1

How can the sequence composed by the number of diagonals ($diagonal = \frac{side(side - 3)}{2}$) of convex regular polygons be described (as Arithmetic Progression, Geometric Progression or other)?

For example, if I consider it as an Arithmetic Progression, the general term formula would be $(a_n = a_1 + (N-1)R)$, where $a_n$ is the general term, $a_1$ is the first term, $N$ is the $N$th term and $R$ is the reason. However, in this sequence, the ratio does not remain constant, varying in an increasing way. Example: $R = a_2-a_1 = 2-0 = 2; \ R = a_3-a_2 = 5-2=3;$ etc.

In the case of a Geometric Progression (where the formula of the general term is $a_n = a_1*q^{(N-1)}$, with $q$ being the ratio), it is the same situation, the ratio is not constant.

Example of the described sequence: $$0(Triangle),\\2(Square),\\5(Pentagon),\\9(Hexagon),\\14(Heptagon),\\...$$

Isosceles
  • 33

1 Answers1

0

It is a convination of an arithmetic and a quadratic progression. The differences between consecutive terms form an arithmetic series as you noted.

$ x_0=0 \ \ , \ \Delta_0=2 \\ x_1=2 \ \ , \ \Delta_1=3 \\ x_2=5 \ \ , \ \Delta_2=4 \\ x_3=9 \ \ , \ \Delta_3=5 \\ x_4=14 $

Therefore we can express a function D in terms of n as

$D(n)=\frac{(n^2 - 3n)}{2} \\ \int{D(n) dn} = (\frac{1}{6}) n^3 - (\frac{3}{4}) n^2 + c \ \forall \ c \in ℝ$

mathsandscience
  • 82
  • I corrected an error referring to my question, the value of the number of diagonals of the heptagon was wrong, therefore, $Δ_3=6$ must be changed to $Δ_3=5$. – Isosceles Jul 04 '23 at 21:15
  • @Isosceles oh yes sorry, I copied the numbers from your question without much thought, already changed it – mathsandscience Jul 05 '23 at 11:54