1+2-3-4+5+6-7-8+9+10-11-12+…… Is it a continuous integer starting from 1? Take two positive integers and two negative integers in turn. What is the sum of them? Fast, complete

1+2-3-4+5+6-7-8+9+10-11-12+…… Is it a continuous integer starting from 1? Take two positive integers and two negative integers in turn. What is the sum of them? Fast, complete

The original formula = (1 + 2-3-4) + (5 + 6-7-8) + (9 + 10-11-12) +... = - 2-2-2 -. = - 2 * m, where m is the number of positive and negative pairs

Integral part of (1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + 1 / 6 + 1 / 7 + 1 / 8 + 1 / 9 + 1 / 10 + 1 / 11 + 1 / 12 + 1 / 13 + 1 / 14 + 1 / 15) * 4004

(1/2+1/3+1/4+1/5+1/6+1/7+1/8)+(1/9+1/10+1/11+1/12)+(1/13+1/14+1/15) =481/280+763/1980+587/2730 ≈2.318228993 (1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/11+1/12+1/13+1/14+1/15)*4004 =(481/280+763/1980+587...

+1. + 2, - 3, - 4, + 5, + 6, - 7, - 8, + 9, + 10... Integers starting from 1
Starting from 1, two integers are positive and two integers are negative. What is the sum of the 2000 integers
Why - 2000? How

-2000