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Other Sequences (Cambridge (CIE) IGCSE International Maths)

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Amber

Written by: Amber

Reviewed by: Dan Finlay

Cambridge (CIE) International Maths: ExtendedCambridge (CIE) Maths: Extended

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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

    • Cubic

      • nth term = a n cubed plus b

    • Exponential (Geometric)

      • nth term = a cross times r to the power of n minus 1 end exponent

  • 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 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

  • You can also use second differences to help find the n th term

    • For the simple cubic sequence, n th term = an3 + b

      • where a  is 1 over 6 of the third difference

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

  • The n th term formula is bold italic a bold cross times bold italic r to the power of bold italic n bold minus bold 1 end exponent

    • where a is the first term

    • r is the common ratio

    • n is the position number of the term

      • E.g. 3, 6, 12, 24, 48, ... has a = 3 and r = 2

      • so the n th term = 3 cross times 2 to the power of n minus 1 end exponent

    • Remember that the power is n minus 1, not n

  • 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

These are constant so the sequence is cubic, with nth term formula a n cubed plus b

Method 1
Use the rule that a is 1 over 6 of the third difference

a equals 1 over 6 cross times 12 equals 2

This means the nth term formula is 2 n cubed plus b
There are different ways to find b
E.g. substitute in n equals 1 to find the first term from the formula, then make it equal to 5 (the first term in the question)

table row cell 2 cross times 1 cubed plus b end cell equals 5 row cell 2 plus b end cell equals 5 row b equals 3 end table

The nth term formula is 2 n cubed plus 3

Method 2

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
The nth term formula is a cross times r to the power of n minus 1 end exponent where a is the first term and r is the common ratio

a equals 4 and r equals 5

The nth term formula is 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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    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths

    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

    Author: Dan Finlay

    Expertise: Maths 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.