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October 5th, 2017, 05:45 AM  #11 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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 n1 will generate n given values exactly.

October 8th, 2017, 03:13 AM  #12 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  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. 
October 8th, 2017, 03:52 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
If the kth differences all equal some nonzero 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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determine, nth, series, term 
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