Sums of Successive Integers

Suppose we want to find 1 + 2 + ... + 100?

Here's a neat trick:

1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
... ... ...
99 + 2 = 101
100 + 1 = 101

So 2*(1 + 2 + ... + 100) = 100*101

That is, 1 + 2 + ... + 100 = 100*101/2

In general, 1 + 2 + ... + N = N*(N+1)/2

This approach is attributed to Gauss, when he was in the fourth grade.