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.