IGCSEFurther MathsEdexcelRevision NotesSeriesBinomial SeriesBinomial Expansion

Binomial Expansion (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Binomial Expansion

What is the Binomial Expansion?

  • The binomial expansion gives a method for expanding a two-term expression in a bracket raised to a power

    • For example  open parentheses a plus b close parentheses to the power of n

    • You may also see it referred to as the binomial theorem

    • In this note n will be a positive integer

      • See the 'General Binomial Expansion' revision note for the general case

  • To expand a bracket with a two-term expression in it:

    • Determine what a and b are for your example

    • Then use the formula for the binomial expansion

      • open parentheses a plus b close parentheses to the power of n space equals space a to the power of n space plus space open parentheses table row n row 1 end table close parentheses space a to the power of n minus 1 space end exponent b space plus space horizontal ellipsis space plus space open parentheses table row n row r end table close parentheses space a to the power of n minus r end exponent b to the power of r space plus space horizontal ellipsis space plus space b to the power of n

    • open parentheses table row n row r end table close parentheses  in the formula is known as the binomial coefficient

      • open parentheses table row n row r end table close parentheses space equals space fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction space equals space 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

      • n factorial ("n factorial") is defined by n factorial equals 1 cross times 2 cross times 3 cross times... cross times n

      • You may also see open parentheses table row n row r end table close parentheses  written as scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n

      • Your calculator should be able to calculate open parentheses table row n row r end table close parentheses for you

      • Or you can use Pascal's triangle (see the next section)

  • To get all the terms

    • Start with r equals 0

    • Then use  r equals 1, r equals 2,...  until you get up to r equals n

    • So there will always be bold italic n bold plus bold 1 terms in the full expansion

  • This version of the binomial expansion formula is not on the exam formula sheet

    • But it is a special case of the Binomial Series formula for open parentheses 1 plus x close parentheses to the power of n which is on the formula sheet

    • See the 'General Binomial Expansion' revision note

  • When expanding something like open parentheses p plus q x close parentheses to the power of n you may only be asked to find the first few terms of an expansion

    • Check whether the question wants ascending or descending powers of x

      • For ascending powers start with the constant term, p to the power of n

      • For descending powers start with the term with x, open parentheses q x close parentheses to the power of n

      • Choosing a and b appropriately will make it easier to follow the formula above

  • If you are not writing the full expansion you can either

    • show that the series continues by putting an ellipsis (…) after your final term

    • or show that the terms you have found are an approximation of the full series by using the 'approximately equals' sign (≈)

Finding binomial coefficients using Pascal's triangle

  • Pascal’s triangle is a way of arranging (and finding!) the binomial coefficients

    • The first row has just the number 1

    • Each row begins and ends with a 1

    • Starting in the third row

      • Each other terms is the sum of the two terms immediately above it

Pascal's Triangle

  • Pascal’s triangle is an alternative way of finding the binomial coefficients open parentheses table row n row r end table close parentheses  (also written scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n )

    • It can be useful for finding the values of the coefficients without a calculator

      • Most useful for smaller values of n

      • For larger values of n it is slow and prone to arithmetic errors

  • Taking the first row as corresponding to n equals 0,

    • each row gives the binomial coefficient values for the corresponding value of n

    • within a row the values run from r equals 0 to r equals n

    • e.g. from the 6th row of the table (n equals 5):

      • open parentheses a plus b close parentheses to the power of 5 equals a to the power of 5 plus bold 5 a to the power of 4 b plus bold 10 a cubed b squared plus bold 10 a squared b cubed plus bold 5 a b to the power of 4 plus b to the power of 5

How do I find the coefficient of a single term?

  • You may just be asked to find the coefficient of a single term, rather than the whole expansion

  • Use the formula for the general term

    • open parentheses table row n row r end table close parentheses space a to the power of n minus r end exponent space b to the power of r

  • To find a particular power of bold italic x term in an expansion

    • Choose which value of bold italic r  you will need to use in the formula

    • The laws of indices can help you decide which value of r  to use:

      • For left parenthesis p plus q x right parenthesis to the power of n,  to find the coefficient of x squared let  a equals p comma space space b equals q x and use r equals 2

      • For left parenthesis p plus q x squared right parenthesis to the power of n,  to find the coefficient of x squared let  a equals p comma space b equals q x squared and use r equals 1

      • For something like open parentheses p x plus q over x close parentheses to the power of n, you need to consider how the powers will cancel each other

        • E.g. for open parentheses p x plus q over x close parentheses to the power of 6 , to find the coefficient of x squared let a equals p x space comma b equals fraction numerator q over denominator x space space end fraction and user r equals 2

        • Because then x to the power of n minus r end exponent space open parentheses 1 over x close parentheses to the power of r equals x to the power of 6 minus 2 space end exponent open parentheses 1 over x close parentheses to the power of 2 space end exponent equals x to the power of 4 open parentheses 1 over x squared close parentheses equals x squared space space

      • There are a lot of variations, so practice is better than trying to memorise formulae for r!

  • If you know the coefficient of a particular term, you can use it to find an unknown in the brackets

    • Use the laws of indices to choose the correct term

    • Then use the general term formula to form and solve an equation

Examiner Tips and Tricks

  • Binomial expansion questions can get messy

    • Use separate lines to keep your working clear

    • And always put terms in brackets

Worked Example

Using the binomial expansion, find the complete expansion of  open parentheses x plus y close parentheses to the power of 4.

Use the formula with a equals x, b equals y and n equals 4
r will run from 0 to 4, so there will be 5 terms

open parentheses x plus y close parentheses to the power of 4 equals open parentheses table row 4 row 0 end table close parentheses x to the power of 4 plus open parentheses table row 4 row 1 end table close parentheses x cubed y plus open parentheses table row 4 row 2 end table close parentheses x squared y squared plus open parentheses table row 4 row 3 end table close parentheses x y cubed plus open parentheses table row 4 row 4 end table close parentheses y to the power of 4

Now just work out the values of the binomial coefficients
You can use the formula, your calculator or Pascal's triangle

Note that  open parentheses table row 4 row 0 end table close parentheses equals open parentheses table row 4 row 4 end table close parentheses equals 1
That's why we usually don't bother writing the binomial coefficients for the first and last terms of an expansion!

stretchy left parenthesis x plus y stretchy right parenthesis to the power of bold 4 bold equals bold italic x to the power of bold 4 bold plus bold 4 bold italic x to the power of bold 3 bold italic y bold plus bold 6 bold italic x to the power of bold 2 bold italic y to the power of bold 2 bold plus bold 4 bold italic x bold italic y to the power of bold 3 bold plus bold italic y to the power of bold 4 

Worked Example

Find the first three terms, in ascending powers of x, in the expansion of left parenthesis 3 minus 2 x right parenthesis to the power of 5.

For ascending powers of x we want to start with the constant term

So we want to use the formula with a equals 3, b equals negative 2 x, and n equals 5

For the first three terms (constant term, x term  and x squared term) we want r from 0 to 2

Substitute those values into the formula

open parentheses 3 minus 2 x close parentheses to the power of 5 equals open parentheses 3 close parentheses to the power of 5 plus open parentheses table row 5 row 1 end table close parentheses open parentheses 3 close parentheses to the power of 4 open parentheses negative 2 x close parentheses plus open parentheses table row 5 row 2 end table close parentheses open parentheses 3 close parentheses cubed open parentheses negative 2 x close parentheses squared plus...

Find the value of the binomial coefficients and bring the powers inside the brackets
Be careful with the minus signs!

 open parentheses 3 minus 2 x close parentheses to the power of 5 equals 243 plus open parentheses 5 close parentheses open parentheses 81 close parentheses open parentheses negative 2 x close parentheses plus open parentheses 10 close parentheses open parentheses 27 close parentheses open parentheses 4 x squared close parentheses plus...

Expand the remaining brackets and write down the final answer

stretchy left parenthesis 3 minus 2 x stretchy right parenthesis to the power of bold 5 bold equals bold 243 bold minus bold 810 bold italic x bold plus bold 1080 bold italic x to the power of bold 2 bold plus bold. bold. bold.

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    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.