Natural numbers/Adding up/Induction/Intersection points/Motivation/Example

We would like to find an easy formula for the sum of the first ${}n$ natural numbers, which equals the maximal number of intersection points of a configuration of ${}n+1$ lines. We claim that

${}\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}\,$

holds. For small numbers ${}n$, this is easy to check just by computing the left-hand and the right-hand side. To prove the identity in general, we try to understand what happens on the left and on the right, if we increase ${}n$ to ${}n+1$, like we have added in example another line to a line configuration. On the left-hand side, we just have the additional summand ${}n+1$. On the right-hand side, we go from ${}{\frac {n(n+1)}{2}}$ to ${}{\frac {(n+1)(n+1+1)}{2}}$. If we can show that the difference between these fractions is ${}n+1$, then the right-hand side behaves like the left-hand side. Then we can conclude: the identity holds for small ${}n$, say for ${}n=1$. By comparing the differences, it also holds for the next ${}n$, so it holds for ${}n=2$, then again for the next ${}n$ and so on. Since this argument always works, and since one arrives at every natural number by taking the successor again and again, the formula holds for every natural number.