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

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

Maths4Scotland Higher Sequences The following questions are on Non-calculator questions will be indicated Click to continue You will need a pencil, paper, ruler and rubber.

Maths4Scotland Higher Hint PreviousNext Quit Put u 1 into recurrence relation Solve simultaneously: 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 2 into recurrence relation (2) – (1) Hence State limit condition -1 < p < 1, so a limit L exists Use formula Limit = 16½

Maths4Scotland Higher Hint PreviousNext Quit Construct a recurrence relation State limit condition -1 < 0.8 < 1, so a limit L exists Use formula Limit = 2.5 metres 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. u n = height at the start of year Use formula again Minimum prune = 25% m = 0.75

Maths4Scotland Higher Hint PreviousNext Quit 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 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. u 0 = 2500 Calculate each term in the recurrence relation 1 Mar u 0 = 2500.00 1 Apr u 1 = 2237.50 1 May u 2 = 1971.06 1 Jun u 3 = 1700.62 1 Jul u 4 = 1426.14 1 Aug u 5 = 1147.53 1 Sept u 6 = 864.74 1 Oct u 7 = 577.71 1 Nov u 8 = 286.38 1 Dec Final payment £290.68

Maths4Scotland Higher Hint PreviousNext Quit Equate the two limits Cross multiply Sequence 1 Since limit exists a  1, so Use formula for each sequence Limit = 25 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. Sequence 2 Simplify Solve Deduction

Maths4Scotland Higher Hint PreviousNext Quit Equate the two limits Cross multiply Sequence 1 Use formula for each sequence Sequence 2 Rearrange Two sequences are defined by the recurrence relations If both sequences have the same limit, express p in terms of q.

Maths4Scotland Higher Hint PreviousNext Quit Sequence 2 Requirement for a limit List terms of 1 st sequence 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 n th term exceeds 1000, and ii)the value of that term. First sequence has no limit since 3 is not between –1 and 1 2 nd sequence has a limit since –1 < 0.3 < 1 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

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