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abver · Dec 16, 2011, 11:19 PM

if have well-known (jednacinu) equation ((n 1) * n) / 2 = 1 2.. n. Now if the difference between the closest numbers smaller, to get ((n 0,1) * n) / (2 * 0,1) = 0,1 0,2.. n. If the difference between the closest numbers the smallest possible, will obtain ((n 0,0.. 1) * n) / (2 * 0,0.. 1) = 0,0.. 1 0,0.. 2.. n or can transform n ^ 2 / 2 = (0,0.. 1 0,0.. 2.. n) / 0,0.. 1 . For example if n = 10 will obtain ((10 1) * 10) / 2 = 1 2.. 10 or ((10 0,1) * 10) / (2 * 0,1) = 0,1 0,2.. 10 or ((10 0,0.. 1) * 10) / (2 * 0,0.. 1) = 0,0.. 1 0,0.. 2.. 10 can transform (10 ^ 2) / 2 = (0,0.. 1 0,0.. 2.. 10) / 0,0.. 1 Question, which is correct ? Srbin

Aurora2000 · Dec 17, 2011, 02:32 AM

The computations are a bit different.

If you are summing from 1 to N, and the difference between 2 numbers is $x$ , then put (assuming (N-1)/x integer):

$a_k=1+kx,\ k=0,1,2,\cdots$

and the last term is $a_{(N-1)/x}=N$ . Thus you have the sum

$a_0+a_1+\cdots+a_{(N-1)/x}=\sum_{k=0}^{(N-1)/x} a_k<br /> = \sum_{k=0}^{(N-1)/x} 1+kx =$
$\frac{N-1}{x}+1 +x \sum_{k=0}^{(N-1)/x}k = \frac{N-1}{x}+1 + \frac{x}{2}\frac{N-1}{x}(\frac{N-1}{x}+1)=\frac{(N-1+x)(N+1)}{2x}$