Equivalence of Definitions of Second Pentagonal Number

Theorem

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

Definition 1

The second pentagonal numbers are the integers obtained by applying the Closed Form for Pentagonal Numbers to negative $n$:

$\map {P'} n = \dfrac {-n \paren {-3 n - 1} } 2$

Definition 2

The second pentagonal numbers are the integers obtained from formula:

$\map {P'} n = \dfrac {n \paren {3 n + 1} } 2$

for $n = 0, 1, 2, \ldots$

Proof

Let $\map {P'} n$ denote the $n$th second pentagonal number.

We have:

\(\ds \map {P'} n\)

\(=\)

\(\ds \dfrac {-n \paren {-3 n - 1} } 2\)

Definition 1 of Second Pentagonal Number

\(\ds \)

\(=\)

\(\ds \dfrac {n \paren {3 n + 1} } 2\)

simplifying

But:

$\map {P'} n = \dfrac {n \paren {3 n + 1} } 2$

is the definition of a second pentagonal Number by definition 2.

Hence the result.

$\blacksquare$