Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ 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".
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: $$S = \frac{n(n+1)}{2}.$$
Well if we graph the numbers $1,2,3,4,.. $ we get a line. Graphing $(1,1)$, $(2,2)$, $(3,3)$ ...I mean. And these numbers occur in a linear fashion so the average of these numbers from $1$ to $n$ is $\frac{n+1}{2}$ and we multiply that by $n$ numbers. Pretty cool huh?

Maybe this one will interest you too:



$$1+3+5+7+9....+(2n-1)=n^2$$