(related to Proposition: Sum of Consecutive Natural Numbers)
We provide a proof by induction for all natural numbers $n.$ * Base case: $n=1$ * $$\sum_{k=0}^1k =0 + 1=1(1+1)/2.$$ * Induction step $n\to n+1$ * Assume, $$\sum_{k=0}^n k=\frac{n(n+1)}2$$ is correct for an $n\ge 1.$ * Then $$\begin{align}\sum_{k=0}^{n+1} k&=\frac{n(n+1)}2+(n+1)\nonumber\\ &=\frac{n(n+1)+2(n+1)}{2}\nonumber\\ &=\frac{(n+1)(n+2)}2.\nonumber\end{align}$$
