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Finding a Formula Ex1 A recurrence relation is defined by the formula u n+1 = au n + b. Given that u 6 = 48, u 7 = 44 and u 8 = 42 then find a & b. u 8.

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Presentation on theme: "Finding a Formula Ex1 A recurrence relation is defined by the formula u n+1 = au n + b. Given that u 6 = 48, u 7 = 44 and u 8 = 42 then find a & b. u 8."— Presentation transcript:

1 Finding a Formula Ex1 A recurrence relation is defined by the formula u n+1 = au n + b. Given that u 6 = 48, u 7 = 44 and u 8 = 42 then find a & b. u 8 = au 7 + b becomes44a + b = 42 u 7 = au 6 + b becomes48a + b = 44 Simult equations Subtract up4a = 2 so a = 0.5 Now put a = 0.5 into 44a + b = 42 to get 22 + b = 42 so b =20

2 Ex2 (non-recurrence relation) The nth 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. Using u n = an + b u 10 = 10a + b becomes10a + b = 25 u 12 = 12a + b becomes 12a + b = 31 Simult equations subtract up 2a = 6 a = 3

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 X 300 - 5 = 895

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

5 2) Geometric Series 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 X 2 0 u 2 = 10 = 5 X 2 1 u 3 = 20 = 5 X 2 2 u 4 = 40 = 5 X 2 3 etc In general u n = u 1 X r (n-1) So for the above u 100 = u 1 X r 99 = 5 X 2 99 = 3.17 X 10 30


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