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October 5th, 2017, 05:45 AM   #11
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This kind of problem has an infinite number of correct answers. Given any finite number of values, there are an infinite number of polynomials that will generate those values. However, a unique polynomial of degree n-1 will generate n given values exactly.
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October 8th, 2017, 03:13 AM   #12
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Originally Posted by skipjack View Post
As the second differences are all 4, one can find a quadratic formula of the form 4n²/2 + bn + c for the nth term. It's now easy to determine the values of b and c.
I also notice that this is the 'recommended' method for solving these problems, but mind sharing a bit of the background theory of where did that equation came from?. Has the third differences were constant then would a cubic equation be formed? How would it be?

Last edited by skipjack; October 8th, 2017 at 03:17 AM.
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October 8th, 2017, 03:52 AM   #13
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If the kth differences all equal some non-zero constant, q, you're looking for a polynomial of degree k,
so the nth term is an^k + terms of lower degree (which don't affect the kth differences).

The coefficient of n^k is given by a = q/k! (to see why, work out the kth differences if the nth term is n^k).
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