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From: stassen@spp2.UUCP (Chris Stassen)
Newsgroups: net.math
Subject: Re: series sum
Message-ID: <854@spp2.UUCP>
Date: Sun, 26-Jan-86 10:43:01 EST
Article-I.D.: spp2.854
Posted: Sun Jan 26 10:43:01 1986
Date-Received: Tue, 28-Jan-86 05:31:03 EST
References: <2265@utcsstat.uucp>
Reply-To: stassen@spp2.UUCP (Chris Stassen)
Organization: TRW, Redondo Beach  CA
Lines: 70


In article <2265@utcsstat.uucp> anthony writes:
>
>       Can anyone deduce the sum of the following series after n terms?
>
>         1 + 3 + 6 + 10 + 15 + 21 + .....

        S(n) =  n ( n^2 + 3n + 2 ) / 6

        I have a simple method that works for all series that can
be expressed as polymonials.  I developed it on my own in High
School, but I am sure that it is an accepted method (and that
someone else has named it).  In fact, there is probably a less
clumsy version of the method, but I haven't read anything about
it.  (Any pointers to books would be accepted - I like this stuff).

1   3   6   10   15   21   28   36   45   55    - Set of terms

1   4  10   20   35   56   - First few sums
  3   6   10   15   21     - Difference between adjacent sums (n^1)
    3   4    5    6        - Difference between adjacent diffs (n^2)
      1   1    1           - Difference between adjacent diffs (n^3)

        When we get constants across a row, then that row is the highest
power of n involved in the expression for the sum  (all rows that follow
would be zero).  The factor of n^m  (in constant row "m")  is  C/(m!),
where C is the constant (the value of all numbers in that row).

        Therefore, n^3/6 is the first part of the polynomial.  Now,
we have to subtract n^3/6 from each sum.

n^3/6

5/6   16/6   33/6   56/6   85/6   120/6  - First few sums (minus n^3/6)
   11/6   17/6   23/6   29/6   35/6      - Difference btw adj sums (n)
       6/6     6/6   6/6    6/6          - Difference btw adj diff (n^2)

        Note that the n^2 row is now the constant row (we have subtracted
out the n^3 part of the polynomial expression).  C=1, and the coefficient
for n^2 is therefore 1/2.  Now, we have to subtract n^2/2 from each sum.

n^2/2

2/6   4/6   6/6   8/6   10/6  - First few sums (less n^3/6 and n^2/2)
   2/6   2/6   2/6    2/6     - Difference between adjacent sums (n)

        Note that the n row is now the constant row (we have subtracted
out the n^3 and n^2 parts of the polynomial expression).  C=1/3, and
the coefficient of n is therefore 1/3.  Now, we subtract n/3 from each
sum.

n/3

0 0 0 0 0 - Sums

        There is no constant in the polynomial.  Therefore, the
expression for the sum is:

                n ( n^2 + 3n + 2 ) / 6

        Let's test it to be sure.

1   4  10   20   35   56   - First few sums

        n = 1 :   S(n)  = 1 ( 1 + 3 + 2 ) / 6         = 1
        n = 2 :   S(n)  = 2 ( 4 + 6 + 2 ) / 6         = 4
        n = 3 :   S(n)  = 3 ( 9 + 9 + 2 ) / 6         = 10
        n = 4 :   S(n)  = 4 (16 +12 + 2 ) / 6         = 20

                        -- Chris