# Equivalence of Definitions of Triangular Number

## Theorem

The following definitions of the concept of Triangular Number are equivalent:

### Definition 1

$T_n = \begin{cases} 0 & : n = 0 \\ n + T_{n-1} & : n > 0 \end{cases}$

### Definition 2

$\ds T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \paren {n - 1} + n$

### Definition 3

$\forall n \in \N: T_n = \map P {3, n} = \begin{cases} 0 & : n = 0 \\ \map P {3, n - 1} + \paren {n - 1} + 1 & : n > 0 \end{cases}$

where $\map P {k, n}$ denotes the $k$-gonal numbers.

## Proof

### Definition 1 implies Definition 2

Let $T_n$ be a triangular number by definition 1.

Let $n = 0$.

By definition:

$T_0 = 0$
$\ds T_0 = \sum_{i \mathop = 1}^0 i = 0$

By definition of summation:

$\ds T_{n - 1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \paren {n - 1}$

and so:

 $\ds T_n$ $=$ $\ds T_{n - 1} + n$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \paren {n - 1} + n$

Thus $T_n$ is a triangular number by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $T_n$ be a triangular number by definition 2.

$\ds T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \paren {n - 1} + n$

Thus:

$\ds T_{n - 1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \paren {n - 1}$

and so:

$T_n = T_{n - 1} + n$

Then:

$\ds T_0 = \sum_{i \mathop = 1}^0 i$

is a vacuous summation and so:

$T_0 = 0$

Thus $T_n$ is a triangular number by definition 1.

$\Box$

### Definition 1 equivalent to Definition 3

We have by definition that $T_n = 0 = \map P {3, n}$.

Then:

 $\ds \forall n \in \N_{>0}: \,$ $\ds \map P {3, n}$ $=$ $\ds \map P {3, n - 1} + \paren {n - 1} + 1$ $\ds$ $=$ $\ds \map P {3, n - 1} + n$

Thus $\map P {3, n}$ and $T_n$ are generated by the same recurrence relation.

$\blacksquare$