Question 1089737
5;6;8;11;15 
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we use the method of common differences
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level 1 is 1, 2, 3, 4
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level 2 is 1, 1, 1
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therefore we know that we have a n^2 term
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the general form is
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an^2 + bn + c
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we have the following
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a(1^2) + b(1) +c = 5
a(2^2) + b(2) +c = 6
a(3^2) + b(3) +c = 8
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so we have 3 equations in 3 unknowns
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 a +b  +c = 5
4a +2b +c = 6
9a +3b +c = 8
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You can use whatever method you like, including using matrices in your graphing calculator
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a = 1/2, b = -1/2, c = 5
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the nth term is (n^2/2) -(n/2) + 5
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