This question already has an answer here:

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

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


marked as duplicate by Jack D'Aurizio, Newb, graydad, k170, apnorton May 2 '15 at 17:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


\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}

  • $\begingroup$ I get lost where you write 2S = ... $\endgroup$ – Joel May 2 '15 at 10:31
  • $\begingroup$ 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? $n-1$. $\endgroup$ – Demosthene May 2 '15 at 10:35
  • $\begingroup$ Oh... you should added the same lists but in opposing ordering. Thanks. $\endgroup$ – Joel May 2 '15 at 10:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.