T Madas 1 st 2 nd 3 rd

We use the method with the zeroth term etc We use trial and error

Finding the n th term of a quadratic sequence © T Madas

1 2 – 2 Find the generating formula of: 1 +0 1, 5, +4

Finding the nth term of a quadratic sequence by trial and error © T

2 1, 5, 13, 25, 41, … +4 +4 +8 +12 +16 +4 +4

a little A level aside: We can work the n th term straight from

Find the generating formula of: 1, 7, 25, 6 18 12 63, 129, 231,

1, 7, 25, 6 18 12 63, 129, 231, … 38 20 102 66

1, 7, 25, 63, 129, 231, … Suppose the nth term was © T

Suppose the n th term of the quadratic part was -2 n 2 ©

1, 7, 25, 63, 129, 231, … Check that it works! © T Madas

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1 st 2 nd 3 rd 4 th 5 th 6 th nth 1 1 1 1 4 x 0 4 x 1 4 x 2 4 x 3 4 x 4 4 x. T 1 4 x. T 2 4 x. T 3 4 x. T 4 1 5 13 25 41 4 x 5 4 x(n – 1) 4 x Tn 61 ? – 2 triangular numbers © T Madas

We use the method with the zeroth term etc We use trial and error © T Madas

Finding the n th term of a quadratic sequence © T Madas

1 2 – 2 Find the generating formula of: 1 +0 1, 5, +4 “zeroth” +4 term +8 +4 13, 25, 41, … +12 +16 +4 +4 1 2 nd differences b = -2 © T Madas

Finding the nth term of a quadratic sequence by trial and error © T Madas

2 1, 5, 13, 25, 41, … +4 +4 +8 +12 +16 +4 +4 © T Madas

2 2 -1 1, 5, 13, 25, 41, … 2 n 2 : 2, 8, 18, 32, 50, … difference: 1, 3, 5, 7, 9, … 2 n - 1 This difference must be due to the “linear bit” bn + c from un = 2 n 2 to un = 2 n 2 + bn + c © T Madas

1 st 2 nd 3 rd 4 th 5 th 6 th nth 1 1 1 1 4 x 0 4 x 1 4 x 2 4 x 3 4 x 4 4 x 5 4 x(n – 1) 4 x. T 1 4 x. T 2 4 x. T 3 4 x. T 4 1 5 13 25 41 4 x Tn 61 – 2 ? © T Madas

a little A level aside: We can work the n th term straight from this table: As before 1 st 2 nd 3 rd 4 th 5 th 6 th nth 1 1 1 1 4 x 0 4 x 1 4 x 2 4 x 3 4 x 4 4 x 5 4 x(n – 1) 4 x. T 1 4 x. T 2 4 x. T 3 4 x. T 4 1 5 13 25 41 4 x Tn 61 ? – 2 triangular numbers © T Madas

1 1 5 5 1 13 5 1 1 5 13 25 41 13 25 5 13 1 5 1 © T Madas

1 © T Madas

1 st 1 2 nd 7 3 rd 25 4 th 63 5 th 129 6 th 231 1 © T Madas

1 1 5 1 © T Madas

1 st 1 2 nd 7 1 1 5 1 3 rd 25 4 th 63 5 th 129 6 th 231 © T Madas

1 st 1 1 2 nd 7 1 5 1 3 rd 25 1 5 13 5 1 4 th 63 5 th 129 1 5 13 25 13 5 1 1 5 13 25 41 25 13 5 1 6 th 231 1 5 13 25 41 61 41 25 13 5 1 © T Madas

1, 7, 25, 6 18 12 63, 129, 231, … 38 20 102 66 28 36 In a cubic sequence the coefficient of n 3 is 1/ of the 3 rd difference 6 8 8 8 3 rd differences © T Madas

1, 7, 25, 63, 129, 231, … Suppose the nth term was © T Madas

2 nd differences © T Madas

Suppose the n th term of the quadratic part was -2 n 2 © T Madas

1, 7, 25, 63, 129, 231, … Check that it works! © T Madas