algebra precalculus - Proof for formula for sum of sequence $1+2+3+ldots+n$?

Apparently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$.

How? What's the proof? Or maybe it is self apparent just looking at the above?

PS: This problem is known as "The sum of the first $n$ positive integers".

Answer

Let $$S = 1 + 2 + \ldots + (n-1) + n.$$ Write it backwards: $$S = n + (n-1) + \ldots + 2 + 1.$$ Add the two equations, term by term; each term is $n+1,$ so $$2S = (n+1) + (n+1) + \ldots + (n+1) = n(n+1).$$ Divide by 2: $$S = \frac{n(n+1)}{2}.$$