26490

Sums of consecutive numbers:
1 + 2 + 3 + 4 ... + n-1 + n

can be rearranged:

(1 + n) + (2 + n - 1) + (3 + n - 2) ... in the form (x + n - y)

and simplify to:

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

The number of ''(1 + n)''s that are added is n/2 since (x + n - y) is always n + 1, as can be derived through induction from such a case:

S = 1 + 2 + 3 + 4 + 5
= (1 + 5) + (2 + 5 - 1) + 3
= 6 + 6 + 3
= 2(5 + 1) + 2.5(5 + 1)
= 2.5(5 + 1)
= 5/2(5 + 1)
= n/2(n + 1)
= (n^2 + n)/2

Therefore,

sum from i=1 to n of i = 1 + 2 + 3 + 4 ... + n-1 + n
= (n^2 + n)/2