How do I expand/solve the following summation?

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

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

Answer

\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}