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

we want to find S

rewrite S backwards:
S = n + (n-1) + (n-2) + ... + 3 + 2 + 1

then add both versions of S to each other...

so S + S = 

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

and notice, if we add each two values above each other together we get:

(n+1) + (n+1) + (n+1) + ... + (n+1)   (n times)

so we have

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

hence

S = n(n+1)/2