Other Sequences (Cambridge (CIE) IGCSE Maths): Revision Note

Exam code: 0580 & 0980

Amber

Written by: Amber

Reviewed by: Dan Finlay

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Types of Sequences

What are common types of sequences?

  • Sequences can follow any rule, but common sequences are

    • Linear

      • nth term = a n plus b

    • Quadratic

      • nth term = a n squared plus b n plus c

    • Cubic

      • nth term = a n cubed plus b n squared plus c n plus d

    • Exponential (Geometric)

      • nth term = a cross times r to the power of n

  • Sequences may also be formed using common numbers

    • Prime numbers

      • 2, 3, 5, 7, 11, ...

    • Triangular numbers

      • 1, 3, 6, 10, 15, ...

What is a cubic sequence?

  • A cubic sequence has an n th term formula that involves n3

  • The third differences are constant (the same)

    • These are the differences between the second differences

    • For example,   4, 25, 82, 193, 376, 649, ...
      1st Differences:  21, 57, 111, 183, 273, ...

      2nd Differences:   36,  54,  72,   90, ...
      3rd Differences:      18,    18,   18, ...

How do I find the nth term formula for a simple cubic sequence?

  • The sequence with the n th term formula of n3 is the cube numbers 

    • 1, 8, 27, 64, 125, ...

      • From 13, 23, 33, 43, ...

  • Finding the n th term formula of other cubic sequences comes from comparing them to the cube numbers, n3

    • 2, 9, 28, 65, 126, ... has the formula n3 + 1

      • Each term is one more than the cube numbers

    • 2, 16, 54, 128, 250, ...  has the formula 2n3

      • Each term is double a cube number

    • 8, 27, 64, 125, ... has the formula (n+1)3

      • They are the cube numbers starting from 23

  • You can also use third differences to help find the n th term a n cubed plus b n squared plus c n plus d

    • The value of a  is 1 over 6 of the third difference

      • e.g. the third difference of 8, 23, 62, 137, 260, ... is 12

    • You can then subtract an3 from the sequence and find the nth term of the differences

      • e.g. (8, 23, 62, 137, 260, ...) - (2, 16, 54, 128, 250, ... ) is (6, 7, 8, 9, 10, ...) which has nth term n + 5

      • So the nth term is 2n3 + n + 5

What is an exponential (geometric) sequence? 

  • An exponential (geometric) sequence is one where you multiply each term by the same number to get the next term

    • E.g. 3, 6, 12, 24, 48, ... is exponential because:

      • terms are multiplied by 2 each time

      • 2 is called the common ratio (or constant multiplier)

      • You can find this by dividing any term by the term immediately before, 6 ÷ 3 or 12 ÷ 6 or 24 ÷ 12 etc

How do I find the nth term formula for a simple exponential sequence?

  • The sequence with the n th term formula of rn is the powers of r

    • e.g. 2, 4, 8, 16, 32, ... has formula 2n

  • Finding the n th term formula of other exponential sequences comes from comparing them to the powers of the multiplier, rn

    • 6, 12, 24, 48, 96, ... has the formula 3× 2n

      • Each term is three times more than a power of 2

    • 4, 8, 16, 32, 64, ... has the formula 2n+1

      • Each term is a power of 2 starting at 22

  • If the common ratio satisfies

    • r greater than 1 then the sequence increases

    • 0 less than r less than 1 then the sequence decreases

      • E.g. r equals 1 half

How can sequences be made harder?

  • You may be given a fraction with two different sequences on the top and bottom

    • E.g. 3 over 1 comma 5 over 8 comma 7 over 27 comma 9 over 64 comma...

      • The numerators are the linear sequence 2 n plus 1

      • The denominators are the cube numbers, n cubed

      • So the n th term formula is fraction numerator 2 n plus 1 over denominator n cubed end fraction

  • You may be asked to find combinations of two different sequences

    • E.g. if sequence U is the prime numbers and sequence V has the n th term formula 4 n squared, find the sequence U + V

      • U = 2, 3, 5, 7, ... and V = 4, 16, 36, 64, ...

      • U + V = (2 + 4), (3 + 16), (5 + 36), (7 + 64), ... = 6, 19, 41, 71, ...

  • Other problems involving setting up and solving equations

    • This may lead to a pair of simultaneous equations

Worked Example

(a) Find the formula for the nth term of the sequence 5, 19, 57, 131, 253, 435, ...

See if the sequence is linear, quadratic or cubic by finding the first, second or third differences

The first differences are

14, 38, 74, 122, 182

These are not constant, so find the second differences

24, 36, 48, 60

These are not constant, so find the third differences

12, 12, 12

Compare 5, 19, 57, 131, 253, 435, ... to the cube numbers 1, 8, 27, 64, 125, ...

Double the cube numbers

2, 16, 54, 128, 250, ...

Add 3

5, 19, 57, 131, 253, ...

The cube numbers (with nth term formula n cubed) are doubled then 3 is added

The nth term formula is 2 n cubed plus 3

(b) Find the formula for the nth term of the sequence 4, 20, 100, 500, 2500, ...

Seeing if the first, second or third differences are constant does not work

The numbers are increasing very fast, suggesting it could be exponential
Check to see if each term is multiplied by the same number each time

4 cross times 5 equals 20 comma space 20 cross times 5 equals 100 comma space 100 cross times 5 equals 500 comma space 500 cross times 5 equals 2500 comma space...

Each term is multiplied by 5 (the "common ratio") to get the next, so it is exponential
Compare the sequence to the sequence 5n

Sequence: 4, 20, 100, 500, 2500, ...
5n: 5, 25, 125, 625, 3125, ...

You need to divide the powers of 5 by 5 and multiply by 4 to get the sequence

The nth term formula is 4 over 5 cross times 5 to the power of n or 4 cross times 5 to the power of n minus 1 end exponent

(c) Write down the formula for the nth term of the sequence

5 over 4 comma 19 over 20 comma 57 over 100 comma 131 over 500 comma 253 over 2500 comma space...

This sequence is a fraction formed by dividing the sequence in part (a) by the sequence in part (b)
Divide their nth term formulas

The nth term formula is fraction numerator 2 n cubed plus 3 over denominator 4 cross times 5 to the power of n minus 1 end exponent end fraction

Worked Example

The first three terms of an exponential sequence are shown below

x minus 1 space space space space space space space space space space 2 x space space space space space space space space space space x squared

By forming and solving an equation, find the common ratio, r, given that x not equal to 0.

As this is an exponential sequence, each term is multiplied by the common ratio, r, to get the next term

Consider the first two terms

table row cell open parentheses x minus 1 close parentheses cross times r end cell equals cell 2 x end cell row r equals cell fraction numerator 2 x over denominator x minus 1 end fraction end cell end table

Considering the next two terms

table row cell 2 x cross times r end cell equals cell x squared end cell row r equals cell fraction numerator x squared over denominator 2 x end fraction end cell end table

This is a pair of simultaneous equations
They can be solved by substituting one into the other (replacing r)

fraction numerator 2 x over denominator x minus 1 end fraction equals fraction numerator x squared over denominator 2 x end fraction

Multiply both sides by 2 x

fraction numerator 4 x squared over denominator x minus 1 end fraction equals x squared

Multiply both sides by x minus 1, and expand

4 x squared equals x squared open parentheses x minus 1 close parentheses
4 x squared equals x cubed minus x squared

Subtract 4 x squared from both sides and factorise

0 equals x cubed minus 5 x squared
0 equals x squared open parentheses x minus 5 close parentheses

Solve

table row cell x squared end cell equals 0 end table so table row x equals 0 end table
or
table row cell open parentheses x minus 5 close parentheses end cell equals 0 end table so table row x equals 5 end table

You are told x not equal to 0 so x equals 5
Substitute this into the original sequence, x minus 1 comma space 2 x comma space x squared

open parentheses 5 minus 1 close parentheses comma space 2 open parentheses 5 close parentheses comma space 5 squared
equals 4 comma space 10 comma space 25

Find the common ratio r (for example, by dividing a term by its previous term)

10 over 4 equals 2.5

r = 2.5

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Amber

Author: Amber

Expertise: Maths Content Creator

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.