Let $n$ be a natural number. Give an explicit formula for the sum $0+1+2+\ldots+n$.
The initialization for $n=0$ is already done (see the judicious choice of $a, b, c$).

Suppose now that there exists a natural number $n$ such that $u_{n}=\frac{n(n+1)}{2}$. Then:

$$
u_{n+1}=u_{n}+(n+1)=\frac{n(n+1)}{2}+n+1=\frac{(n+1)(n+2)}{2}
$$

so $\mathcal{P}_{n+1}$ is true, which concludes by the principle of induction.