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Topic: 1 + 2 + ... + n a polynomial how?
Replies: 5   Last Post: Apr 2, 2013 11:55 AM

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David Bernier

Posts: 3,276
Registered: 12/13/04
Re: 1 + 2 + ... + n a polynomial how?
Posted: Apr 2, 2013 12:52 AM
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On 04/01/2013 12:18 PM, Jussi Piitulainen wrote:
> Is it obvious that 1 + 2 + ... + n is a polynomial of degree 2? How?
>
> I mean the sum of the first n positive integers. I would like to see
> that it is a polynomial of degree 2 _without using_ the fact that it
> is equal to n(n + 1)/2. Zeilberger (his new Opinion 129) says Gauss
> could have used the polynomiality of the sum to support the equality,
> rather than the other way around.
>
> Thanks for any insight.
>


X O O O
O X O O
O O X O
O O O X


n = 4;

n^2 - n = 16 - 4 = number of O's in diagram above.

(n^2 - n)/2 = (16-4)/2 = 12/2 = 6 = number of O's below the diagonal,
by symmetry.


Count of O's by columns:
number of O's below the diagonal = 3 + 2 + 1, or more
generally (n-1) + (n-2) + ... + 2 +1.

So,
1 + 2 + 3 + ... (n-2) + (n-1)
= (n-1) + ... + 3 + 2 +1 =
number of O's below the diagonal
=
(n^2 - n)/2.

dave

see also Brian Hayes:
http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning

Gauss and Sophie Germain had a love child? mhhh..
could it be the April 1st date of Opinion 129?



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$apr1$LJgyupye$GZQc9jyvrdP50vW77sYvz1



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