[JosAH]

We don't need no steenkin' loops; the old Greeks even knew that; here's how: suppose you want to add the numbers 1 ... 6; you can visualize it like this:

*
**
***
****
*****
******

... now add that same tirangle to the original one but upside down:

*######
**#####
***####
****###
*****##
******#

So twice the number we want to know (let the number be 'n') equals n*(n+1),
so the sum of the numbers 1 ... n equals n*(n+1)/2

That is easy enough to turn into a little method:

int sum(int n) {
   return n*(n+1)/2;

If we want to know the sum of the numbers m ... n we actually want to know the sum of the numbers 1 ... n minus the sum of the numbers 1 ... m-1. In Java that makes:

int sum(int m, int n) {
   return sum(n)-sum(m-1);
}

Now that wasn't too difficult was it?

kind regards,

Jos