 and this gives us the following sequence…. . .

Linear Recurrence Relations Higher www. mathsrevision. com A 0 = 150 A 1 =

Linear Recurrence Relations Higher www. mathsrevision. com A 0 = 150 A 1 = 0. 4 X 150 + 150 = 210 A 2 = 0. 4 X 210 + 150 = 234 A 3 = 0. 4 X 234 + 150 = 243. 6 A 10 = 0. 4 X 249. 974 + 150 = 249. 990 When amount of waste reaches 250 kg it stays at this. Check: If An = 250 then An+1 = 0. 4 X 250 + 150 = 250 This is below the danger level so factory could be allowed to continue dumping. We say that the sequence CONVERGES to a LIMIT of 250.

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 3 Q 1 to

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 3 Q 1 to Q 5 Page 23

Finding a Formula www. mathsrevision. com Higher Example A recurrence relation is defined by

Finding a Formula www. mathsrevision. com Higher Example A recurrence relation is defined by the formula un+1 = aun + b Given that u 6 = 48 , u 7 = 44 and u 8 = 42 then find a & b. u 8 = au 7 + b becomes 44 a + b = 42 Sim. u 7 = au 6 + b becomes 48 a + b = 44 equations Subtract up 4 a = 2 so a = 0. 5 Now put a = 0. 5 into 44 a + b = 42 to get 22 + b = 42 so b = 20

Finding a Formula www. mathsrevision. com Higher Example The nth term in a sequence

Finding a Formula www. mathsrevision. com Higher Example The nth term in a sequence is given by the formula un = an + b Given that u 10 = 25 and u 12 = 31 then find a & b. Hence find u 300 - the 300 th term. Using un = an + b u 10 = 10 a + b becomes 10 a + b = 25 u 12 = 12 a + b becomes 12 a + b = 31 subtract up 2 a = 6 Sim. equations a = 3

Finding a Formula Outcome 4 www. mathsrevision. com Higher Now put a = 3

Finding a Formula Outcome 4 www. mathsrevision. com Higher Now put a = 3 into 10 a + b = 25 This gives us 30 + b = 25 So The actual formula is So b = -5 un = 3 n - 5 u 300 = 3 X 300 - 5 = 895

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 3 Q 6 onwards

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 3 Q 6 onwards Page 23

Divergence / Convergence/Limits Higher www. mathsrevision. com Consider the following linear recurrence relations (a)

Divergence / Convergence/Limits Higher www. mathsrevision. com Consider the following linear recurrence relations (a) un+1 = 2 un + 4 with u 0 = 3 As n un u 1 = 10 u 2 = 24 and we say that the sequence DIVERGES. u 3 = 52 u 10 = 7164 u 20 = 7340028

Divergence / Convergence/Limits Higher www. mathsrevision. com (b) un+1 = 0. 5 un +

Divergence / Convergence/Limits Higher www. mathsrevision. com (b) un+1 = 0. 5 un + 4 with u 0 = 3 u 1 = 5. 5 As n un 8 u 2 = 6. 75 we say that the sequence u 3 = 7. 375 CONVERGES to a limit of 8. u 10 = 7. 995 U 20 = 7. 999…. . Check: if un = 8 un+1 = 0. 5 X 8 + 4 = 8

Divergence / Convergence/Limits Higher www. mathsrevision. com (c) un+1 = -2 un + 4

Divergence / Convergence/Limits Higher www. mathsrevision. com (c) un+1 = -2 un + 4 with u 0 = 3 u 1 = -2 u 2 = 8 u 3 = -12 u 10 = 1708 (c) (d) u 20 = 1747628 u 21 = -3495252 As n un ± and we say that the sequence DIVERGES.

Divergence / Convergence/Limits Higher www. mathsrevision. com (d) un+1 = -0. 5 un +

Divergence / Convergence/Limits Higher www. mathsrevision. com (d) un+1 = -0. 5 un + 4 with u 0 = 3 As n u n 2 2/ 3 u 1 = 2. 5 we say that the sequence u 2 = 2. 75 CONVERGES to a limit of 22/3. u 3 = 2. 625 u 10 = 2. 666 u 20 = 2. 666 Check: if un = 22/3 un+1 =- 0. 5 X 22/3 + 4 = 22/3

Divergence / Convergence/Limits Higher www. mathsrevision. com Conclusions The linear recurrence relation un+1 =

Divergence / Convergence/Limits Higher www. mathsrevision. com Conclusions The linear recurrence relation un+1 = aun + b converges to a limit if either -1 < a < 0 or 0 < a < 1 This is usually written as 0 < a < 1 If a > 1 ie a < -1 or a > 1 Then we say that the sequence diverges.

Conclusion: Divergence / Convergence/Limits www. mathsrevision. com Higher if un+1 = aun + b

Conclusion: Divergence / Convergence/Limits www. mathsrevision. com Higher if un+1 = aun + b converges to a limit then changing b changes the limit. Other Factors un+1 = 0. 5 un + 4 with u 0 = 3 (e) compare this with (b) un+1 = 0. 5 un + 10 with u 0 = 3 This is clearly heading to u 1 = 11. 5 a limit of 20 u 2 = 15. 75 Check: if un = 20 u 3 = 17. 875 …. . un+1 = 0. 5 x 20 +10 = 20 u 10 = 19. 98. . . …… u 20 = 19. 99….

Conclusion: Divergence / Convergence/Limits Higher www. mathsrevision. com (f) if un+1 = aun +

Conclusion: Divergence / Convergence/Limits Higher www. mathsrevision. com (f) if un+1 = aun + b converges to a limit then changing u 0 + 4 un+1 = 0. 5 u n does not withlimit. u =3 compare this with (b)affect the 0 un+1 = 0. 5 un + 4 with u 0 = 200 u 1 = 104 Again this is heading u 2 = 56 to a limit of 8 u 3 = 32 u 10 = 8. 1875 u 20 = 8. 0001….

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 4 Page 25

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 4 Page 25

Find the Limit Higher www. mathsrevision. com Proof Suppose a limit exists for the

Find the Limit Higher www. mathsrevision. com Proof Suppose a limit exists for the recurrence relation un+1 = aun + b let the limit be L, then we have L = a. L + b Re arranging L – a. L = b L(1 – a) = b

Limit www. mathsrevision. com Higher Example : Find the limit of U = 0.

Limit www. mathsrevision. com Higher Example : Find the limit of U = 0. 2 U + 12 n+1 n Limit exists since -1 < 0. 2 < 1 Limit is 15

Limit Higher Example : www. mathsrevision. com Find b given Un+1 = -0. 4

Limit Higher Example : www. mathsrevision. com Find b given Un+1 = -0. 4 Un + b U 0 = 5 and recurrence relation limit of 50 b = 70 Un+1 = -0. 4 Un + 70

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 5 Page 26

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 5 Page 26

Applications Conclusion: www. mathsrevision. com Higher the level. Outcome of drug in 4 the

Applications Conclusion: www. mathsrevision. com Higher the level. Outcome of drug in 4 the patients system will never exceed 50 mg under Since un+1 = these 0. 3 un +conditions. 35 this is below the danger level it This sequence would has a limit since 0 < 0. 3 < 1 be safe to continue indefinitely. If we call the limit L then at this limit we have un+1 = un = L The equation un+1 = 0. 3 un + 35 now becomes L = 0. 3 L + 35 0. 7 L = 35 0. 7= 350 7 = 50

Applications www. mathsrevision. com Higher Example 1 A hospital patient is put on medication

Applications www. mathsrevision. com Higher Example 1 A hospital patient is put on medication which is taken once per day. The dose is 35 mg 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 54 mg then the consequences could be fatal. Is it safe for the patient to take the medication indefinitely? We need to create a recurrence relation. First dose = u 0 = 35 Burning off 70% leaves behind 30% or 0. 3 After this another 35 mg is taken so we have …. .

Applications www. mathsrevision. com Higher Conclusion: the level of drug in the patients system

Applications www. mathsrevision. com Higher Conclusion: the level of drug in the patients system will never exceed 50 mg under these conditions. Since = 0. 3 uthe n+1 below n + 35 thisuis danger level it would safesince to continue This sequence has be a limit 0 < 0. 3 < 1 indefinitely. If we call the limit L then at this limit we have un+1 = un = L The equation un+1 = 0. 3 un + 35 now becomes L = 0. 3 L + 35 0. 7 L = 35 0. 7 = 350 7 = 50

Applications www. mathsrevision. com Higher Example 2 The brake fluid reservoir in a car

Applications www. mathsrevision. com Higher Example 2 The brake fluid reservoir in a car is leaky. Each day it loses 3. 14% of its contents. To compensate for this daily loss the driver “tops up” once per week with 50 ml of fluid. For safety reasons the level of fluid in the reservoir should always be between 200 ml & 260 ml. Initially it has 255 ml. (a) Find a recurrence relation to describe the above. (b) Determine the fluid levels after 1 week and 4 weeks. (c. ) Is the process effective in the long run?

Applications Higher www. mathsrevision. com (a) Problem 3. 14% daily loss = ? Weekly

Applications Higher www. mathsrevision. com (a) Problem 3. 14% daily loss = ? Weekly loss. Losing 3. 14% daily leaves behind 96. 86% or 0. 9686. Amount remaining after 1 week = (0. 9686)7 X A 0 = 0. 799854 X A 0 = 0. 80 X A 0 or 80% of A 0 This means that the car is losing 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

Applications Outcome 4 Higher www. mathsrevision. com (b) Using An+1 = 0. 8 An

Applications Outcome 4 Higher www. mathsrevision. com (b) Using An+1 = 0. 8 An + 50 with A 0 = 255 we get A 0 = 255 ml A 1 = 254 ml 1 st week A 2 = 253. 2 ml A 3 = 252. 6 ml A 4 = 252. 0 ml 4 th week NB : even before adding the 50 ml the level is above 200 ml

Applications www. mathsrevision. com Higher (c) considering An+1 = 0. 8 An + 50

Applications www. mathsrevision. com Higher (c) considering An+1 = 0. 8 An + 50 Since 0 < 0. 8 < 1 then a limit must exist and at this An+1 = An = L so An+1 = 0. 8 An + 50 ie L = 0. 8 L + 50 or 0. 2 L = 50 or L = 50 0. 2 = 500 2 = 250 In the long run the weekly level will be 250 ml and won’t fall below 200 ml so the driver should be OK with this routine.

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 6 Page 27

Linear Recurrence Relations www. mathsrevision. com Higher HG Ex 2. 6 Page 27

www. maths 4 scotland. co. uk Higher Maths Strategies Sequences Click to start

www. maths 4 scotland. co. uk Higher Maths Strategies Sequences Click to start

Maths 4 Scotland Higher The following questions are on Sequences Non-calculator questions will be

Maths 4 Scotland Higher The following questions are on Sequences Non-calculator questions will be indicated You will need a pencil, paper, ruler and rubber. Click to continue

Maths 4 Scotland Higher A recurrence relation is defined by where -1 < p

Maths 4 Scotland Higher A recurrence relation is defined by 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 1 into recurrence relation Put u 2 into recurrence relation Solve simultaneously: (2) – (1) Hence State limit condition -1 < p < 1, so a limit L exists Limit = 16½ Use formula Previous Quit Hint Next

Maths 4 Scotland Higher A man decides to plant a number of fast-growing trees

Maths 4 Scotland Higher 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 un = height at the start of Construct a recurrence relation assurance year that the trees will grow no taller than 2 metres. What is the minimum percentage that -1 < 0. 8 < 1, so a limit L State limit condition the trees exists will need to be trimmed each year so as to meet this condition. Use formula Limit = 2. 5 metres Use formula again Previous m = 0. 75 Quit Minimum prune = 25% Next Hint

Maths 4 Scotland Higher On the first day of March, a bank loans a

Maths 4 Scotland Higher 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. u 0 = Construct a recurrence relation Let un and un+1 and represent the amounts that he owes at the starts of two successive months. 2500 Calculate each term in the recurrence relation Write down a recurrence relation involving un and un+1 1 Mar u 0 = 2500. 00 1 Aug u 5 = 1147. 53 b) Find the date and amount of the final payment. 1 Apr u 1 = 2237. 50 1 Sept u 6 = 864. 74 1 May u 2 = 1971. 06 1 Oct u 7 = 577. 71 1 Jun u 3 = 1700. 62 1 Nov u 8 = 286. 38 1 Jul u 4 = 1426. 14 Previous Quit 1 Dec Quit payment Final £ 290. 68 Hint Next

Maths 4 Scotland Higher Two sequences are generated by the recurrence relations and The

Maths 4 Scotland Higher Two sequences are generated by the recurrence relations and The two sequences approach the same limit as n . Determine the value of a and evaluate the limit. Use formula for each sequence Sequence 1 Sequence 2 Cross multiply Equate the two limits Simpli fy Solve Deductio n Previous Since limit exists a 1, so Quit Limit = 25 Quit Hint Next

Maths 4 Scotland Higher Two sequences are defined by the recurrence relations If both

Maths 4 Scotland Higher Two sequences are defined by the recurrence relations If both sequences have the same limit, express p in terms of q. Use formula for each sequence Sequence 1 Sequence 2 Cross multiply Equate the two limits Rearrange Hint Previous Quit Next

Maths 4 Scotland Higher Two sequences are defined by these recurrence relations a) Explain

Maths 4 Scotland Higher Two sequences are defined by these recurrence relations 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 nth term exceeds 1000, and ii) the value of that term. Requirement for a First sequence has no limit since 3 is not between – 1 and 1 limit 2 nd sequence has a limit since – 1 < 0. 3 < 1 Sequence 2 List terms of 1 st sequence u 0 = 1 u 1 = 2. 6 u 3 = 21. 8 u 6 = 583. 4 u 7 = 1749. 8 u 4 = 65 u 2 = 7. 4 = Smallest value of n isu 58; value of 8 th term 194. 6 = 1749. 8 Previous Quit Hint Next

Maths 4 Scotland Higher You have completed all 6 questions in this presentation Previous

Maths 4 Scotland Higher You have completed all 6 questions in this presentation Previous Quit Back to start

Are you on Target ! www. mathsrevision. com Outcome 4 • • Update you

Are you on Target ! www. mathsrevision. com Outcome 4 • • Update you log book Make sure you complete and correct ALL of the Recurrence Relations questions in the past paper booklet.