Equivalence of Definitions of Triangular Number

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

By definition of summation:


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

and so:

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

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.

Definition 1 equivalent to Definition 3
We have by definition that $T_n = 0 = P \left({3, n}\right)$.

Then:

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