Revision NotesPast Papers
IGCSEFurther MathsEdexcelRevision NotesSeriesBinomial SeriesGeneral Binomial Expansion

General Binomial Expansion (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

General Binomial Expansion

What is the general binomial expansion?

  • The general binomial expansion lets us write open parentheses 1 plus x close parentheses to the power of n as a binomial series

    • It is valid for any n element of straight rational numbers

      • straight rational numbers is the set of all rational numbers

      • So n can be negative or a fraction

    • If n is a positive integer

      • Then the series has a finite number of terms

      • For this case see the 'Binomial Expansion' revision note

  • If n element of straight rational numbers is not a positive integer

    • Then the series has an infinite number of terms

    • I.e. 'it goes on forever'

    • An exam question will only ask for the first few terms of the expansion

  • A general binomial expansion is found using the binomial series formula

    • open parentheses 1 plus x close parentheses to the power of n equals 1 plus n x plus fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 factorial end fraction x squared plus... plus fraction numerator n open parentheses n minus 1 close parentheses... open parentheses n minus r plus 1 close parentheses over denominator r factorial end fraction x to the power of r plus... space space space space for space open vertical bar x close vertical bar less than 1 comma space n element of straight rational numbers

    • This formula is on the exam formula sheet

      • So you don't need to remember it

      • But you do need to know how to use it

  • The expansion is only valid for open vertical bar x close vertical bar less than 1

    • This means negative 1 less than x less than 1

    • This is known as the interval of convergence

    • For values of x inside the interval of convergence

      • the (infinite) expansion on the right-hand side of the formula

      • is exactly equal to the function on the left-hand side

How do I use the binomial series formula?

  • Usually you will be asked to expand something in the form open parentheses p plus q x close parentheses to the power of n 

  • But the formula only works if the constant term is a 1

    • So start by pulling out a factor of p

      • open parentheses p plus q x close parentheses to the power of n equals open parentheses p open parentheses 1 plus q over p x close parentheses close parentheses to the power of n equals p to the power of n open parentheses 1 plus q over p x close parentheses to the power of n

    • Then expand  open parentheses 1 plus q over p x close parentheses to the power of n

      • Substitute q over p x everywhere that x is in the formula

      • The interval of convergence becomes   open vertical bar p over q x close vertical bar less than 1

    • Don't forget to multiply everything by p to the power of n again at the end! 

  • Be sure you can recognise a negative or fractional power

    • The expression may be in the denominator of a fraction

      • 1 over open parentheses p plus q x close parentheses to the power of k equals open parentheses p plus q x close parentheses to the power of negative k end exponent

    • Or inside a square root

      • square root of p plus q x end root equals open parentheses p plus q x close parentheses to the power of 1 half end exponent

    • Or be written as a more complex root

      • m-th root of open parentheses p plus q x close parentheses to the power of k end root equals open parentheses p plus q x close parentheses to the power of k over m end exponent

Examiner Tips and Tricks

  • Remember the formula is on the formula sheet 

  • Be especially careful with

    • negative numbers

    • subtracting 1 from fractions

  • Use brackets to separate things out

  • Don't rush!

Worked Example

(a) Expand  fraction numerator 1 over denominator square root of 9 blank minus blank 3 x end root end fraction  in ascending powers of x up to and including the term in x cubed and simplifying each term as far as possible.

Start by rewriting using laws of indices

fraction numerator 1 over denominator square root of 9 minus 3 x end root end fraction equals open parentheses 9 minus 3 x close parentheses to the power of negative 1 half end exponent

Now pull out a factor to make the constant term inside the brackets a 1

table row cell open parentheses 9 minus 3 x close parentheses to the power of negative 1 half end exponent end cell equals cell open parentheses 9 to the power of negative 1 half end exponent close parentheses open parentheses 1 minus 3 over 9 x close parentheses to the power of negative 1 half end exponent end cell row blank equals cell 1 third open parentheses 1 minus x over 3 close parentheses to the power of negative 1 half end exponent end cell end table

Now use the binomial series formula to expand open parentheses 1 minus x over 3 close parentheses to the power of negative 1 half end exponent

Use n equals 1 half and substitute  negative x over 3 everywhere that x appears in the formula

table row cell open parentheses 1 minus x over 3 close parentheses to the power of negative 1 half end exponent end cell equals cell 1 plus open parentheses negative 1 half close parentheses open parentheses negative x over 3 close parentheses plus fraction numerator open parentheses negative 1 half close parentheses open parentheses negative 1 half minus 1 close parentheses over denominator 2 factorial end fraction open parentheses negative x over 3 close parentheses squared plus fraction numerator open parentheses negative 1 half close parentheses open parentheses negative 1 half minus 1 close parentheses open parentheses negative 1 half minus 2 close parentheses over denominator 3 factorial end fraction open parentheses negative x over 3 close parentheses cubed plus... end cell row blank equals cell 1 plus 1 over 6 x plus fraction numerator open parentheses negative 1 half close parentheses open parentheses negative 3 over 2 close parentheses over denominator 2 end fraction open parentheses x squared over 9 close parentheses plus fraction numerator open parentheses negative 1 half close parentheses open parentheses negative 3 over 2 close parentheses open parentheses negative 5 over 2 close parentheses over denominator 6 end fraction open parentheses negative x cubed over 27 close parentheses plus... end cell row blank equals cell 1 plus 1 over 6 x plus fraction numerator open parentheses 3 over 4 close parentheses over denominator 2 end fraction open parentheses x squared over 9 close parentheses plus fraction numerator open parentheses negative 15 over 8 close parentheses over denominator 6 end fraction open parentheses negative x cubed over 27 close parentheses plus... end cell row blank equals cell 1 plus 1 over 6 x plus open parentheses 3 over 8 close parentheses open parentheses x squared over 9 close parentheses plus open parentheses negative 5 over 16 close parentheses open parentheses negative x cubed over 27 close parentheses plus... end cell row blank equals cell 1 plus 1 over 6 x plus 1 over 24 x squared plus 5 over 432 x cubed plus... end cell end table

Now don't forget to multiply by 1 third (factorised out earlier) to get the final answer!

fraction numerator 1 over denominator square root of 9 minus 3 x end root end fraction equals 1 third open parentheses table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 plus 1 over 6 x plus 1 over 24 x squared plus 5 over 432 x cubed plus... end cell end table close parentheses

fraction numerator bold 1 over denominator square root of bold 9 bold minus bold 3 bold x end root end fraction bold equals bold 1 over bold 3 bold plus bold 1 over bold 18 bold italic x bold plus bold 1 over bold 72 bold italic x to the power of bold 2 bold plus bold 5 over bold 1296 bold italic x to the power of bold 3 bold plus bold. bold. bold.

(b) Find the interval of convergence for the expansion in part (a).

Remember that we used  negative x over 3 in place of x when we used the binomial series formula

We also need to substitute  negative x over 3 into the standard convergence interval  open vertical bar x close vertical bar less than 1

open vertical bar negative x over 3 close vertical bar less than 1 1 third open vertical bar x close vertical bar less than 1 minus 1 less than 1 third x less than 1

bold minus bold 3 bold less than bold italic x bold less than bold 3

Last updated:

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Did this page help you?

    Previous:Binomial ExpansionNext:Applications of Binomial Expansion
    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    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.