1. HIGHER MATHEMATICS Unit 1 - Outcome 4 Sequences

2. A car depreciates in value by 15% each year. Its value, C n, at the end of each year is given by the formula: C n+1 = 0.85C n ; C 0 = 12000 (a)Find an explicit formula for C n in terms of n. (b) After how many years has it lost 70% of its original value? POINTS TO NOTE Value decreases by 15%, so 85% of value remains, hence 0.85 in formula. Original value of car is £12000, hence C o = 12000 Ex.4 Question 1

3. (a) C 0 = 12000 C 1 = 0.85 x 12000 C 2 = 0.85 2 x 12000 C 3 = 0.85 3 x 12000 C n = 0.85 n x 12000 (b) 70% lost, so 30% of value remains 30% of £12000 = £3600 C 1 = 0.85 x 12000 = £10200 C 4 = 0.85 4 x 12000 = £6264.08 C 7 = 0.85 7 x 12000 = £3846.93 The car loses 70% of its value after 8 years. C 8 = 0.85 8 x 12000 = £3269.89

4. POINTS TO NOTE Population increases by 3%, so population will be 103% of the population the previous year. Starting population, P 0, is 700000. In 1999, the population of Glasgow was 700000. The net rate of increase of the population, taking account of births, deaths and migration patterns, is 3%. Set up a recurrence relation to describe this situation, and calculate, to the nearest 100, the population of Glasgow at the end of 2004. Ex.4 Question 2

5. P n+1 = 1.03P n ; P 0 = 700000 (1999) 2000 P 1 = 1.03(700000) 2001 P 2 = 1.03 2 x 700000 2002 P 3 = 1.03 3 x 700000 2003 P 4 = 1.03 4 x 700000 2004 P 5 = 1.03 5 x 700000 = 811492 At the end of 2004, the population of Glasgow was 811500.

6. POINTS TO NOTE Starting population, u 0, is 3000. 40% lost, so 60% (0.6) remains Ex.4 Question 3 Three thousand fish are introduced to a man made loch. Each day thereafter, 40% of the fish are caught by anglers. At the end of each day, 200 new fish are put into the loch. Set up a recurrence relation, u n+1, to describe this situation, and calculate how many fish there will be after 5 days.

7. u n+1 =0.6u n + 200 u 0 = 3000 u 1 = 0.6(3000) + 200 = 2000 u 2 = 0.6(2000) + 200 = 1400 u 3 = 0.6(1400) + 200 = 1040 u 4 = 0.6(1040) + 200 = 824 u 5 = 0.6(824) + 200 = 694.4 After 5 days, there will be 694 fish.

8. POINTS TO NOTE Starting weight, w 0, is 80kg. 5% lost, so 95% of weight remains. Ex.4 Question 4 Chubby Bunny weighs 80kg. He is on a diet, and he thinks this will result in a weekly weight loss of 5% of his body weight. Sadly, he is addicted to Scotch pies, and this results in a weekly gain of 5kg. Set up a recurrence relation, w n+1, to describe his weight, and calculate Chubby Bunny’s weight four weeks into this diet.

9. w n+1 =0.95w n + 5w 0 = 80 w 1 = 0.95(80) + 5 = 81 w 2 = 0.95(81) + 5 = 81.95 w 3 = 0.95(81.95) + 5 = 82.85 w 4 = 0.95(82.85) + 5 = 83.70 After 4 weeks, Chubby Bunny's weight is 83.7kg(!)

10. The population of Edinburgh at the start of 1992 was 500000. Migration studies suggest that, each year, 2% of people born in Edinburgh leave to live somewhere else. During this same period, 7000 new people arrive to settle in the city. If this situation persists, what will eventually happen to the population of Edinburgh? Ex.4 Question 5 P n+1 =0.98P n + 7000P 0 =500000 Since -1 < 0.98 < 1, a limit exists L = 0.98L + 7000 0.02L = 7000 L = 350000 Eventually, the population of Edinburgh will stabilise (or settle) at 350000 (it will not drop below 350000).

11. Chubby Bunny – who now weighs 90kg - has abandoned his old diet, and started the Atkin’s diet. He now loses one fifth of his body weight each month. Sadly, he is now hopelessly addicted to haggis suppers from his local chip shop, and this results in a monthly gain in weight of 15kg. If his weight drops below 76kg, he is going to buy himself an Aston Martin Vanquish. Will he get his car? Ex.4 Question 6 w n+1 =0.8w n + 15w 0 = 90 Since -1 < 0.8 < 1, a limit exists L = 0.8L + 15 0.2L = 15 L = 75 L < 76, so Chubby’s weight will eventually drop below target weight. He DOES get his car.

12. A recurrence relation is defined by the formula u n+1 = au n + b. Given that u 2 = 4, u 3 = 10 and u 4 = 28, find the values of a and b. Ex.4 Question 7 u n+1 = au n + b u 2 = 4 u 3 = 10 u 4 = 28 u 3 = au 2 + bu 4 = au 3 + b 10 = a(4) + b 28 = a(10) + b 10 = 4a + b28 = 10a + b Now use simultaneous equations to find the values of a and b.

13. 4a + b = 10 10a + b = 28 6a = 18 a = 3 b = -2 u n+1 = 3u n - 2