Closed Form for Triangular Numbers

Theorem:

$$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$

Direct Proof
$$\sum_{i=1}^{n}i = 1 + 2 + ... + n$$.

Consider $$2\sum_{i=1}^{n}i$$ $$= 2(1 + 2 + ... + n)$$ $$= (1 + 2 + ... + n) + (1 + 2 + ... + n)$$ $$=(1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1)$$ by commutativity and associativity. $$= (n + 1)_{1} + (n + 1)_{2} + ... + (n + 1)_{n}$$ $$= n(n+1)$$.

Therefore, $$2\sum_{i=1}^{n}i = n(n+1)$$ $$\implies \sum_{i=1}^{n}i=\frac{n(n+1)}{2}$$

QED