Using result R5644: Complement formula for the sum of initial natural numbers, we have
\begin{equation}
\begin{split}
2 \sum_{n = 0}^N n
& = \sum_{n = 0}^N n + \sum_{n = 0}^N (N - n) \\
& = \sum_{n = 0}^N (n + N - n) \\
& = \sum_{n = 0}^N N \\
& = (N + 1) N
\end{split}
\end{equation}
Dividing both sides by $2$, we then obtain
\begin{equation}
\begin{split}
\sum_{n = 0}^N n
= \frac{(N + 1) N}{2}
\end{split}
\end{equation}
as desired. $\square$
