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

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
= \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}