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$\sum\limits_{i=1}^{n-1} i$.

I know the answer is $\frac{1}{2}(n-1)n$ but I don't quite understand it how to get there.

Joel
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1 Answers1

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\begin{align} S=\sum_{i=1}^{n-1}i&=1+2+3+\ldots+n-1\\ S=\sum_{i=1}^{n-1}i&=n-1+n-2+n-3+\ldots+1 \end{align} \begin{align} 2S&=(1+n-1)+(2+n-2)+(3+n-3)+\ldots+(n-1+1)\\ 2S&=n+n+n+\ldots+n\\ 2S&=n(n-1)\\ \therefore S&=\dfrac{n(n-1)}{2} \end{align}

Demosthene
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  • I get lost where you write 2S = ... – Joel May 02 '15 at 10:31
  • Sum up term by term the two previous lines, i.e. the $1$ with the $n-1$, the $2$ with the $n-2$ and so forth. The addition always yields $n$. Now, how many times are we summing up? $n-1$. – Demosthene May 02 '15 at 10:35
  • Oh... you should added the same lists but in opposing ordering. Thanks. – Joel May 02 '15 at 10:52