Equivalence of Definitions of Pentagonal Number

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Definition 1

$P_n = \begin{cases}

0 & : n = 0 \\ P_{n - 1} + 3 n - 2 & : n > 0 \end{cases}$

Definition 2

$\ds P_n = \sum_{i \mathop = 1}^n \paren {3 i - 2} = 1 + 4 + \cdots + \paren {3 \paren {n - 1} - 2} + \paren {3 n - 2}$

Definition 3

$\forall n \in \N: P_n = \map P {5, n} = \begin {cases}

0 & : n = 0 \\ \map P {5, n - 1} + 3 \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 $P_n$ be a pentagonal number by definition 1.

Let $n = 0$.

By definition:

$P_0 = 0$

By vacuous summation:

$P_0 = \ds \sum_{i \mathop = 1}^0 \paren {3 i - 2} = 0$


By definition of summation:

\(\ds P_{n - 1}\) \(=\) \(\ds \sum_{i \mathop = 1}^{n - 1} \paren {3 i - 2}\)
\(\ds \) \(=\) \(\ds 1 + 4 + \cdots + 3 \paren {n - 1} - 2\)

and so:

\(\ds P_n\) \(=\) \(\ds P_{n - 1} + 3 n - 2\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {3 i - 2}\)
\(\ds \) \(=\) \(\ds 1 + 4 + \cdots + 3 n - 2\)

Thus $P_n$ is a pentagonal number by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $P_n$ be a pentagonal number by definition 2.

Then:

\(\ds P_n\) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {3 i - 2}\)
\(\ds \) \(=\) \(\ds 1 + 4 + \cdots + \paren {3 \paren {n - 1} - 2} + \paren {3 n - 2}\)
\(\ds \) \(=\) \(\ds P_{n - 1} + 3 n - 2\)


Then:

$P_0 = \ds \sum_{i \mathop = 1}^0 \paren {3 n - 2}$

is a vacuous summation and so:

$P_0 = 0$

Thus $P_n$ is a pentagonal number by definition 1.

$\Box$


Definition 1 equivalent to Definition 3

We have by definition that $P_n = 0 = \map P {5, n}$.

Then:

\(\ds \forall n \in \N_{>0}: \, \) \(\ds \map P {5, n}\) \(=\) \(\ds \map P {5, n - 1} + \paren {5 - 2} \paren {n - 1} + 1\)
\(\ds \) \(=\) \(\ds \map P {5, n - 1} + 3 \paren {n - 1} + 1\)
\(\ds \) \(=\) \(\ds \map P {5, n - 1} + 3 n - 2\)

Thus $\map P {5, n}$ and $P_n$ are generated by the same recurrence relation.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Pentagonal_Number&oldid=544222"