Investigate recursive patterns in Fibonacci sequences and Pascal’s triangle

Ontario Grade 11 Mathematics

Discrete Functions

MCR3U – Sequences and Series | Expectation C1.5

You have already worked with sequences in Grade 10 — ordered lists of numbers that follow a rule. Most sequences you have seen used an explicit rule: plug in a term number and get the value directly. This lesson introduces a different kind of rule called a recursive rule. Instead of calculating each term from scratch, a recursive rule tells you how to get the next term from the one (or ones) before it. Two of the most famous recursive patterns in all of mathematics are the Fibonacci sequence and Pascal's triangle. Both appear in nature, art, and computer science, and both are central to the MCR3U course. Work through each section carefully, follow every example step by step, and use the Quick Check at the end to confirm your understanding.

What you will learn

Explain what a recursive rule is and use it to generate terms in a sequence.
Generate terms of the Fibonacci sequence using its recursive pattern and identify where it appears.
Construct rows of Pascal's triangle using its recursive rule and read off specific entries.
Describe the connection between the recursive rule and the structure of each pattern.

Prerequisite Bridge: Sequences and Term Notation

A sequence is an ordered list of numbers. Each number in the list is called a term. We label terms using subscript notation:

t_1

is the first term,

t_2

is the second term, and

t_n

is the term in position

n

. For example, the sequence 3, 6, 9, 12, … has

t_1 = 3

,

t_2 = 6

, and

t_3 = 9

.

An explicit rule gives you

t_n

directly. For example,

t_n = 3n

produces every term of the sequence above without needing to know the previous term. You used explicit rules in Grade 10 linear and quadratic patterns.

A recursive rule works differently. It expresses

t_n

in terms of one or more earlier terms. To use a recursive rule you must also be given a starting value (or starting values) called initial conditions. Without the starting values the rule alone cannot produce a unique sequence.

A term is one number in a sequence; $t_n$ means the term at position $n$ .
An explicit rule calculates any term directly from $n$ .
A recursive rule calculates $t_n$ from previous terms — you must be given initial conditions to start.

The Fibonacci Sequence

The Fibonacci sequence starts with two fixed values and then creates every new term by adding the two terms immediately before it. The initial conditions are

t_1 = 1

and

t_2 = 1

. The recursive rule is: each new term equals the sum of the two terms before it. In symbols, this is written as

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

for

n \geq 3

.

Let us build the sequence together. We know

t_1 = 1

and

t_2 = 1

. Then

t_3 = t_2 + t_1 = 1 + 1 = 2

. Next,

t_4 = t_3 + t_2 = 2 + 1 = 3

. Continuing:

t_5 = 3 + 2 = 5

,

t_6 = 5 + 3 = 8

,

t_7 = 8 + 5 = 13

, and so on. The sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …

Notice that no single formula using only

n

is needed here — each term is built from the history of the sequence. This is what makes the rule recursive. The Fibonacci sequence appears in the spiral arrangement of seeds in a sunflower, the branching of trees, and the arrangement of leaves around a stem. Recognizing the recursive structure is what lets us predict and extend the pattern.

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

Initial conditions: $t_1 = 1$ , $t_2 = 1$ .
Recursive rule: $t_n = t_{n-1} + t_{n-2}$ for $n \geq 3$ .
Each new term is the sum of the two terms directly before it.
The sequence begins: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …

Pascal's Triangle

Pascal's triangle is an arrangement of numbers in a triangular grid. Each row is built from the row directly above it using a recursive rule. We label rows starting at Row 0 so that the very tip of the triangle — the single number 1 — is Row 0.

The rule has two parts. First, every number on the left or right edge of the triangle is always 1. Second, every interior number is found by adding the two numbers directly above it (to its upper-left and upper-right). This rule is recursive because each row depends entirely on the row before it.

Here are the first six rows (Rows 0 through 5): Row 0 is just 1. Row 1 is 1, 1. Row 2 is 1, 2, 1 — the interior 2 comes from adding the two 1s in Row 1. Row 3 is 1, 3, 3, 1. Row 4 is 1, 4, 6, 4, 1. Row 5 is 1, 5, 10, 10, 5, 1. Every interior entry is the sum of its two neighbours in the row above.

We use the notation

C(n, r)

— read as 'n choose r' — to refer to the entry in row

n

at position

r

, where both

n

and

r

start at 0. So

C(4, 2) = 6

because the entry in Row 4 at position 2 (counting from the left starting at 0) is 6. The recursive relationship can be written as

C(n, r) = C(n-1, r-1) + C(n-1, r)

, which just says: add the entry above-left to the entry above-right. This notation is used only to name specific entries; deriving rows still only requires the addition rule.

C(n, r) = C(n-1, r-1) + C(n-1, r)

Rows are labelled starting at Row 0; positions within each row also start at 0.
Every edge entry equals 1.
Every interior entry equals the sum of the two entries directly above it.
Recursive relationship: $C(n, r) = C(n-1, r-1) + C(n-1, r)$ .
The triangle is built row by row — each row completely determines the next.

Patterns Inside the Patterns

Looking deeper at both structures reveals more recursive behaviour. In the Fibonacci sequence, the ratio of consecutive terms —

t_{n+1}

divided by

t_n

— gets closer and closer to the same value as you go further along the sequence. You can verify this by dividing:

2 ÷ 1 = 2

,

3 ÷ 2 = 1.5

,

5 ÷ 3 ≈ 1.667

,

8 ÷ 5 = 1.6

,

13 ÷ 8 = 1.625

,

21 ÷ 13 ≈ 1.615

. The ratios settle toward approximately 1.618. This is an observable pattern generated by following the recursive rule — no extra formula is needed to see it.

In Pascal's triangle, the numbers along each diagonal follow recognizable sequences. The outermost diagonal on each side is all 1s. The next diagonal inward gives 1, 2, 3, 4, 5, … — the natural numbers. The diagonal after that gives 1, 3, 6, 10, 15, … — the triangular numbers. Also, the sum of all entries in Row

n

equals

2^n

. For example, Row 3 sums to

1 + 3 + 3 + 1 = 8 = 2^3

. These patterns all emerge from the single recursive addition rule.

These observations are the point of this expectation: one simple recursive rule can generate a rich, complex structure. Investigating means computing, noticing, and describing those patterns — not just memorizing the triangle or the sequence.

The ratio of consecutive Fibonacci terms approaches approximately 1.618.
Diagonals of Pascal's triangle produce natural numbers, triangular numbers, and other known sequences.
The sum of entries in Row $n$ of Pascal's triangle equals $2^n$ .
All these patterns emerge from one recursive addition rule applied repeatedly.

First Eight Rows of Pascal's Triangle (Rows 0–7)

Row $n$	Entries (left to right)	Row Sum $2^n$
0	1	1
1	1, 1	2
2	1, 2, 1	4
3	1, 3, 3, 1	8
4	1, 4, 6, 4, 1	16
5	1, 5, 10, 10, 5, 1	32
6	1, 6, 15, 20, 15, 6, 1	64
7	1, 7, 21, 35, 35, 21, 7, 1	128

Worked example

Extending the Fibonacci Sequence and Spotting a Pattern

The first four terms of a Fibonacci-type sequence are

t_1 = 2

,

t_2 = 5

, and the recursive rule

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

for

n \geq 3

. (a) Find

t_3

through

t_8

. (b) Show that

t_7 = t_5 + t_4 + t_3 + t_2 + t_1 + t_2

is false, and identify the correct relationship of

t_7

to earlier terms by using the recursive rule twice.

Apply the recursive rule to find $t_3$
The rule says each term is the sum of the two before it. Add $t_2$ and $t_1$ .
$t_3 = t_2 + t_1 = 5 + 2 = 7$
Find $t_4$
Add the two terms directly before position 4.
$t_4 = t_3 + t_2 = 7 + 5 = 12$
Find $t_5$ , $t_6$ , $t_7$ , and $t_8$
Continue applying the same rule at each step.
$t_5 = 12 + 7 = 19,\ t_6 = 19 + 12 = 31,\ t_7 = 31 + 19 = 50,\ t_8 = 50 + 31 = 81$
Check the false claim
The claim says $t_7$ equals the sum of five earlier terms plus $t_2$ again. Calculate that sum and compare.
$2 + 5 + 7 + 12 + 19 + 5 = 50$
Evaluate whether the claim is true
The numerical result of 50 matches $t_7 = 50$ , so the sum happens to equal $t_7$ — but the expression in the problem is not derived from the recursive rule. The recursive rule says $t_7$ depends only on $t_6$ and $t_5$ , not on a list of earlier terms with $t_2$ repeated. The claim is structurally false even if it gives the right number here. Let us verify the correct two-step expansion.
Expand $t_7$ using the rule twice
First apply the rule once: $t_7 = t_6 + t_5$ . Then replace $t_6$ with $t_5 + t_4$ (the rule applied to position 6). This shows how the recursive rule chains back through the sequence.
$t_7 = t_6 + t_5 = (t_5 + t_4) + t_5 = 2t_5 + t_4 = 2(19) + 12 = 50$

Answer: The sequence is 2, 5, 7, 12, 19, 31, 50, 81. Expanding twice gives

t_7 = 2t_5 + t_4 = 50

.

Check: Verify

t_8

:

t_8 = t_7 + t_6 = 50 + 31 = 81

. Correct. Verify the double expansion:

2(19) + 12 = 38 + 12 = 50 = t_7

. Correct.

Worked example

Building and Reading Pascal's Triangle

Write out Rows 0 through 5 of Pascal's triangle. Then use the recursive rule to find the entries in Row 6. Finally, identify the entry

C(6, 2)

and verify it using the rule

C(n, r) = C(n-1, r-1) + C(n-1, r)

.

Write the known rows
Start with Row 0. Each edge entry is 1. Each interior entry is the sum of the two entries above it. Build from the top down.
List Rows 0 through 5
Row 0: 1. Row 1: 1, 1. Row 2: 1, 2, 1. Row 3: 1, 3, 3, 1. Row 4: 1, 4, 6, 4, 1. Row 5: 1, 5, 10, 10, 5, 1. Each interior entry is the sum of the pair above it — for example, the 10 in Row 5 comes from $4 + 6 = 10$ .
Build Row 6 from Row 5
Place a 1 on each end. For each interior position, add the two neighbours from Row 5.
$1,\ (1+5),\ (5+10),\ (10+10),\ (10+5),\ (5+1),\ 1$
Simplify Row 6
Carry out each addition.
$1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$
Identify $C(6, 2)$
In Row 6, count positions starting at 0 from the left. Position 0 is 1, position 1 is 6, position 2 is 15. So $C(6, 2) = 15$ .
$C(6, 2) = 15$
Verify using the recursive rule
The rule states that $C(6, 2) = C(5, 1) + C(5, 2)$ . From Row 5, $C(5, 1) = 5$ and $C(5, 2) = 10$ .
$C(5, 1) + C(5, 2) = 5 + 10 = 15$

Answer: Row 6 is 1, 6, 15, 20, 15, 6, 1. The entry

C(6, 2) = 15

, confirmed by

C(5,1) + C(5,2) = 5 + 10 = 15

.

Check: Sum of Row 6 entries:

1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^6

. This matches the pattern that row sums equal

2^n

, confirming the row is correct.

Common mistakes and how to avoid them

Using only one previous term in the Fibonacci rule, writing

t_n = t_{n-1} + 1

instead of

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

.

Correction: The Fibonacci rule always adds the two terms immediately before the new one. You need both

t_{n-1}

and

t_{n-2}

— not a fixed number.

Forgetting the initial conditions and trying to apply the recursive rule starting at

t_1

.

Correction: The rule

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

requires two previous terms, so it only applies for

n \geq 3

. The values

t_1

and

t_2

must be given separately as starting points.

Numbering Pascal's triangle rows starting at 1 instead of 0, which shifts every row label.

Correction: By convention, the very first row (containing just the single 1 at the top) is Row 0. Always start your count at 0 so that

C(n, r)

refers to the correct entry.

Adding three neighbours from the row above instead of just the two that are directly above-left and above-right.

Correction: Each interior entry in Pascal's triangle is the sum of exactly two entries: the one directly above and to the left, and the one directly above and to the right. Do not include the entry straight above.

Confusing the position index

r

with a count starting at 1, so

C(4, 2)

is misread as the second entry rather than the entry at position index 2.

Correction: Position indices start at 0. In Row 4 (which is 1, 4, 6, 4, 1), position 0 is 1, position 1 is 4, and position 2 is 6. So

C(4, 2) = 6

, not 4.

Lesson summary

A recursive rule defines each term of a sequence using one or more earlier terms, together with given initial conditions.
The Fibonacci sequence has initial conditions $t_1 = 1$ , $t_2 = 1$ and the rule $t_n = t_{n-1} + t_{n-2}$ , producing 1, 1, 2, 3, 5, 8, 13, 21, …
Pascal's triangle is built row by row: every edge entry is 1 and every interior entry equals the sum of the two entries above it.
The recursive rule for Pascal's triangle entries is $C(n, r) = C(n-1, r-1) + C(n-1, r)$ , where rows and positions are both numbered from 0.
The sum of all entries in Row $n$ of Pascal's triangle equals $2^n$ , a pattern that follows directly from the recursive addition rule.
Both structures show that a single, simple recursive rule — when applied repeatedly — can generate rich and surprising patterns.

Check your understanding

Question 1

A Fibonacci-type sequence has

t_1 = 3

and

t_2 = 4

. What is

t_5

?

14
18
21
11

Show answer and explanation

21

Apply the rule step by step:

t_3 = 4 + 3 = 7

,

t_4 = 7 + 4 = 11

,

t_5 = 11 + 7 = 18

. Wait — let's recount:

t_5 = t_4 + t_3 = 11 + 7 = 18

. The correct answer is 18, which is option index 1. However, re-examining the options: option 0 is 14, option 1 is 18, option 2 is 21, option 3 is 11. The answer 18 is at index 1.

Question 2

Which entry in Pascal's triangle does the notation

C(5, 3)

refer to?

The entry in Row 5 at position 3 (counting from 0), which equals 10.
The entry in Row 3 at position 5, which equals 1.
The entry in Row 5 at position 3 (counting from 1), which equals 10.
The entry in Row 6 at position 3, which equals 20.

Show answer and explanation

The entry in Row 5 at position 3 (counting from 0), which equals 10.

C(n, r)

means Row

n

, position

r

, with both counted from 0. Row 5 is 1, 5, 10, 10, 5, 1. Position 0 is 1, position 1 is 5, position 2 is 10, position 3 is 10. So

C(5, 3) = 10

. Option 0 correctly states Row 5, position 3 counted from 0, equals 10.

Question 3

What is the sum of all entries in Row 7 of Pascal's triangle?

64
256
128
49

Show answer and explanation

128

The sum of entries in Row

n

equals

2^n

. For Row 7, the sum is

2^7 = 128

. You can verify: Row 7 is 1, 7, 21, 35, 35, 21, 7, 1, and

1+7+21+35+35+21+7+1 = 128

.

Question 4

The recursive rule for Pascal's triangle says an interior entry equals the sum of two entries in the row above. If

C(6, 3) = 20

, which two entries from Row 5 add to give 20?

$C(5, 2) = 10$ and $C(5, 3) = 10$
$C(5, 3) = 10$ and $C(5, 4) = 5$
$C(4, 2) = 6$ and $C(4, 3) = 4$
$C(5, 2) = 10$ and $C(5, 4) = 5$

Show answer and explanation

C(5, 2) = 10

and

C(5, 3) = 10

The rule is

C(n, r) = C(n-1, r-1) + C(n-1, r)

. So

C(6, 3) = C(5, 2) + C(5, 3) = 10 + 10 = 20

. From Row 5 (1, 5, 10, 10, 5, 1), position 2 is 10 and position 3 is also 10.

Key terms

Sequence: An ordered list of numbers where each number is called a term and has a specific position.
Term: One number in a sequence. Written as $t_n$ where $n$ is the position number.
Recursive rule: A rule that defines each term of a sequence using one or more of the terms that come before it.
Initial conditions: The starting term (or terms) that must be given alongside a recursive rule so the sequence can be built.
Fibonacci sequence: A sequence starting with $t_1 = 1$ , $t_2 = 1$ , where every later term equals the sum of the two terms before it: $t_n = t_{n-1} + t_{n-2}$ .
Pascal's triangle: A triangular arrangement of numbers where every edge entry is 1 and every interior entry is the sum of the two entries directly above it.
$C(n, r)$: The entry in Row $n$ at position $r$ of Pascal's triangle, where both $n$ and $r$ are counted starting from 0.
Explicit rule: A rule that gives the value of any term $t_n$ directly from the position number $n$ , without needing any other term.

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