Higher Maths Sequences Strategies Click to start

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Higher Maths Sequences Strategies Click to start www.maths4scotland.co.uk Higher Maths Strategies Sequences Click to start

Sequences The following questions are on Maths4Scotland 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

Limit = 16½ Maths4Scotland Higher A recurrence relation is defined by 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 u1 into recurrence relation Put u2 into recurrence relation Solve simultaneously: (2) – (1) Hence State limit condition -1 < p < 1, so a limit L exists Hint Use formula Limit = 16½ Previous Quit Quit Next

un = height at the start of year Maths4Scotland 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. If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run. 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. un = height at the start of year Construct a recurrence relation State limit condition -1 < 0.8 < 1, so a limit L exists Use formula Limit = 2.5 metres Use formula again Hint m = 0.75 Minimum prune = 25% Previous Quit Quit Next

u0 = 2500 Maths4Scotland Higher Construct a recurrence relation 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. u0 = 2500 Construct a recurrence relation 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 Hint Previous Quit Quit Next

Limit = 25 Maths4Scotland 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 Equate the two limits Cross multiply Simplify Solve Hint Deduction Since limit exists a  1, so Limit = 25 Previous Quit Quit Next

Two sequences are defined by the recurrence relations Maths4Scotland 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 Equate the two limits Cross multiply Rearrange Hint Previous Quit Quit Next

Two sequences are defined by these recurrence relations Maths4Scotland 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. First sequence has no limit since 3 is not between –1 and 1 Requirement for a limit 2nd sequence has a limit since –1 < 0.3 < 1 Sequence 2 u0 = 1 u1 = 2.6 u2 = 7.4 u3 = 21.8 u4 = 65 u5 = 194.6 u6 = 583.4 u7 = 1749.8 List terms of 1st sequence Hint Smallest value of n is 8; value of 8th term = 1749.8 Previous Quit Quit Next

You have completed all 6 questions in this presentation Maths4Scotland Higher You have completed all 6 questions in this presentation Previous Quit Quit Back to start

Maths4Scotland Higher Hint Previous Quit Quit Next