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I need a help regarding my question I searched a lot Somewhere I found the answer of the sequence is $\frac{n(n-1)}{2}.$

Thomas Andrews
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Rubat Majeed
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  • the sum of all integers up to and including $ n $ is $ S_n = \frac{n(n+1)}{2} $, that is because in case of even n, we look at pairs, $(1 $ and $ n)$, $(2 $ and $ (n-1))$, ... $((\frac{n}{2}) - 1 $ and $(\frac{n}{2}) +1)$. there are $ \frac{n}{2} $ such pairs, the sum of a pair is always $ n+1 $ so the total sum is $ \frac{n}{2} $ times $ n+1 $. in the case of odd n, you have $ \frac{n}{2} - 1 $ pairs of sum $ n+1 $ and than you add the number in the middle which is $ \frac{n}{2} $ and get the same answer – Donat Pants Feb 24 '18 at 05:36

1 Answers1

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$S_n=1+2+\dots+(n-1)+n$

$S_n=n+(n-1)+\dots+2+1$

$2S_n=n(n+1)$, from where you get the desired formula.

alans
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