find the sum

The sum of the first N integers is always:

(N^2 + N)
------------
2

i.e. n (n+1) / 2
i.e. 500 x 1001 = 500500

Use the solution of the boy Gauss
Put the 1000 aside and add 999+1 then 998+2 continuously until you get to 501+499. That results in 499x1000=499000. Now you have to account for the 1000 you put aside and the 500 not yet counted. Add then up to 1500, and add them to the product of the rest. That sum would be the total all integers less than and equal to 1000 (49900+1500=500500).

So the sum is 500500.

There is also a derivation of an general formulae for the sum of the first n integers using triangular numbers. This formulae for this sum of the first n integers is n(n+1)/2, which for 1000 is 1000x1001/2=500500. Nice to know that the boy Gauss solved that problem.

Rap

I was surprised when I discovered that in my math book earlier this year, because I discovered the same thing in third grade, altough I never wrote the formula down until sometime about a year ago. Too bad I'll never actually be as good at math as he was...