Equivalence of Definitions of Triangular Number

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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

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

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:

$T_0 = 0$

By vacuous summation:

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


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.

$\blacksquare$