Sum of the first n natural numbers

Statement

For every natural number $n$, $$0 + 1 + 2 + \dots + n = \sum_{k=0}^{n} k = \frac{n(n + 1)}{2}.$$

Remarks

The number $n(n + 1)/2$ is the $n$th triangular number. For $n = 1, 2, 3, 4$ it is $1, 3, 6, 10$. A proof by induction uses $\frac{n(n+1)}{2} + (n + 1) = \frac{(n+1)(n+2)}{2}$. An equivalent form without division is $2 \sum_{k=0}^{n} k = n(n + 1)$.

Explanations

L1 human

Write the sum $1 + 2 + \dots + n$ two times, once forward and once backward. Each column then adds up to $n + 1$, and there are $n$ columns. So two copies of the sum give $n(n + 1)$, and one copy is half of that.