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Here are the first five terms of a sequence. 4, 11, 22, 37, 56 Find an expression, in terms of , for the th term of this sequence.

User Zzen
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1 Answer

3 votes

Answer:


a_n = 2n^2 + n + 1

Explanation:

4, 11, 22, 37, 56

11 - 4 = 7

22 - 11 = 11

37 - 22 = 15

56 - 37 = 19

After the first difference, 11 - 4 = 7, each difference is 4 more than the previous difference.

Difference of differences:

11 - 7 = 4

15 - 11 = 4

19 - 15 = 4

Since we need the difference of differences to find a constant, this must be a second degree function.


a_1 = 4 = 2^2 + 1(0)


a_2 = 11 = 3^2 + 2 = 3^2 + 2(1)


a_3 = 22 = 4^2 + 6 = 4^2 + 3(2)


a_4 = 37 = 5^2 + 12 = 5^2 + 4(3)


a_5 = 56 = 6^2 + 20 = 6^2 + 5(4)


a_n = (n + 1)^2 + (n)(n - 1)


a_n = (n + 1)^2 + (n)(n - 1)


a_n = n^2 + 2n + 1 + n^2 - n


a_n = 2n^2 + n + 1

User Kalana Demel
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