Closed Form for Triangular Numbers/Proof using Cardinality of Set

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Theorem

The closed-form expression for the $n$th triangular number is:

$\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$


Proof

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

Let $A = \set {\tuple {a, b}: a \le b, a, b \in \N_n^*}$

Let $B = \set {\tuple {a, b}: a \ge b, a, b, \in \N_n^*}$


Let $\phi: A \to B$ be the mapping:

$\map \phi {x, y} = \tuple {y, x}$

By definition of dual ordering, $\phi$ is a bijection:

$(1): \quad \size A = \size B$

We have:

\(\displaystyle A \cup B\) \(=\) \(\displaystyle \set {\tuple {a, b}: a, b \in \N_n^*}\)
\(\displaystyle A \cap B\) \(=\) \(\displaystyle \set {\tuple {a, b}: a = b: a, b \in \N_n^*}\)

Thus:

\(\displaystyle \size A + \size B\) \(=\) \(\displaystyle \size {A \cup B} + \size {A \cap B}\) Inclusion-Exclusion Principle
\(\displaystyle \) \(=\) \(\displaystyle n^2 + n\) Count of a finite set

Combined with $\left({1}\right)$ this yields:

$\size A = \dfrac {n^2 + n} 2 = \dfrac {n \paren {n + 1} } 2$


It remains to prove that:

$T_n = \size A$
\(\displaystyle T_n\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n i\) Definition of $T_n$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop \in \N_n^*} i\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop \in \N_n^*} \size {\set {a: a \in \N_i^*} }\) Count of Finite Set,Definition of Initial Segment of One-Based Natural Numbers
\(\displaystyle \) \(=\) \(\displaystyle \size {\set {\tuple {a, i} :a\in \N_i^*, i \in \N_n^*} }\) Inclusion-Exclusion Principle, sets are mutually exclusive as their second argument in the ordered pair are different
\(\displaystyle \) \(=\) \(\displaystyle \size {\set {\tuple {a, b}: a \le b, a, b \in \N_n^*} }\) Change of Variable, Definition of Initial Segment of One-Based Natural Numbers
\(\displaystyle \) \(=\) \(\displaystyle \size A\) Definition of $A$

$\blacksquare$