Saturday, February 27, 2016

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: $$S = \frac{n(n+1)}{2}.$$


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