Equivalence of Definitions of Triangular Number
Jump to navigation
Jump to search
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.
$\blacksquare$