# 3: Number Patterns

- Page ID
- 4792

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

THING OUT LOUD

Consider the following sequence of numbers in which only first two terms are given: 1,3,⋯,⋯. Create four different number patterns having first two terms as 1,3, by writing out next four terms. In each case explain the rule for your pattern. What happens if the first four terms are given as \(1,3, 5,7,\cdots\). How many possibilities are there?

THING OUT LOUD

What is the perimeter of the design by joining \(n\) regular hexagons in a row? How can you prove your prediction?

Numbers can be organized into many different sequences. These sequences have patterns which can be used to predict the next number in the pattern. Misunderstandings may occur when we list a few numbers in the sequence. For example, \(3,5,7..\), the next term could be either \(9\) (sequence of odd integers) or \(11\) (sequence of prime numbers). Therefore it is wise to define sequences in terms of an explicit formula for the \(n\)^th term.

There are many types of patterns, but we will be looking at the following:

- Arithmetic sequences
- Finite sums of arithmetic sequences
- Geometric sequences
- Finite sums of geometric sequences

All sequences, regardless of how they progress, have **terms**. To denote which term we wish to consider, we use \(n\). So, if we say that \(n = 3\), we are considering the third term in a sequence. The first term in a sequence is given by \(a\). So, if we say that \(a = 23\), the first term in the given sequence is 23.

### Proof by Induction

In mathematics, we use induction to prove mathematical statements involving integers. There are two types of induction: regular and strong. The steps start the same but vary at the end. Here are the steps. In mathematics, we start with a statement of our assumptions and intent:

Let \(p(n) \forall n \geq n_0, \, n, \, n_0 \in \mathbb{Z}\) be a statement. We would show that p(n) is true for all possible values of n.

- Show that p(n) is true for the smallest possible value of n: In our case \(p(n_0)\).
**AND** **For regular Induction:**Assume that the statement is true for \(n = k,\) for some integer \(k \geq n_0\). Show that the statement is true for n = k + 1.

**OR**

**For Strong Induction: **Assume that the statement p(r) is true for all integers r, where \(n_0 ≤ r ≤ k \) for some \(k ≥ n_0\). Show that p(k+1) is true.

If these steps are completed and the statement holds, we are saying that, by mathematical induction, we can conclude that the statement is true for all values of \(n \geq n_0.\)

We shall use the following template for proof by induction:

Template for proof by induction

In order to prove a mathematical statement involving integers, we may use the following template:

Suppose \(p(n) \forall n \geq n_0, \, n, \, n_0 \in \mathbb{Z}\) be a statement.

**For regular Induction:**

**Base Case:**__Show that__p(n) is true for the smallest possible value of n: In our case \(p(n_0)\).**Induction Hypothesis**:__Assume that__the statement \(p(n)\) is true for \(n = k,\) for some integer \(k \geq n_0\).**Inductive Step:**__Show tha__t the statement \(p(n)\) is true for \(n=k+1.\).

**For strong Induction:**

**Base Case:**__Show that__p(n) is true for the smallest possible value of n: In our case \(p(n_0)\).**Induction Hypothesis**:__Assume that__the statement \(p(n)\) is true for all integers r, where \(n_0 ≤ r ≤ k \) for some \(k ≥ n_0\).**Inductive Step:**__Show tha__t the statement \(p(n)\) is true for \(n=k+1.\).

If these steps are completed and the statement holds, by mathematical induction, we can conclude that the statement is true for all values of \( n\geq n_0.\)

So, without further ado, let's be off!

**New Notation & Definitions**

Terms: the numbers in a sequence

- When considering a specific term: \(n = x\), where x is a whole number.
- The first term in a sequence: \(a\)

*Thumbnail: Derivation of triangular numbers from a left-justified Pascal's triangle. Image used with permission 9Cc BY-SA 4.0; Cmglee).*

*Thanks to Thomas Thangarajah for sharing his hexagonal drawing.*

## Contributor

Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)