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secondary 3 | E Maths
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Date Posted: 5 years ago
Views: 199
n² + 4n + 13
18 = 13 + 5 = 13 + 1 x 5 = 13 + (1)(1+4)
25 = 13 + 12 = 13 + 2 x 6 = 13 + (2)(2+4)
34 = 13 + 21 = 13 + 3 x 7 = 13 + (3)(3+4)
45 = 13 + 32 = 13 + 4 x 8 = 13 + (4)(4+4)
So nth term = 13 + n(n+4) = 13 + n² + 4n
18 = 13 + 5 = 13 + 1 x 5 = 13 + (1)(1+4)
25 = 13 + 12 = 13 + 2 x 6 = 13 + (2)(2+4)
34 = 13 + 21 = 13 + 3 x 7 = 13 + (3)(3+4)
45 = 13 + 32 = 13 + 4 x 8 = 13 + (4)(4+4)
So nth term = 13 + n(n+4) = 13 + n² + 4n
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If you are unable to see any obvious pattern, this is a longer method, but allows you to solve almost all number patterns. Also helps you better understand and visualise patterns in the long term.
Date Posted:
5 years ago
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