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 November 10th, 2014, 12:44 PM #1 Member   Joined: Jan 2013 Posts: 47 Thanks: 0 Super simple sequence question Right so I've got myself in a right pickle on this question: Find the nth term of -2,-1,2,11 So I've tried using the standard an^2 + bn + c But that hasn't worked .. Any help?? Last edited by skipjack; November 10th, 2014 at 04:03 PM.
 November 10th, 2014, 12:52 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,087 Thanks: 2360 Math Focus: Mainly analysis and algebra You will be able to make a cubic equation fit.
 November 10th, 2014, 01:02 PM #3 Member   Joined: Jan 2013 Posts: 47 Thanks: 0 How would you do that?
 November 10th, 2014, 01:25 PM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms You can fit any n+1 points to an n-th degree equation -- this is the 'boring' solution. In your case I get (2n^3 - 9n^2 + 16n - 15)/3 which you could get by solving an^3 + bn^2 + cn + d = y for each of the (x, y) pairs you have. It's a bit tiresome.
 November 10th, 2014, 01:33 PM #5 Member   Joined: Jan 2013 Posts: 47 Thanks: 0 got 4/6n^3 -3n^2 + 14 and 2/3 times n - 29/3 Any good?
 November 10th, 2014, 01:57 PM #6 Member   Joined: Jan 2013 Posts: 47 Thanks: 0 Anyone else got an answer?
 November 10th, 2014, 02:35 PM #7 Member   Joined: Jan 2013 Posts: 47 Thanks: 0 Someone must have a solution and answer!
 November 10th, 2014, 02:40 PM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,658 Thanks: 964 Math Focus: Elementary mathematics and beyond You run out of room with differences ... so assume it's a cubic! Solve the system a + b + c + d = -2 8a + 4b + 2c + d = -1 27a + 9b + 3c + d = 2 64a + 16b + 4c + d = 11 where a, b, c and d are the coefficients of the generating function. By the way, please stop bumping.
 November 10th, 2014, 03:08 PM #9 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Another way, compute the differences. Code: -2 -1 2 11 1 3 9 2 6 4 Now, the interpolating polynomial is of the form $\displaystyle a\cdot {n \choose 3} + b\cdot {n \choose 2}+c \cdot {n \choose 1}+d\cdot {n \choose 0}$ Then, look at the left 'column', having numbers -2, 1, 2 and 4. These are the values for a, b, c, d; (a, b, c, d) = (4, 2, 1, -2) to get $\displaystyle 4\cdot {n \choose 3} + 2\cdot {n \choose 2}+1 \cdot {n \choose 1}-2\cdot {n \choose 0}$, which can be proved by induction. You don't need to compute all difference to get these. Now, x in {0, 1, 2, 3} give the resulting elements from the sequence. An appropriate translation gives a desired offset (n->n-1). Alternatively, the first differences are of the form 3^k, for k in {0, 1, 2}. Summing them, with adding an extra constant, gives (3^n-1)/2 - 2. n in {0, 1, 2, 3} produce the sequence. Last edited by skipjack; November 10th, 2014 at 05:34 PM.
 November 10th, 2014, 05:17 PM #10 Global Moderator   Joined: Dec 2006 Posts: 18,235 Thanks: 1437 Or 2n! + n - 4, for n = 0, 1, 2 and 3.

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