Question 666237
In general, the sum of the first n numbers is  n(n+1)/2.  Then 1000(1000+1)/2 =  500*1001 = 500500.

If you don't believe the result, here's a nice proof:

Let S be the sum of the first n numbers.

S = 1+2+3+...+n 

Also, notice that we can use the commutative property to write S in a different way.

S = n+n-1+n-2+n-3+...+1

If we add these two equations we get

2S = n+1 + n+1 + n+1 +...+ n+1

How many times do we get n+1?  n times.

2S = n*(n+1)

S = n*(n+1)/2  :)

So, this form will work for any sum 1 to #whatever.