One thing not mentioned in whizkid's writeup above is that there's nothing magic about that formula: you can rewrite (b + a)(b - a + 1) / 2 as:
b(b + 1) a(a + 1)
-------- - -------- + a
meaning that the sum of the numbers between and including two integers a < b is just the sum of the numbers from 1 to b, minus the sum of the numbers from 1 to a, plus a again (since we subtracted it in the sum from 1 to a).
The point is that there's really no extra work to be done to accomodate the fact that a, or b, or both could be negative. The easiest proof of n(n + 1)/2 is by induction (in fact it is done in one of the writeups for induction). While induction is at its core a demonstration that something is equivalent to the natural numbers (an "inductive set"), the negative integers are equivalent to the natural numbers as well (by trivially multiplying by -1) so induction works equally well there.
Since you then automatically know that the sum of the numbers from -1 to -n (or -n to -1 if you prefer) is -n(n + 1)/2, i.e. just the negative of the sum from 1 to n where n is positive, the more general result follows from the ordinary laws of addition.
It's probably also worth mentioning that Gauss, being one of the greatest mathematicians that ever lived, has his name on a mind-bogglingly large number of formulas, theorems and ideas in mathematics, so the name of this node is kind of misleadingly general, although I grant you that "Gauss's Consecutive Integer Sum Formula" is kind of wordy...