Closed Form for Triangular Numbers/Proof using Cardinality of Set

Theorem
The closed-form expression for the $n$th triangular number is:
 * $\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \left({n+1}\right)} 2$

Proof
Let $\N_n^*=\left\{ {1,2,3,\cdots,n}\right\}$ be the initial segment of natural numbers.

Let $A=\left\{ {\left({a,b}\right):a\le b,a,b\in \N_n^*}\right\}$

Let $B=\left\{ {\left({a,b}\right):a\ge b,a,b,\in \N_n^*}\right\}$

Let $\phi:A\to B$ be the mapping:
 * $\phi\left({x,y}\right)=\left({y,x}\right)$

By definition of dual ordering, $\phi$ is a bijection, i.e. $\vert A\vert = \vert B \vert \cdots\left({1}\right)$

$A\cup B=\left\{ {\left({a,b}\right):a,b\in\N_n^*}\right\}$

$A\cap B=\left\{ {\left({a,b}\right):a=b:a,b\in\N_n^*}\right\}$

Combined with $\left({1}\right)$ yields $\vert A \vert = \dfrac {n^2 + n} 2 = \dfrac {n\left({n+1}\right)} 2$

It remains to prove that $T_n = \vert A \vert$