Theorem 1. For all n ≥ 1,

1 + 2 + ⋯ + n = \frac{n(n + 1)}{2}. (1)

Proof. (Version #1)

n = 1 is clear.

1 + 2 + ⋯ + n + (n + 1) = \frac{n(n + 1)}{2} + (n + 1)
= \frac{n(n + 1)}{2} + \frac{2n + 2}{2}
= \frac{n^{2} + 3n + 2}{2}
= \frac{(n + 1)((n + 1) + 1)}{2}

Proof. (Version #2)

We use induction on n ≥ 1. For the base case, when n = 1, we need to show

that 1 = \frac{1(1 + 1)}{2}. Since \frac{1(1 + 1)}{2} = \frac{1 · 2}{2} = \frac{2}{2} = 1, we have verified the base case.

Now suppose, as our inductive hypothesis, that 1 + 2 + ⋯ + n = \frac{n(n + 1)}{2} for some fixed n. We need to show

1 + 2 + ⋯ + n + (n + 1) = \frac{(n + 1)((n + 1) + 1)}{2}. (2)

Computing the right hand side of (??) we see that

\frac{(n + 1)((n + 1) + 1)}{2} = \frac{n^{2} + 3n + 2}{2}
= \frac{(n^{2} + n) + (2n + 2)}{2}
= \frac{n(n + 1) + 2(n + 1)}{2}
= \frac{n(n + 1)}{2} + (n + 1)

But by the inductive hypothesis, \frac{n(n + 1)}{2} = 1 + 2 + ⋯ + n, and so we have

\frac{(n + 1)((n + 1) + 1)}{2} = 1 + 2 + ⋯ + n + (n + 1).

This proves (??) and hence the theorem.

Proof. (Version #3)

The identity (??) is clearly true when n = 1. Assuming it holds for a fixed n, we prove it for n + 1 as follows:

1 + 2 + ⋯ + n + (n + 1) = \frac{n(n + 1)}{2} + (n + 1)
= \frac{n(n + 1)}{2} + \frac{2n + 2}{2}
= \frac{n^{2} + 3n + 2}{2}
= \frac{(n + 1)((n + 1) + 1)}{2}.

Hence, by induction, identity (??) is true for all n ≥ 1.