Sequences nth Term and Recurrence Relations — Edexcel A-level Mathematics (Pure)

SPEC 4.2 - SEQUENCES AND SERIES

Sequences: nth Term and Recurrence Relations

Understanding position-to-term and term-to-term rules for generating sequences

What is a Sequence?

Definition

A sequence is an ordered list of numbers that follow a particular pattern or rule. Each number in the sequence is called a term.

We use the following notation:

u_{1}

= first term,

u_{2}

= second term,

u_{3}

= third term, etc.

u_{n}

= the nth term (the general term)

The sequence can be written as

u_{1}, u_{2}, u_{3}, ..., u_{n}, ...

Two Ways to Define a Sequence

1. Position-to-term rule (nth term formula) - gives any term directly from its position

2. Term-to-term rule (recurrence relation) - gives each term from the previous term(s)

Position-to-Term Rules

A position-to-term rule gives a formula for the nth term

u_{n}

in terms of

n

.

Formula

u_{n} = f (n)

This allows us to find any term directly without calculating all previous terms.

Sequencenth Term FormulaCheck:

u_{5}

2, 4, 6, 8, 10, ...

u_{n} = 2 n

2 (5) = 10

✔

1, 4, 9, 16, 25, ...

u_{n} = n^{2}

5^{2} = 25

✔

3, 5, 7, 9, 11, ...

u_{n} = 2 n + 1

2 (5) + 1 = 11

✔

Advantage

We can find

u_{100}

or

u_{1000}

instantly without finding all previous terms!

Worked Example01

Find the nth term formula for the sequence: 5, 8, 11, 14, 17, ...

1

Identify the pattern:

8 - 5 = 3

,

11 - 8 = 3

,

14 - 11 = 3

. Common difference

d = 3

.

2

For a linear sequence:

u_{n} = an + b

where

a

= common difference. So

u_{n} = 3 n + b

.

3

Find

b

using the first term: When

n = 1

,

u_{1} = 5

. So

3 (1) + b = 5

, giving

b = 2

.

Answer

u_{n} = 3 n + 2

. Check:

u_{4} = 3 (4) + 2 = 14

✔

02Worked Example

A sequence is defined by $u_{n} = n^{2} - 3 n + 5$ .

(a) Find

u_{1}

,

u_{2}

,

u_{3}

, and

u_{4}

. (b) Find

u_{20}

. (c) Which term has value 47?

(a)

u_{1} = 1 - 3 + 5 = 3

,

u_{2} = 4 - 6 + 5 = 3

,

u_{3} = 9 - 9 + 5 = 5

,

u_{4} = 16 - 12 + 5 = 9

(b)

u_{20} = 400 - 60 + 5 = 345

(c)

Solve

n^{2} - 3 n + 5 = 47

gives

n^{2} - 3 n - 42 = 0

, so

(n - 7) (n + 6) = 0

. Since

n > 0

:

n = 7

.

Answers: (a) 3, 3, 5, 9 (b) 345 (c) The 7th term

Term-to-Term Rules

A term-to-term rule tells us how to get from one term to the next.

Definition

A recurrence relation defines each term of a sequence using the previous term(s).

ApproachPosition-to-TermTerm-to-Term

Formula

u_{n} = 3 n + 2

u_{n + 1} = u_{n} + 3

,

u_{1} = 5

MeaningSubstitute n to find termAdd 3 to get next term

AdvantageFind any term directlyEasy to understand

Note

Both rules describe the same sequence: 5, 8, 11, 14, 17, ...

Recurrence Relations

A recurrence relation has two essential parts:

Two Parts

1. The iterative formula:

u_{n + 1} = f (u_{n})

2. The starting value(s):

u_{1} = a

Important

You must have both parts! The recurrence relation alone does not uniquely define a sequence without the starting value.

Common notation:

u_{n + 1} = u_{n} + 3

means "add 3 to get the next term"

u_{n + 1} = 2 u_{n}

means "multiply by 2 to get the next term"

u_{n + 1} = u_{n}^{2} - 1

means "square and subtract 1"

Alternative

u_{n + 1} = u_{n} + 3

can also be written as

u_{n} = u_{n - 1} + 3

Worked Example03

A sequence is defined by $u_{n + 1} = 3 u_{n} - 2$ , where $u_{1} = 4$ . Find the first five terms.

1

u_{1} = 4

(given)

2

u_{2} = 3 (4) - 2 = 12 - 2 = 10

3

u_{3} = 3 (10) - 2 = 30 - 2 = 28

4

u_{4} = 3 (28) - 2 = 84 - 2 = 82

5

u_{5} = 3 (82) - 2 = 246 - 2 = 244

AnswerThe first five terms are 4, 10, 28, 82, 244

Worked Example04

A sequence begins: 2, 6, 18, 54, 162, ... Write down a recurrence relation for this sequence.

1

Look at the relationship:

\frac{6}{2} = 3

,

\frac{18}{6} = 3

,

\frac{54}{18} = 3

2

Each term is 3 times the previous term.

(This is a geometric sequence with first term 2 and common ratio 3.)

Answer

u_{n + 1} = 3 u_{n}

, where

u_{1} = 2

Increasing and Decreasing Sequences

Definitions

A sequence is increasing if

u_{n + 1} > u_{n}

for all

n

.

A sequence is decreasing if

u_{n + 1} < u_{n}

for all

n

.

TypeExampleTest

Increasing2, 5, 8, 11, 14, ...

u_{n + 1} - u_{n} = 3 > 0

for all

n

Decreasing100, 50, 25, 12.5, ...

u_{n + 1} - u_{n} < 0

for all

n

How to Prove

Increasing: Show that

u_{n + 1} - u_{n} > 0

for all valid

n

.

Decreasing: Show that

u_{n + 1} - u_{n} < 0

for all valid

n

.

05Worked Example

A sequence is defined by $u_{n} = n^{2} - 10 n + 30$ for $n \geq 1$ .

(a) Find the first five terms. (b) Is the sequence increasing, decreasing, or neither?

(a)

u_{1} = 1 - 10 + 30 = 21

,

u_{2} = 4 - 20 + 30 = 14

,

u_{3} = 9 - 30 + 30 = 9

,

u_{4} = 16 - 40 + 30 = 6

,

u_{5} = 25 - 50 + 30 = 5

(b)

Find

u_{n + 1} - u_{n} = [(n + 1)^{2} - 10 (n + 1) + 30] - [n^{2} - 10 n + 30]

•

= n^{2} + 2 n + 1 - 10 n - 10 + 30 - n^{2} + 10 n - 30 = 2 n - 9

•

For

n \leq 4

:

2 n - 9 < 0

(decreasing). For

n \geq 5

:

2 n - 9 > 0

(increasing).

Answer(a) 21, 14, 9, 6, 5 (b) Neither - the sequence decreases then increases

Periodic Sequences

Definition

A sequence is periodic if the terms repeat in a cycle. The number of terms in one complete cycle is called the period (or order).

If a sequence has period

k

, then:

Formula

u_{n + k} = u_{n}

for all

n

Examples:

1, 2, 3, 1, 2, 3, 1, 2, 3, ... has period 3

5, -5, 5, -5, 5, -5, ... has period 2

Common Sources

Periodic sequences often arise from recurrence relations like

u_{n + 1} = \frac{1}{u _{n}}

or

u_{n + 1} = - u_{n}

.

06Worked Example

A sequence is defined by $u_{n + 1} = \frac{1}{1 - u _{n}}$ , where $u_{1} = 2$ .

(a) Find

u_{2}

to

u_{6}

. (b) State the period. (c) Find

u_{100}

.

(a)

u_{1} = 2

,

u_{2} = \frac{1}{1 - 2} = - 1

,

u_{3} = \frac{1}{1 - ( - 1 )} = \frac{1}{2}

,

u_{4} = \frac{1}{1 - \frac{1}{2}} = 2

,

u_{5} = - 1

,

u_{6} = \frac{1}{2}

(b)

Sequence:

2, - 1, \frac{1}{2}, 2, - 1, \frac{1}{2}, ...

repeats every 3 terms. Period = 3

(c)

100 \div 3 = 33

remainder

1

. So

u_{100}

is in position 1 of the cycle.

Answer(c)

u_{100} = 2

Convergent and Divergent Sequences

Convergent

A sequence is convergent if its terms approach a fixed value (the limit) as

n

tends to infinity. We write

u_{n} \to L

as

n \to \infty

.

Divergent

A sequence is divergent if it does not converge to a finite limit. This includes sequences that tend to infinity, tend to negative infinity, or oscillate without settling.

07Worked Example

A sequence is defined by $u_{n + 1} = 2 u_{n} + 3$ , where $u_{1} = 1$ .

(a) Find

u_{2}

,

u_{3}

,

u_{4}

to 3 d.p. (b) Given the sequence converges to

L

, find

L

.

(a)

u_{2} = 2 (1) + 3 = 5 = 2.236

,

u_{3} = 2 (2.236) + 3 = 2.733

,

u_{4} = 2 (2.733) + 3 = 2.910

(b)

At the limit:

L = 2 L + 3

. Square both sides:

L^{2} = 2 L + 3

.

•

Rearrange:

L^{2} - 2 L - 3 = 0

, so

(L - 3) (L + 1) = 0

, giving

L = 3

or

L = - 1

.

•

Since all terms are positive,

L = - 1

is rejected.

Answer(a) 2.236, 2.733, 2.910 (b)

L = 3

Practice

Question 1: A sequence is defined by

u_{n} = \frac{n ^{2} + 1}{n}

for

n \geq 1

.

(a) Find the first four terms of the sequence.

(b) Find the value of

n

for which

u_{n} = 10.1

.

(a)

u_{1} = 2

,

u_{2} = 2.5

,

u_{3} = 3. \overline{3}

,

u_{4} = 4.25

(b) Solve

\frac{n ^{2} + 1}{n} = 10.1

. Multiply:

n^{2} + 1 = 10.1 n

, so

10 n^{2} - 101 n + 10 = 0

.

Factorising:

(10 n - 1) (n - 10) = 0

, giving

n = 0.1

or

n = 10

. Since

n

is a positive integer: $n = 10$

Practice

Question 2: A sequence is defined by

u_{n + 1} = u_{n}^{2} - 4 u_{n} + 5

, where

u_{1} = 3

.

(a) Find

u_{2}

,

u_{3}

,

u_{4}

, and

u_{5}

.

(b) Describe the behaviour of this sequence.

(a)

u_{2} = 9 - 12 + 5 = 2

,

u_{3} = 4 - 8 + 5 = 1

,

u_{4} = 1 - 4 + 5 = 2

,

u_{5} = 4 - 8 + 5 = 1

(b) After

u_{1} = 3

, the sequence becomes 2, 1, 2, 1, 2, 1, ...

AnswerThe sequence is periodic with period 2 (after the first term).

Practice

Question 3: A sequence is defined by

u_{n + 1} = \frac{6}{u _{n}}

, where

u_{1} = 2

.

(a) Show that

u_{3} = 2

.

(b) Hence find the period of this sequence.

(c) Find the value of

r = 1 \sum 50 u_{r}

.

(a)

u_{1} = 2

,

u_{2} = \frac{6}{2} = 3

,

u_{3} = \frac{6}{3} = 2

✔

(b) Since

u_{3} = u_{1}

, the sequence repeats: 2, 3, 2, 3, ... Period = 2

(c) Odd terms (25 of them) = 2, Even terms (25 of them) = 3. Sum =

25 \times 2 + 25 \times 3 = 50 + 75 = 125

Summary

Key Concepts

ConceptDescription

Position-to-term rule

u_{n} = f (n)

- gives any term directly from its position

Recurrence relation

u_{n + 1} = f (u_{n})

with

u_{1} = a

- gives each term from the previous

Increasing

u_{n + 1} > u_{n}

for all

n

Decreasing

u_{n + 1} < u_{n}

for all

n

Periodic (order k)

u_{n + k} = u_{n}

for all

n

Convergent to limit LSolve

L = f (L)

Key Exam Skills

Generate terms from nth term formula or recurrence relation

Find the nth term formula for a linear sequence

Prove a sequence is increasing or decreasing

Identify periodic sequences and find the period

Find the limit of a convergent sequence