Proof for the sum of the first n integers

Let,

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

Now add,

2S = (n + 1) + (n + 1) + (n + 1) + ...+ (n + 1) <--- there are n terms here
2S = n*(n + 1)
S = n*(n + 1)/2

Done!