We denote Pn as the property « ∑k=0nk=0+1+2+⋯+n=2n(n+1)».
- Initialization: 0=20×1, so P0 is verified.
- Heredity: Let n be an integer such that Pn is verified, and let's show that Pn+1 is also verified. 0+1+2+⋯+n+(n+1)=2n(n+1)+(n+1) by the induction hypothesis.
Therefore, 0+1+2+⋯+n+(n+1)=(n+1)(2n+1)=2(n+1)(n+2):Pn+1 is then verified.
By the principle of induction, we have then shown that 0+1+2+⋯+n=2n(n+1).