Let's say we have n number of elements. Now you have two possibilites:
1.) 2n/2 is an integral, (n = 2 or 4, 6 and so on)
then the left side of the equation can be put as a number of sums, example: (1-2) + (3-4) + ... + [n - ( n - 1 ) ] + ...
there are n/2 sums like this one, each gives the result '-1', this gives (-1)*(n/2)=-n/2, as the left side of the equation.
2.) second possibility is that 2n/2 is NOT an integral. Then the left side of the equation is equal to:
(1-2) + (3-4) + ... + [ n - ( n + 1 ) ... + n.
in the end I add 'n' since 'n' is an odd number,
this sums up to [(n/2)*(-1) + n] = -n/2 + 2n/2 = n/2
So I have two possibilites, assuming n->infinity.
Both possibilites are equally probable, therefore truth is the average value of the sum, which is: