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Re: 1 + 2 + ... + n a polynomial how?
Posted:
Apr 2, 2013 12:52 AM


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 = (164)/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 (n1) + (n2) + ... + 2 +1.
So, 1 + 2 + 3 + ... (n2) + (n1) = (n1) + ... + 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/gausssdayofreckoning
Gauss and Sophie Germain had a love child? mhhh.. could it be the April 1st date of Opinion 129?
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