Triangle Series

Date: 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/