4.1 Patterning, Sequences & Functional Rules

4.1 Patterning, Sequences & Functional Rules

Introduction to Patterning in the Ontario Curriculum

Patterning and algebraic reasoning form a core pillar of the Ontario Mathematics Curriculum from the junior grades through Grade 9 (MTH1W). The Ontario Mathematics Proficiency Test (MPT) assesses candidates' ability to analyze numerical and visual patterns, transition between concrete visual representations and abstract algebraic models, and generalize relationships using functional rules. Educator candidates must not only solve sequence problems efficiently but also understand the pedagogical progression from informal skip-counting to formal linear equations ($t_n = dn + c$).

A sequence is an ordered list of numbers where each number is called a term. We designate the position of a term using a natural number index $n$ (where $n = 1, 2, 3, \dots$), and the numerical value of the term at position $n$ is denoted as $t_n$ or $f(n)$.

---

Linear vs. Non-Linear Patterns

Understanding whether a pattern is linear or non-linear is the critical first step in identifying its functional rule.

Linear Patterns

A pattern is linear if the difference between consecutive terms—known as the first difference ($\Delta$)—is constant throughout the entire sequence. Graphically, if term values ($t_n$) are plotted against their position numbers ($n$), the resulting points lie on a straight line with a constant slope.

For example, consider the sequence: $4, 7, 10, 13, 16, \dots$

• First differences: $7 - 4 = 3$, $10 - 7 = 3$, $13 - 10 = 3$, $16 - 13 = 3$.
• Because the first difference is constant ($+3$), this sequence represents a linear relationship.

Non-Linear Patterns

A pattern is non-linear if the first differences between consecutive terms are not constant. Non-linear patterns tested on the MPT frequently fall into two distinct categories:

1. Quadratic Patterns: The first differences change, but the second differences (the differences between consecutive first differences) are constant. Example: $1, 4, 9, 16, 25, \dots$ (square numbers).
   • First differences: $3, 5, 7, 9$
   • Second differences: $2, 2, 2$ (constant $+2$)
2. Geometric/Exponential Patterns: Consecutive terms are generated by multiplying by a constant factor (common ratio, $r$). Example: $3, 6, 12, 24, 48, \dots$ (multiplying by 2).

---

Representing Patterns: Visual Tile Structures & Tables of Values

In Ontario classrooms, algebraic concepts are introduced visually using manipulative tiles, toothpicks, or geometric dot arrangements. On the MPT, visual patterns are presented as a sequence of diagrams representing Frame 1, Frame 2, Frame 3, etc.

Visual Analysis Technique

To analyze a visual pattern systematically:

1. Count the total elements (tiles, dots, lines) in each frame.
2. Construct a table of values linking the Frame Number ($n$) to the Number of Tiles ($t_n$).
3. Identify the fixed component (base tiles that never change) and the growing component (tiles added with each new frame).

Example Visual Pattern Analysis:

Suppose Frame 1 has 5 tiles, Frame 2 has 8 tiles, Frame 3 has 11 tiles, and Frame 4 has 14 tiles.

---

Term-to-Term Rules vs. Position-to-Term Rules

A critical conceptual distinction on the MPT is between term-to-term rules (recursive) and position-to-term rules (explicit).

1. Term-to-Term Rule (Recursive)

A term-to-term rule explains how to compute the next term in a sequence using the current term:

• Verbal Description: "Start at 5 and add 3 to each subsequent term."
• Algebraic Notation: $t_n = t_{n-1} + 3$, with $t_1 = 5$.
• Limitation: To find the 100th term ($t_{100}$), you must calculate all 99 preceding terms step-by-step. This is highly inefficient for large values of $n$.

2. Position-to-Term Rule (Explicit General Term)

A position-to-term rule calculates the term value $t_n$ directly from its position number $n$, without relying on previous terms:

• Algebraic Notation: $t_n = 3n + 2$.
• Advantage: Allows instant computation of any term. For $n = 100$, $t_{100} = 3(100) + 2 = 302$.

---

Deriving the General Linear Term Formula $t_n = dn + c$

For any linear sequence, the position-to-term explicit formula takes the linear functional form: $t_n = dn + c$ where:

• $d$ is the constant first difference (the common difference between consecutive terms).
• $n$ is the position index ($n = 1, 2, 3, \dots$).
• $c$ is the constant adjustment factor, which represents the theoretical "zero-th term" ($t_0 = t_1 - d$).

Step-by-Step Method to Derive $t_n = dn + c$:

1. Find $d$: Calculate the difference between consecutive terms: $d = t_2 - t_1$.
2. Set up the base expression: Write $dn$ (e.g., if $d = 4$, write $4n$).
3. Determine $c$: Substitute $n = 1$ and $t_1$ into $t_1 = d(1) + c$, then solve for $c$: $c = t_1 - d$
4. Assemble the final formula: Combine $dn$ and $c$ to state $t_n = dn + c$.
5. Verify: Test the formula with $n = 2$ and $n = 3$.

---

Worked Step-by-Step Examples

Example 1: Visual Growing Tile Pattern Generalization

Problem: A tile pattern has 7 tiles in Design 1, 11 tiles in Design 2, and 15 tiles in Design 3. (a) Determine the explicit position-to-term formula $t_n$. (b) How many tiles are required to construct Design 50?

Step-by-Step Solution:

• Step 1: Identify given terms. $t_1 = 7$, $t_2 = 11$, $t_3 = 15$.
• Step 2: Find common difference $d$. $d = t_2 - t_1 = 11 - 7 = 4$ Check with $t_3 - t_2$: $15 - 11 = 4$. Constant first difference $d = 4$.
• Step 3: Calculate constant $c$. $c = t_1 - d = 7 - 4 = +3$
• Step 4: Formulate $t_n$. $t_n = 4n + 3$
• Step 5: Solve for Design 50 ($n = 50$). $t_{50} = 4(50) + 3 = 200 + 3 = 203\text{ tiles}$

Example 2: Finding a Missing Position Given a Term Value

Problem: Given the linear sequence $11, 17, 23, 29, 35, \dots$, which position $n$ contains the term value $t_n = 143$?

Step-by-Step Solution:

• Step 1: Find common difference $d$. $d = 17 - 11 = 6$.
• Step 2: Calculate constant adjustment $c$. $c = t_1 - d = 11 - 6 = 5$.
• Step 3: Write explicit formula. $t_n = 6n + 5$.
• Step 4: Set $t_n = 143$ and solve for $n$. $6n + 5 = 143$ $6n = 143 - 5$ $6n = 138$ $n = \frac{138}{6} = 23$ Conclusion: The 23rd term in the sequence is 143.

Example 3: Shrinking (Decreasing) Linear Sequences

Problem: A shrinking sequence is given as $85, 78, 71, 64, \dots$ (a) Write the general formula $t_n$. (b) Determine the first term position $n$ where the term value becomes negative.

Step-by-Step Solution:

• Step 1: Calculate common difference $d$. $d = 78 - 85 = -7$
• Step 2: Calculate constant adjustment $c$. $c = t_1 - d = 85 - (-7) = 85 + 7 = 92$
• Step 3: State formula. $t_n = -7n + 92$.
• Step 4: Set up inequality for negative term value ($t_n < 0$). $-7n + 92 < 0$ $-7n < -92$ Remember: Dividing by negative numbers flips the inequality sign! $n > \frac{-92}{-7} \implies n > 13.14$ Since $n$ must be a positive integer, the smallest integer greater than 13.14 is $n = 14$.
• Step 5: Verify $t_{14}$. $t_{14} = -7(14) + 92 = -98 + 92 = -6$ Conclusion: The 14th term is the first negative term, with a value of $-6$.