2

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:

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

Michael Hardy
  • 1
  • 31
  • 294
  • 591
Austin
  • 680
  • 8
  • 17
  • 7
    This question is [basically the same as this previously asked question](https://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn). It's just that you stop at $n-1$ and this question stops at $n$. – Noble Mushtak Feb 13 '16 at 01:33
  • 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](http://math.stackexchange.com/questions/28885/anecdotes-about-famous-mathematicians-or-physicists). Once you see the trick, you will find it easier to remember than the formula itself. – hardmath Feb 13 '16 at 01:41
  • A neat proof of this formula can be obtained by induction. – em29 Feb 13 '16 at 05:08

2 Answers2

3

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

Michael Hardy
  • 1
  • 31
  • 294
  • 591
Kerr
  • 1,856
  • 8
  • 18
2

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

Michael Hardy
  • 1
  • 31
  • 294
  • 591