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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 n-1 will generate n given values exactly. October 8th, 2017, 03:13 AM   #12
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Math Focus: Calculus Quote:
 Originally Posted by skipjack 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. October 8th, 2017, 03:52 AM #13 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 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). Tags determine, nth, series, term Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bongantedd Algebra 2 February 26th, 2014 04:38 PM Polaris84 Real Analysis 3 October 26th, 2013 02:12 PM RuneBoggler Algebra 2 May 10th, 2011 11:14 AM reddd Algebra 3 May 20th, 2010 12:50 PM ammarkhan123 Real Analysis 0 April 16th, 2010 07:52 AM

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