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From: mouse@mcgill-vision.UUCP (der Mouse)
Newsgroups: net.math
Subject: Re: series sum
Message-ID: <361@mcgill-vision.UUCP>
Date: Thu, 30-Jan-86 06:00:42 EST
Article-I.D.: mcgill-v.361
Posted: Thu Jan 30 06:00:42 1986
Date-Received: Sat, 1-Feb-86 07:46:06 EST
References: <2265@utcsstat.uucp>, <854@spp2.UUCP>
Organization: McGill University, Montreal
Lines: 48

>>	Can anyone deduce the sum of the following series after n terms?
>>	  1 + 3 + 6 + 10 + 15 + 21 + .....

>	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).  (Any pointers to books would be
> accepted - I like this stuff).
> 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)

     Ah, but would pointers to books be welcomed?  (:-)

     I  saw this  in one  of Martin Gardner's books, probably one of his
[N, N=1, 2, ...] th Book of Scientific American Mathematical Puzzles and
Diversions, or some such title.  You can probably find it by looking him
up  as an author  (Books  In  Print, or the  Author section of  the card
catalog).   It was  called the  Calculus of  Finite  Differences, and an
anecdote was told:

     "When I was [n, n~=10  to 15] years old, I noticed that if you took
the differences between the squares of the numbers, and  then again, you
got a constant.  I  showed this to my  father,  and  in his  good Quaker
English, he said, 'Why John, thee hast discovered the calculus of finite
differences.'" -- I forget who the speaker is in this quote.

     I did something similar:  I realized that  the  C of F D guaranteed
that  a cubic would be sufficient, but not remembering  the formula  and
not feeling like deriving it, I wrote out

  a +   b +  c + d = 1		[ from an^3+bn^2+cn+d = S, n=1 ]
 8a +  4b + 2c + d = 4		[ ... and n=2 ... ]
27a +  9b + 3c + d = 10		[ n=3 ]
64a + 16b + 4c + d = 20		[ n=4 ]

and solved....
-- 
					der Mouse

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