C1.4 Represent sequences recursively, explicitly, and with function notation

Ontario Grade 11 Mathematics

Discrete Functions

MCR3U — Unit C: Sequences and Series (Expectation C1.4)

Imagine you are saving money each week. You start with CAD 20, and every week you add CAD 15. After one week you have CAD 35, after two weeks CAD 50, and so on. That ordered list of amounts — 20, 35, 50, 65, … — is a sequence. In this lesson you will learn three different but equivalent ways to describe any sequence precisely: a recursive formula (each term depends on the one before it), an explicit formula (a rule that jumps straight to any term you want), and function notation (which connects sequences to the function ideas you already know from Grade 10). Understanding all three gives you flexibility — sometimes one form is easier to build, and another is easier to use for calculations.

What you will learn

Describe what a sequence is and identify its terms using proper notation.
Write a recursive formula that uses a previous term to generate the next term.
Write an explicit formula that gives any term directly from its position number.
Express sequences using function notation and connect all three representations.

Prerequisite Bridge: Functions and Ordered Lists

In Grade 10 you learned that a function is a rule that assigns exactly one output to each input. You wrote functions as equations such as

f(x) = 3x + 1

and evaluated them by substituting a number for

x

. That skill transfers directly to sequences.

A sequence is simply an ordered list of numbers. Each number in the list is called a term. The position of a term in the list is called its index (or term number). By convention, the first term is at index 1 unless the problem states otherwise.

Notation check: the symbol

t_n

means 'the term at position

n

.' So

t_1

is the first term,

t_2

is the second term, and so on. You will also see

f(n)

used the same way once we switch to function notation.

A sequence is an ordered list; each element is called a term.
The index $n$ records the position of a term in the list.
$t_n$ and $f(n)$ are two common ways to name the term at position $n$ .

Recursive Representation

A recursive formula defines a term by referring to one or more terms that came before it. It always has two parts: a starting value (called the initial condition) and a recurrence rule that says how to get from one term to the next.

Think of the savings example: you always add CAD 15 to whatever you had last week. In math symbols, if

t_n

is the amount after week

n

, then

t_{n+1} = t_n + 15

. The starting value is

t_1 = 20

. Together, those two pieces form the complete recursive formula.

Recursive formulas are great for showing the pattern step by step. Their main limitation is that to find, say,

t_{100}

, you would have to calculate all 99 terms before it — which takes a long time. That is why we also need the other two representations.

General structure: write the initial condition first, then the recurrence rule. Always specify both — a recurrence rule without a starting value does not uniquely determine a sequence.

t_{n+1} = t_n + d, t_1 = a

A recursive formula has two parts: an initial condition and a recurrence rule.
The recurrence rule expresses $t_{n+1}$ (or $t_n$ ) in terms of a previous term.
Every term after the first is found by applying the rule to the term before it.
Recursive form is easy to build but slow for finding terms far along the sequence.

Explicit Representation

An explicit formula gives

t_n

directly as an expression in

n

, with no reference to earlier terms. You substitute the position number and the formula produces the term — no stepping through the sequence required.

For the savings example, notice that

t_1 = 20

,

t_2 = 35 = 20 + 15

,

t_3 = 50 = 20 + 2(15)

,

t_4 = 65 = 20 + 3(15)

. The pattern is clear: each term is the starting value plus

(n-1)

copies of the common difference 15. So the explicit formula is

t_n = 20 + (n-1)(15)

, which simplifies to

t_n = 15n + 5

.

You can verify:

t_1 = 15(1) + 5 = 20

✓,

t_4 = 15(4) + 5 = 65

✓. This formula lets you jump straight to

t_{100} = 15(100) + 5 = 1505

without listing every term in between.

For an arithmetic sequence with first term

a

and common difference

d

, the general explicit formula is

t_n = a + (n-1)d

. For a geometric sequence with first term

a

and common ratio

r

, it is

t_n = a \cdot r^{n-1}

. Recognising the sequence type points you to the right template.

t_n = a + (n-1)d

An explicit formula expresses $t_n$ purely in terms of $n$ — no previous terms needed.
Arithmetic sequences follow $t_n = a + (n-1)d$ ; geometric sequences follow $t_n = a \cdot r^{n-1}$ .
Explicit formulas are efficient for calculating terms far down a sequence.
Always verify by substituting $n = 1$ and at least one other known term.

Function Notation for Sequences

Because a sequence assigns exactly one term to each position number, it is a function whose domain is the positive integers (1, 2, 3, …). Writing it in function notation makes this connection visible.

Instead of

t_n

, write

f(n)

. The explicit formula

t_n = 15n + 5

becomes

f(n) = 15n + 5

, where the domain is

\{1, 2, 3, \ldots\}

. Evaluating

f(4)

gives the fourth term:

f(4) = 15(4) + 5 = 65

.

The recursive form also works in function notation. The initial condition becomes

f(1) = 20

and the recurrence rule becomes

f(n+1) = f(n) + 15

for

n \geq 1

.

Stating the domain is important. The formula

f(n) = 15n + 5

looks like a straight line, but it is only defined at the positive integers — not for every real number. This is what makes it a sequence rather than a continuous linear function.

f(n) = a + (n-1)d, n ∈ \{1, 2, 3, \ldots\}

A sequence is a function with domain $\{1, 2, 3, \ldots\}$ (or sometimes starting at 0).
Replace $t_n$ with $f(n)$ to shift into function notation — the algebra stays the same.
Always state the domain when writing a sequence as a function.
The recursive rule in function notation: $f(n+1) = f(n) + d$ , $f(1) = a$ .

Connecting All Three Representations

The three representations describe the same sequence — they are just different lenses. The recursive form shows the building process step by step. The explicit form gives a fast calculation shortcut. The function notation makes the input-output structure of a sequence precise and connects sequences to the broader idea of functions.

When you are given a sequence as a list of terms, your first job is to spot the pattern. Ask: is each term found by adding the same number (arithmetic) or multiplying by the same number (geometric)? That answer determines which explicit template to use and what the recurrence rule looks like.

Moving between representations is a key skill. Start with the explicit formula, substitute two consecutive values of

n

to find two terms, and the pattern of the recurrence rule becomes visible. Or start with a recursive formula, compute several terms, and look for the explicit shortcut.

All three representations describe the same underlying sequence.
Identify the sequence type (arithmetic or geometric) before choosing a template.
You can convert between representations by generating terms and spotting the pattern.
Function notation emphasises that a sequence is a function with a restricted domain.

Comparing the Three Representations

Feature	Recursive Formula	Explicit Formula	Function Notation
What it shows	Step-by-step building rule	Direct formula for any term	Input-output function structure
Requires previous term?	Yes	No	No (explicit form)
Fast for large $n$ ?	No — must step through all terms	Yes — substitute $n$ directly	Yes — substitute $n$ directly
Arithmetic example	$t_1=7,\ t_{n+1}=t_n+4$	$t_n = 4n+3$	f(n)=4n+3,\ n∈\{1,2,3,\ldots\}
Geometric example	$f(1)=3,\ f(n+1)=2f(n)$	$t_n = 3\cdot 2^{n-1}$	f(n)=3· 2^{n-1},\ n∈\{1,2,3,\ldots\}

Worked example

Example 1 — Arithmetic Sequence: Three Representations

A sequence begins 7, 11, 15, 19, … Write (a) a recursive formula, (b) an explicit formula, and (c) the explicit formula in function notation. Then find the 20th term.

Identify the pattern
Check whether a constant number is added each time. From 7 to 11 is $+4$ , from 11 to 15 is $+4$ , from 15 to 19 is $+4$ . The common difference is $d = 4$ and the first term is $a = 7$ . This is an arithmetic sequence.
$d = 11 - 7 = 4$
Write the recursive formula
State the initial condition $t_1 = 7$ , then write the recurrence rule: each term equals the previous term plus 4. Include the condition $n \geq 1$ so the domain is clear.
$t_1 = 7, t_{n+1} = t_n + 4 \text{ for } n ≥ 1$
Write the explicit formula
Use the arithmetic template $t_n = a + (n-1)d$ with $a = 7$ and $d = 4$ . Substitute and simplify.
$t_n = 7 + (n-1)(4) = 7 + 4n - 4 = 4n + 3$
Write the function notation form
Replace $t_n$ with $f(n)$ and state the domain as the positive integers.
$f(n) = 4n + 3, n ∈ \{1, 2, 3, \ldots\}$
Find the 20th term
Substitute $n = 20$ into the explicit formula. This is much faster than listing all 20 terms.
$f(20) = 4(20) + 3 = 80 + 3 = 83$
Verify with the recursive check
Confirm $f(1) = 4(1) + 3 = 7$ ✓ and $f(2) = 4(2) + 3 = 11$ ✓. Both match the given sequence.
$f(1) = 7, f(2) = 11$

Answer: Recursive:

t_1 = 7

,

t_{n+1} = t_n + 4

. Explicit:

t_n = 4n + 3

. Function notation:

f(n) = 4n + 3

, n ∈ \{1, 2, 3, \ldots\}. The 20th term is

f(20) = 83

.

Check: Using the recursive rule from

t_1 = 7

: add 4 nineteen times, giving

7 + 19(4) = 7 + 76 = 83

✓.

Worked example

Example 2 — Geometric Sequence: Three Representations

A sequence is defined recursively by

f(1) = 3

and

f(n+1) = 2 \cdot f(n)

for

n \geq 1

. (a) List the first four terms. (b) Write the explicit formula. (c) Use function notation to find

f(8)

.

Generate the first four terms from the recursive rule
Start at $f(1) = 3$ . Each new term is twice the previous one. Apply the rule three more times.
$f(1)=3,\ f(2)=6,\ f(3)=12,\ f(4)=24$
Identify the sequence type
Check the ratio between consecutive terms: $6 \div 3 = 2$ , $12 \div 6 = 2$ , $24 \div 12 = 2$ . The common ratio is $r = 2$ , confirming this is a geometric sequence with first term $a = 3$ .
$r = \frac{f(2)}{f(1)} = \frac{6}{3} = 2$
Write the explicit formula
Use the geometric template $t_n = a \cdot r^{n-1}$ with $a = 3$ and $r = 2$ .
$t_n = 3 · 2^{n-1}$
Write in function notation with domain
Replace $t_n$ with $f(n)$ and state the domain.
$f(n) = 3 · 2^{n-1}, n ∈ \{1, 2, 3, \ldots\}$
Evaluate $f(8)$
Substitute $n = 8$ into the explicit formula. Compute $2^7 = 128$ first, then multiply by 3.
$f(8) = 3 · 2^{7} = 3 · 128 = 384$
Verify two known terms
Check $f(1) = 3 \cdot 2^0 = 3 \cdot 1 = 3$ ✓ and $f(4) = 3 \cdot 2^3 = 3 \cdot 8 = 24$ ✓. Both match the terms generated in Step 1.
$f(4) = 3 · 2^{3} = 24$

Answer: First four terms: 3, 6, 12, 24. Explicit:

f(n) = 3 \cdot 2^{n-1}

, n ∈ \{1, 2, 3, \ldots\}. The eighth term is

f(8) = 384

.

Check: Applying the recursive rule eight times from 3: 3 \to 6 \to 12 \to 24 \to 48 \to 96 \to 192 \to 384 ✓.

Common mistakes and how to avoid them

Writing a recursive formula without an initial condition, e.g., only writing

t_{n+1} = t_n + 4

with no starting value.

Correction: A recurrence rule alone does not pin down a unique sequence. Always pair it with an initial condition such as

t_1 = 7

.

Using

(n)

instead of

(n-1)

in the explicit formula, giving

t_n = a + nd

instead of

t_n = a + (n-1)d

.

Correction: When

n = 1

you want the first term

a

, so you need zero copies of

d

. That means the multiplier on

d

must be

(n-1)

, not

n

. Always verify by substituting

n = 1

.

Forgetting to state the domain when writing a sequence in function notation.

Correction: Write n ∈ \{1, 2, 3, \ldots\} (or the appropriate starting value). Without the domain,

f(n) = 4n + 3

looks like a continuous linear function, not a sequence.

Confusing common difference (arithmetic) with common ratio (geometric), leading to the wrong formula type.

Correction: Check whether consecutive terms differ by a constant amount (arithmetic, use

t_n = a + (n-1)d

) or by a constant multiplier (geometric, use

t_n = a \cdot r^{n-1}

).

Evaluating

f(n)

for a non-integer or zero when the domain starts at 1, e.g., computing

f(0)

and treating it as a valid term.

Correction: The domain of a sequence is restricted to the positive integers (or non-negative integers if the sequence starts at index 0). Only substitute allowed values of

n

.

Lesson summary

A sequence is an ordered list of numbers; each element is a term and its position is the index $n$ .
A recursive formula has two parts: an initial condition and a recurrence rule that expresses each term using the previous one.
An explicit formula gives $t_n$ directly in terms of $n$ : $t_n = a + (n-1)d$ for arithmetic sequences and $t_n = a \cdot r^{n-1}$ for geometric sequences.
Function notation rewrites $t_n$ as $f(n)$ and highlights that a sequence is a function whose domain is the positive integers.
All three representations describe the same sequence; choose the one that best fits your purpose — recursive for showing steps, explicit or function notation for efficiency.
Always verify any formula by substituting $n = 1$ and at least one other known term.

Check your understanding

Question 1

A sequence begins 5, 8, 11, 14, … Which recursive formula correctly represents it?

$t_1 = 5$ , $t_{n+1} = t_n + 3$ for $n \geq 1$
$t_1 = 5$ , $t_{n+1} = 3t_n$ for $n \geq 1$
$t_1 = 3$ , $t_{n+1} = t_n + 5$ for $n \geq 1$
$t_1 = 5$ , $t_{n+1} = t_n + 8$ for $n \geq 1$

Show answer and explanation

t_1 = 5

,

t_{n+1} = t_n + 3

for

n \geq 1

The sequence increases by 3 each time (8 − 5 = 3, 11 − 8 = 3), so the common difference is 3, and the first term is 5. The correct recursive formula pairs the initial condition

t_1 = 5

with the rule

t_{n+1} = t_n + 3

. Option B multiplies instead of adds. Options C and D use wrong starting values or differences.

Question 2

What is the explicit formula for the sequence 2, 6, 18, 54, …?

$t_n = 2n + 4$
$t_n = 2 \cdot 3^n$
$t_n = 2 \cdot 3^{n-1}$
$t_n = 3 \cdot 2^{n-1}$

Show answer and explanation

t_n = 2 \cdot 3^{n-1}

Each term is multiplied by 3 (6 ÷ 2 = 3, 18 ÷ 6 = 3), so this is geometric with

a = 2

and

r = 3

. The template gives

t_n = 2 \cdot 3^{n-1}

. Option B uses

3^n

instead of

3^{n-1}

, which gives

t_1 = 6

not 2. Option D swaps

a

and

r

. Option A is a linear (arithmetic) formula, which does not fit.

Question 3

A sequence is defined by

f(n) = 5n - 2

with domain n ∈ \{1, 2, 3, \ldots\}. What is

f(6)

?

28
30
32
27

Show answer and explanation

28

Substitute

n = 6

:

f(6) = 5(6) - 2 = 30 - 2 = 28

. Option B forgets to subtract 2. Option C adds 2 instead of subtracting. Option D subtracts 3 instead of 2.

Question 4

Why must a recursive formula always include an initial condition?

Because without it the formula produces negative terms.
Because without it the recurrence rule could generate infinitely many different sequences.
Because the initial condition replaces the recurrence rule for the first term.
Because function notation requires a starting value to define the domain.

Show answer and explanation

Because without it the recurrence rule could generate infinitely many different sequences.

A recurrence rule like

t_{n+1} = t_n + 4

tells you how to move from one term to the next, but it does not say where to start. Without the initial condition, the rule could describe a sequence starting at any value — there would be infinitely many possibilities. The initial condition pins down the one specific sequence you intend.

Key terms

Sequence: An ordered list of numbers, where each number occupies a specific position.
Term: A single number in a sequence. The term at position $n$ is written $t_n$ or $f(n)$ .
Index: The position number of a term in a sequence, usually starting at 1.
Recursive formula: A two-part rule that defines a sequence: an initial condition giving the first term and a recurrence rule expressing each new term from the previous one.
Explicit formula: A formula that gives the term $t_n$ directly as an expression in $n$ , without needing earlier terms.
Common difference ( $d$ ): The constant amount added between consecutive terms of an arithmetic sequence.
Common ratio ( $r$ ): The constant multiplier between consecutive terms of a geometric sequence.
Domain of a sequence: The set of allowed input values for a sequence written as a function, typically the positive integers $\{1, 2, 3, \ldots\}$ .

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation C1.4. It is a study resource, not an official curriculum publication.

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C1.4 · Represent sequences recursively, explicitly, and with function notation

Ontario Grade 11 Mathematics

MCR3U — Unit C: Sequences and Series (Expectation C1.4)

What you will learn

Prerequisite Bridge: Functions and Ordered Lists

Recursive Representation

Explicit Representation

Function Notation for Sequences

Connecting All Three Representations

Comparing the Three Representations

Example 1 — Arithmetic Sequence: Three Representations

Example 2 — Geometric Sequence: Three Representations

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson