Proof 1+2+3+4+\cdots+n = \frac{n\times(n+1)}21+2+3+4+\cdots+n = \frac{n\times(n+1)}2

Apparently $$1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}21+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$$.

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 $$nn$$ positive integers”.

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