# 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

$\displaystyle T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \left({n-1}\right) + n$

### Definition 3

$\forall n \in \N: T_n = P \left({3, n}\right) = \begin{cases} 0 & : n = 0 \\ P \left({3, n - 1}\right) + \left({n - 1}\right) + 1 & : n > 0 \end{cases}$

where $P \left({k, n}\right)$ 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:

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

By definition of summation:

$\displaystyle T_{n-1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \left({n - 1}\right)$

and so:

 $\displaystyle T_n$ $=$ $\displaystyle T_{n-1} + n$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \left({n - 1}\right) + n$

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.

$\displaystyle T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \left({n - 1}\right) + n$

Thus:

$\displaystyle T_{n-1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \left({n - 1}\right)$

and so:

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

Then:

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

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 $T_n = 0 = P \left({3, n}\right)$.

Then:

 $\, \displaystyle \forall n \in \N_{>0}: \,$ $\displaystyle P \left({3, n}\right)$ $=$ $\displaystyle P \left({3, n - 1}\right) + \left({n - 1}\right) + 1$ $\displaystyle$ $=$ $\displaystyle P \left({3, n - 1}\right) + n$

Thus $P \left({3, n}\right)$ and $T_n$ are generated by the same recurrence relation.

$\blacksquare$