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)}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}.

Attribution
Source : Link , Question Author : b1_ , Answer Author : Community

Leave a Comment