Here are the first 4 terms of a quadratic sequence. 7 18 33 52 Find an expression, in terms of n, for the nth term of the sequence.

Question

Question

Here are the first 4 terms of a quadratic sequence. 7 18 33 52 Find an expression, in terms of n, for the nth term of the sequence.

Answer 1

2n² + 5n + 7

Answer 2

A quadratic sequence is a sequence of numbers in which the second difference is constant. They have the form an^2 + bn + c, where coefficients a, b and c satisfy certain conditions. Given a sequence 7, 18, 33, 52, we can find the first difference ("delta 1") by subtracting each term from the term that follows it. Delta 1 is 11 (18-7), 15 (33 - 18) and 19(52 - 33). That means the rate of change between terms is not constant hence quadratic not linear. To prove it's indeed quadratic, let's find second difference ("delta 2") by also subtracting each term of delta 1 from the following one, in the same way as done with the sequence elements. That calculates into 4 and 4 which is constant therefore confirming a quadratic sequence. Now the n-th term an^2 + bn + c can be expressed from measurements. Since the coefficient 'a' of quadratic sequence is half of the "delta2" that is a = 4/2 = 2. To find 'b', subtract first term of Delta1 sequence by 3a = 11 - 3*2 making b = 5 Final 'c' is simply the very the first term of the sequence = 7 All together, the formula yn= an^2 + bn + c, after replacing coefficients a=2 b=5 c=7 gives yn=2n^2 + 5n + 7

Question

Here are the first 4 terms of a quadratic sequence. 7 18 33 52 Find an expression, in terms of n, for the nth term of the sequence.

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Answer

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