Olympiad Maths Prep

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Problem 853

AIME late
Algebra Difficulty 5.6 Prove it

Show that k=0nk=0+1+2++n=n(n+1)2\sum_{k=0}^{n} k=0+1+2+\cdots+n=\frac{n(n+1)}{2}.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

We denote Pn\mathcal{P}_{n} as the property « k=0nk=0+1+2++n=n(n+1)2»\sum_{k=0}^{n} k=0+1+2+\cdots+n=\frac{n(n+1)}{2} ».

- Initialization: 0=0×120=\frac{0 \times 1}{2}, so P0\mathcal{P}_{0} is verified.
- Heredity: Let nn be an integer such that Pn\mathcal{P}_{n} is verified, and let's show that Pn+1\mathcal{P}_{n+1} is also verified. 0+1+2++n+(n+1)=n(n+1)2+(n+1)0+1+2+\cdots+n+(n+1)=\frac{n(n+1)}{2}+(n+1) by the induction hypothesis.

Therefore, 0+1+2++n+(n+1)=(n+1)(n2+1)=(n+1)(n+2)2:Pn+10+1+2+\cdots+n+(n+1)=(n+1)\left(\frac{n}{2}+1\right)=\frac{(n+1)(n+2)}{2}: \mathcal{P}_{n+1} is then verified.

By the principle of induction, we have then shown that 0+1+2++n=n(n+1)20+1+2+\cdots+n=\frac{n(n+1)}{2}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.