Find the nth term of the quadratic sequences In the form n^2 + b 1) 8, 11, 16, 23, 32 2) 0, 3, 8, 15, 24               3) -8, -5, 0, 7, 16 In the form an^2 + b 1) 11, 17, 27, 41, 59 2) 7, 16, 31, 52, 79 3) 6, 18, 38, 66, 102 4) 0, 9, 24, 45, 72 5) 0, 12, 32, 60, 96

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Find the nth term of the quadratic sequences

The formulae for quadratic sequences are established by taking second differences between the list of terms. If the sequence is indeed quadratic (of order two) then the second differences will be constant and equal to `2a ` where `a ` is the coefficient of the quadratic term (term in `n^2 `).

We are told that the first set of sequences A1) - A3) are of the form `S_n = n^2 +b `  , where `S_n ` defines the `n `th term of the sequence, so that we should find that the second differences are constant and equal to twice the coefficient of the `n^2 ` term, namely 2.

So, for the first set of questions, take the (first) differences between the sequence terms and then the second differences (these are the differences between the sequence of first differences). Assume the first term of the sequence corresponds to  `n=1 `

A1) Sequence: 8, 11, 16, 23, 32

1st differences:  3, 5, 7, 9.  2nd differences: 2, 2, 2  (as expected)

Therefore `S_n = n^2 + 7 `

so that the...

(The entire section contains 2 answers and 652 words.)

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