Let \(S_n\) be:
\[S_n = 1 + 2 + 3 + \ldots + n\]
If we write the same thing in reverse:
\[S_n = \textcolor {red} n + \textcolor{red}{n-1} + \textcolor{red}{n-2} + \ldots + \textcolor{red}2 + \textcolor{red}1\]
Adding the corresponding terms gives:
\[2S_n = \left( \textcolor {red} n+1 \right) + \left( \textcolor{red}{n-1}+2 \right) + \left( \textcolor{red}{n-2}+3 \right) + \ldots + \left( \textcolor{red}2+n-1 \right) + \left( \textcolor{red}1+n \right)\]
Simplify:
\[2S_n = \left( n+1 \right) + \left( n+1 \right) +\ldots + \left( n+1 \right) = n \left(n+1 \right)\]
This means:
\[S_n = \frac{n \left( n+1 \right)}{2}\]