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I know this is simple, but I don't know very much at all about series, and I'm wondering how it's shown that:

$$ 1 + 2 + 3 + \cdots + (n - 1) = \frac{n(n - 1)}{2} $$

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    $\begingroup$ This question is basically the same as this previously asked question. It's just that you stop at $n-1$ and this question stops at $n$. $\endgroup$ Feb 13, 2016 at 1:33
  • $\begingroup$ For inspiration on this sum, look up the famous anecdote about Gauss as a young boy doing such a sum, alluded to in this Question. Once you see the trick, you will find it easier to remember than the formula itself. $\endgroup$
    – hardmath
    Feb 13, 2016 at 1:41
  • $\begingroup$ A neat proof of this formula can be obtained by induction. $\endgroup$
    – em29
    Feb 13, 2016 at 5:08

2 Answers 2

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The simple way to understand it is to sum the following sums: $$S=1+2+...+(n-2)+(n-1)$$ $$S=(n-1)+(n-2)+\cdots+2+1$$ These sums are the same, just flipped.

The result is: $$2S=(1+n-1)+(2+n-2)+(n-1+1)=n+n+\cdots+n$$ The number of terms $n$ in the $2S$ is $(n-1)$, so $2S=n(n-1)$, and hence: $$S=1+2+\cdots+(n-1)=\frac{n(n-1)}{2}$$

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The average of $1$ and $n-1$ is $\dfrac n 2$.

The average of $2$ and $n-2$ is $\dfrac n 2$.

The average of $3$ and $n-3$ is $\dfrac n 2$.

The average of $4$ and $n-4$ is $\dfrac n 2$.

and so on $\ldots$

So the average of all of them is $\dfrac n 2$: $$ \frac{1+2+3+\cdots+(n-1)}{n-1} = \frac n 2. $$

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  • $\begingroup$ I like this one, thanks. $\endgroup$
    – Austin
    Feb 13, 2016 at 1:45

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