Triangle SeriesDate: 07/09/99 at 14:55:40 From: Simer Singh Subject: Triangular Series Dear Dr. Math, I am doing a project on a formula that calculates the sum of the nth row of the following triangle: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 I found the formula by trial-and-error, for example 1 = 2/2 = 1x2/2 or (1+1)/2 or 1(1^2+1)/2, hence the formula n(n^2+1)/2 My teacher has asked me use a method that will apply for all the other series too. Thank you for your help. Simer Singh Date: 07/09/99 at 19:02:42 From: Doctor Anthony Subject: Re: Triangular Series Make up a difference table giving the sum for each row n = 1 2 3 4 5 6 7 f(n) = 1 5 15 34 65 111 175 1st Diff. 4 10 19 31 46 64 2nd diff. 6 9 12 15 18 3rd diff. 3 3 3 3 With 3rd differences constant the expression for f(n) will be a cubic polynomial in n. So we assume f(n) = an^3 + bn^2 + cn + d Now substitute in the first few values so we can solve for a, b, c and d. n = 1: a + b + c + d = 1 n = 2: 8a + 4b + 2c + d = 5 n = 3: 27a + 9b + 3c + d = 15 n = 4: 64a + 16b + 4c + d = 34 Solving these four equations we get a = 1/2, b = 0, c = 1/2, d = 0 So we get f(n) = n^3/2 + n/2 = n(n^2+1)/2 Check with n = 7 f(7) = 7(49+1)/2 = 175 and this checks, so we can be confident that our formula is correct. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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