Describe recursive procedures that generate sequences

Ontario Grade 11 Mathematics

Discrete Functions

MCR3U · Strand C: Sequences and Series · Expectation C1.2

Have you ever followed a set of instructions where each new step depends on what you just did? That is exactly the idea behind a recursive procedure. In mathematics, a recursive procedure is a rule that uses one or more previous terms of a sequence to produce the next term. Before you can apply the rule even once, you need a starting value — called the initial term. Together, the initial term and the recursive rule completely describe the sequence. This lesson builds that idea from the ground up: we start with a plain-English description, move to a table of values, and finish with proper mathematical notation.

What you will learn

Explain what a recursive procedure is and how it differs from an explicit rule.
Use a recursive formula to generate the terms of a sequence step by step.
Write a recursive formula when given a sequence or a description of a pattern.
Identify the role of the initial term and the recursive rule in defining a sequence completely.

Prerequisite Bridge: What Is a Sequence?

A sequence is an ordered list of numbers. Each number in the list is called a term. Terms are labelled by their position: the first term is

t_1

, the second term is

t_2

, and so on. The term at position

n

is written

t_n

.

You have already seen sequences in Grade 10: for example,

2, 4, 6, 8, \ldots

(even numbers) or

1, 4, 9, 16, \ldots

(perfect squares). In those cases, you could find any term directly from its position using a formula such as

t_n = 2n

. That type of rule is called an explicit formula because it gives

t_n

directly without needing earlier terms.

This lesson focuses on a different kind of rule — one where each new term is calculated from the term (or terms) that came before it. That is the recursive approach.

A sequence is an ordered list; each entry is a term, labelled $t_1, t_2, t_3, \ldots$
An explicit formula gives $t_n$ directly from $n$ .
A recursive formula gives $t_n$ by referring to one or more earlier terms.
Both types of formula describe the same sequence — they are just different representations.

What Makes a Procedure Recursive?

A recursive procedure has exactly two parts working together. First, you state the initial condition — this is the value (or values) you are given to start. Without at least one known term, the recursive rule has nothing to build from. Second, you state the recursive rule itself — a mathematical instruction that expresses

t_n

in terms of at least one earlier term such as

t_{n-1}

.

Think of it like a staircase. You cannot step onto stair 3 without first being on stair 2, and you cannot be on stair 2 without starting at stair 1. The initial condition is stair 1; the recursive rule tells you how high each next stair is.

In notation, a recursive formula looks like this: you write the initial term separately, then write a rule for

t_n

that uses

t_{n-1}

(or sometimes

t_{n-2}

, etc.). For example, the plain-English rule 'each term is 5 more than the previous term, and the first term is 3' becomes:

t_1 = 3

and

t_n = t_{n-1} + 5

for

n \geq 2

.

Notice the condition

n \geq 2

. This is important: the recursive rule only makes sense from the second term onward, because

t_1

has no predecessor. Always include this domain restriction when writing a recursive formula.

t_n = t_{n-1} + d, n ≥ 2

A recursive procedure needs: (1) an initial condition and (2) a recursive rule.
The initial condition gives the starting value(s); the rule builds every subsequent term.
The recursive rule is only valid for $n \geq 2$ (or higher if more initial terms are needed).
Without the initial condition, the recursive rule alone cannot generate a unique sequence.

Generating Terms from a Recursive Formula

To generate terms, you substitute known values into the recursive rule one step at a time. You always work from left to right through the sequence — you must know

t_{n-1}

before you can find

t_n

.

Consider the formula

t_1 = 2

and

t_n = 3 \cdot t_{n-1}

for

n \geq 2

. Starting from

t_1 = 2

: substitute

n = 2

to get

t_2 = 3 \cdot t_1 = 3 \cdot 2 = 6

; substitute

n = 3

to get

t_3 = 3 \cdot t_2 = 3 \cdot 6 = 18

; and so on. Each step uses the most recently calculated term.

This one-at-a-time approach is the defining feature of a recursive procedure. You cannot skip to

t_{10}

without first finding

t_2

through

t_9

. That is a limitation compared with an explicit formula, but recursive formulas are often much easier to write when you can see a pattern between consecutive terms but cannot immediately see how to express

t_n

directly in terms of

n

.

Some sequences require two previous terms. For those, you must be given two initial conditions. The Fibonacci-style sequence defined by

t_1 = 1

,

t_2 = 1

, and

t_n = t_{n-1} + t_{n-2}

for

n \geq 3

is a classic example: each term is the sum of the two terms before it.

t_n = t_{n-1} + t_{n-2}, n ≥ 3

Generate terms in order: find $t_2$ , then $t_3$ , then $t_4$ , and so on.
Substitute the previously found value every time — never skip a term.
If the rule uses two previous terms, you need two initial conditions.
List your substitutions clearly so errors are easy to spot.

Writing a Recursive Formula from a Given Sequence

Sometimes you are given a list of terms and asked to describe the recursive procedure that produces them. The strategy is to look at what changes from one term to the next and express that change as a rule.

Step 1 — Find the pattern between consecutive terms. Calculate

t_2 - t_1

,

t_3 - t_2

, etc. (for a constant difference) or

t_2 \div t_1

,

t_3 \div t_2

, etc. (for a constant ratio). Also consider whether the change itself follows a pattern.

Step 2 — Write the initial condition. State

t_1

(and

t_2

if needed).

Step 3 — Write the recursive rule using the pattern you found, with the appropriate domain restriction.

For example, given the sequence

5, 8, 11, 14, \ldots

: the difference between consecutive terms is always

3

, so the rule is

t_n = t_{n-1} + 3

for

n \geq 2

, and the initial condition is

t_1 = 5

. The complete recursive formula is:

t_1 = 5

and

t_n = t_{n-1} + 3

for

n \geq 2

.

Look at the change between consecutive terms to find the pattern.
Check whether the difference or ratio is constant before writing the rule.
Always state the initial condition and the domain restriction explicitly.
Verify your formula by generating the first few terms and checking them against the original list.

Why Recursive Descriptions Matter

Recursive procedures appear naturally whenever a quantity depends on its own history. Population models, compound interest structures, and patterns in nature can all be described recursively. In each case, knowing the current state and a rule for change is enough to build the entire future of the sequence.

In this course, understanding recursive procedures builds the foundation for working with arithmetic and geometric sequences later in Strand C. Arithmetic sequences have a constant difference added at each step; geometric sequences have a constant ratio multiplied at each step. Both are recursive by nature, and the recursive formula is often the most natural way to first describe them.

Recursive thinking also sharpens your algebraic reasoning. Reading a recursive formula carefully — identifying the initial term(s), the rule, and the domain — is a skill you will use throughout MCR3U and beyond.

Real-world patterns — such as population growth and payment schedules — are naturally recursive.
Arithmetic sequences (constant difference) and geometric sequences (constant ratio) are both recursive.
Identifying the initial term, the rule, and the domain are the three key reading skills for any recursive formula.
Recursive formulas are a first step toward understanding more complex sequence behaviour.

Side-by-Side: Explicit vs. Recursive Formula

Feature	Explicit Formula	Recursive Formula
What it looks like	$t_n = 3n + 1$	$t_1 = 4,\; t_n = t_{n-1} + 3$
How you find $t_n$	Substitute $n$ directly — no earlier terms needed	Must know $t_{n-1}$ first
Easy to find $t_{100}$ ?	Yes — one substitution	No — must find $t_2$ through $t_{99}$ first
Easy to write from a pattern?	Requires finding a formula in $n$	Often easier — just describe the change between terms
Both give the same sequence?	Yes	Yes

Worked example

Example 1 — Generating Terms from a Recursive Formula

A sequence is defined by

t_1 = 4

and

t_n = 2 \cdot t_{n-1} - 1

for

n \geq 2

. List the first five terms of the sequence and describe the type of growth you observe.

Identify the initial condition and the rule
The initial condition gives us $t_1 = 4$ . The rule says: to find any term, multiply the previous term by $2$ and then subtract $1$ . We will apply this rule four times to produce $t_2$ through $t_5$ .
Find the second term
Substitute $n = 2$ into the rule $t_n = 2 \cdot t_{n-1} - 1$ . The previous term is $t_1 = 4$ .
$t_2 = 2 · 4 - 1 = 8 - 1 = 7$
Find the third term
Substitute $n = 3$ . The previous term is now $t_2 = 7$ .
$t_3 = 2 · 7 - 1 = 14 - 1 = 13$
Find the fourth term
Substitute $n = 4$ . The previous term is $t_3 = 13$ .
$t_4 = 2 · 13 - 1 = 26 - 1 = 25$
Find the fifth term
Substitute $n = 5$ . The previous term is $t_4 = 25$ .
$t_5 = 2 · 25 - 1 = 50 - 1 = 49$
Describe the growth
The five terms are CAD 4, 7, 13, 25, 49. The differences between consecutive terms are CAD 3, 6, 12, 24 — each difference doubles. This means the terms grow faster and faster (accelerating growth), which happens because the rule multiplies by $2$ at every step.

Answer: The first five terms are CAD 4, 7, 13, 25, 49. The sequence grows at an accelerating rate because each term is more than double the size of the previous difference.

Check: Work backwards:

49 = 2(25) - 1

✓,

25 = 2(13) - 1

✓,

13 = 2(7) - 1

✓,

7 = 2(4) - 1

✓. All terms are confirmed.

Worked example

Example 2 — Writing a Recursive Formula from a Sequence

The sequence

80, 40, 20, 10, 5, \ldots

is given. Write a complete recursive formula that describes this sequence, then use it to find

t_6

.

Find the pattern between consecutive terms
Compare each term with the one before it by dividing: $40 \div 80 = 0.5$ , $20 \div 40 = 0.5$ , $10 \div 20 = 0.5$ , $5 \div 10 = 0.5$ . The ratio is constant at $\frac{1}{2}$ , so each term is half the previous term.
$\frac{t_n}{t_{n-1}} = \frac{1}{2}$
Rearrange to write the recursive rule
Multiplying both sides of the ratio relationship by $t_{n-1}$ gives the recursive rule directly.
$t_n = \frac{1}{2} · t_{n-1}$
State the complete recursive formula
The initial condition is $t_1 = 80$ . The recursive rule applies for $n \geq 2$ . Writing both parts together gives the complete description of the sequence.
$t_1 = 80, t_n = \frac{1}{2} · t_{n-1}, n ≥ 2$
Use the formula to find the sixth term
We already know $t_5 = 5$ from the given list. Substitute $n = 6$ into the recursive rule.
$t_6 = \frac{1}{2} · 5 = 2.5$
Verify the formula generates the given terms
Start from $t_1 = 80$ and apply the rule repeatedly: $\frac{1}{2}(80) = 40$ , $\frac{1}{2}(40) = 20$ , $\frac{1}{2}(20) = 10$ , $\frac{1}{2}(10) = 5$ . These match the original sequence exactly, confirming the formula is correct.

Answer: The recursive formula is

t_1 = 80

and

t_n = \frac{1}{2} \cdot t_{n-1}

for

n \geq 2

. The sixth term is

t_6 = 2.5

.

Check:

t_6 = \frac{1}{2} \times 5 = 2.5

✓. The sequence CAD 80, 40, 20, 10, 5, 2.5 consistently halves at every step.

Common mistakes and how to avoid them

Writing only the recursive rule without an initial condition, e.g., writing just

t_n = t_{n-1} + 4

with no

t_1

stated.

Correction: A recursive formula is incomplete without the initial condition. Always write both: the starting value(s) and the rule together.

Forgetting the domain restriction and writing

t_n = t_{n-1} + 4

as if it applies for all

n

, including

n = 1

.

Correction: The recursive rule refers to

t_{n-1}

, which does not exist when

n = 1

. Always write 'for

n \geq 2

' (or a higher starting index if two initial terms are needed).

Using the original

t_1

value in every step instead of the most recently found term, for example computing

t_3 = 2 \cdot t_1 - 1

instead of

t_3 = 2 \cdot t_2 - 1

.

Correction: Each step must use the immediately preceding term. Update your working value at every step.

Confusing the ratio and difference: dividing consecutive terms to check for a constant difference, or subtracting to check for a constant ratio.

Correction: To check for a constant difference, subtract:

t_n - t_{n-1}

. To check for a constant ratio, divide:

t_n \div t_{n-1}

. Use the right operation for the pattern you are investigating.

Assuming a recursive formula with two initial terms only needs one, then getting stuck because

t_{n-2}

is undefined for

n = 2

.

Correction: If the rule uses

t_{n-2}

, you must provide both

t_1

and

t_2

as initial conditions before the rule can be applied starting at

n = 3

.

Lesson summary

A recursive procedure defines a sequence using two parts: an initial condition (the starting term or terms) and a recursive rule that expresses each new term in terms of one or more previous terms.
The recursive rule always carries a domain restriction (such as $n \geq 2$ ) because the rule refers to a term that must already exist.
To generate terms, work in order from $t_1$ onward, substituting the most recently calculated term into the rule at each step.
To write a recursive formula from a given sequence, find the consistent change between consecutive terms (constant difference, constant ratio, or another pattern), state $t_1$ , and express the rule using $t_{n-1}$ .
If a rule depends on two previous terms, two initial conditions are required, and the rule applies starting at $n \geq 3$ .
Recursive formulas are especially natural for arithmetic sequences (constant difference) and geometric sequences (constant ratio), which are central topics in Strand C of MCR3U.

Check your understanding

Question 1

A sequence is defined by

t_1 = 6

and

t_n = t_{n-1} + 7

for

n \geq 2

. What is

t_4

?

$t_4 = 27$
$t_4 = 34$
$t_4 = 41$
$t_4 = 48$

Show answer and explanation

t_4 = 27

Apply the rule three times:

t_2 = 6 + 7 = 13

,

t_3 = 13 + 7 = 20

,

t_4 = 20 + 7 = 27

. The answer is

27

.

Question 2

Which of the following is a complete recursive formula for the sequence

3, 9, 27, 81, \ldots

?

$t_n = 3^n$
$t_1 = 3$ and $t_n = 3 \cdot t_{n-1}$ for $n \geq 2$
$t_1 = 3$ and $t_n = t_{n-1} + 6$ for $n \geq 2$
$t_n = 3 \cdot t_{n-1}$ for all $n$

Show answer and explanation

t_1 = 3

and

t_n = 3 \cdot t_{n-1}

for

n \geq 2

Each term is three times the previous term, so the rule is

t_n = 3 \cdot t_{n-1}

. The formula in option B is the only one that includes both the correct initial condition

t_1 = 3

and the correct rule with a domain restriction. Option A is an explicit formula, not recursive. Option C uses an incorrect additive rule. Option D is missing the required initial condition.

Question 3

A recursive formula is given as

t_1 = 1

,

t_2 = 1

, and

t_n = t_{n-1} + t_{n-2}

for

n \geq 3

. What is

t_5

?

$t_5 = 4$
$t_5 = 5$
$t_5 = 6$
$t_5 = 8$

Show answer and explanation

t_5 = 5

Generate each term:

t_3 = t_2 + t_1 = 1 + 1 = 2

,

t_4 = t_3 + t_2 = 2 + 1 = 3

,

t_5 = t_4 + t_3 = 3 + 2 = 5

. The answer is

5

.

Question 4

Why must a recursive formula always include an initial condition alongside the recursive rule?

The initial condition makes the formula look more complete on paper.
Without it, the recursive rule has no starting value to work from and cannot produce any terms.
The initial condition replaces the recursive rule for the first term only and is otherwise unused.
The recursive rule already contains the initial condition inside it.

Show answer and explanation

Without it, the recursive rule has no starting value to work from and cannot produce any terms.

The recursive rule expresses

t_n

in terms of a previous term. If no starting value is given, there is nothing to substitute into the rule — the sequence cannot begin. The initial condition is not optional; it is an essential part of the definition.

Key terms

Sequence: An ordered list of numbers, where each number is called a term and is identified by its position.
Term: A single number in a sequence. The term at position $n$ is written $t_n$ .
Recursive procedure: A method of defining a sequence in which each new term is calculated from one or more of the previous terms.
Initial condition: The starting value (or values) that must be given so that the recursive rule can produce the first term and all subsequent terms.
Recursive rule: The mathematical instruction that expresses $t_n$ in terms of earlier terms such as $t_{n-1}$ or $t_{n-2}$ .
Explicit formula: A rule that expresses $t_n$ directly as a function of $n$ , without needing any previous terms.
Domain restriction: The condition (e.g., $n \geq 2$ ) that specifies which values of $n$ the recursive rule applies to.
Constant ratio: A fixed number by which each term is multiplied to produce the next term; characteristic of a geometric sequence.

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