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IBMathsDPAnalysis & Approaches (AA)SLRevision Notes1. Number & AlgebraBinomial TheoremBinomial Theorem

Binomial Theorem (DP IB Analysis & Approaches (AA)): Revision Note

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Written by: Amber

Reviewed by: Mark Curtis

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Binomial Theorem

What is the binomial theorem?

  • The binomial theorem gives you the expansion of open parentheses a plus b close parentheses to the power of n for different positive integer powers of n:

table row cell open parentheses a plus b close parentheses to the power of n end cell equals cell a to the power of n plus scriptbase straight C subscript 1 end scriptbase presubscript blank presuperscript n a to the power of n minus 1 end exponent b plus... plus scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n a to the power of n minus r end exponent b to the power of r plus... plus b to the power of n end cell row cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n end cell equals cell fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction end cell end table

  • where C presubscript blank presuperscript n subscript r is the binomial coefficient

    • and the factorial symbol factorial means, for example:

      • 4 factorial equals 4 cross times 3 cross times 2 cross times 1 equals 24

Examiner Tips and Tricks

The binomial theorem and binomial coefficient formula are given in the formula booklet.

Examiner Tips and Tricks

You can find the values of C presubscript blank presuperscript n subscript r on your GDC.

How do I use the binomial theorem?

  • An example of using the binomial theorem is to expand open parentheses a plus b close parentheses to the power of 5

    • Substitute in n equals 5

    • open parentheses a plus b close parentheses to the power of 5 equals a to the power of 5 plus C presubscript blank presuperscript 5 subscript 1 a to the power of 4 b plus C presubscript blank presuperscript 5 subscript 2 a cubed b squared plus C presubscript blank presuperscript 5 subscript 3 a squared b cubed plus C presubscript blank presuperscript 5 subscript 4 a b to the power of 4 plus b to the power of 5

    • where

      • C presubscript blank presuperscript 5 subscript 1 equals fraction numerator 5 factorial over denominator 1 factorial open parentheses 5 minus 1 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 4 factorial end fraction equals fraction numerator 5 cross times up diagonal strike 4 cross times up diagonal strike 3 cross times up diagonal strike 2 cross times up diagonal strike 1 over denominator up diagonal strike 4 cross times up diagonal strike 3 cross times up diagonal strike 2 cross times up diagonal strike 1 end fraction equals 5

      • C presubscript blank presuperscript 5 subscript 2 equals fraction numerator 5 factorial over denominator 2 factorial open parentheses 5 minus 2 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 2 factorial 3 factorial end fraction equals... equals 10

      • C presubscript blank presuperscript 5 subscript 3 equals fraction numerator 5 factorial over denominator 3 factorial open parentheses 5 minus 3 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 3 factorial 2 factorial end fraction equals... equals 10

      • C presubscript blank presuperscript 5 subscript 4 equals fraction numerator 5 factorial over denominator 4 factorial open parentheses 5 minus 4 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 4 factorial 1 factorial end fraction equals... equals 5

    • giving

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

  • The total of the powers of each term must equal the power of the binomial, n

    • e.g. the total of the power of 10 a cubed b squared is 3 plus 2 equals 5

  • The binomial theorem saves you from having to expand by hand

    • open parentheses a plus b close parentheses to the power of 5 equals open parentheses a plus b close parentheses open parentheses a plus b close parentheses open parentheses a plus b close parentheses open parentheses a plus b close parentheses open parentheses a plus b close parentheses

Examiner Tips and Tricks

You only need to find half of the C presubscript blank presuperscript n subscript r coefficients, as the other half can be found by symmetry, e.g. 5 comma space 10 comma space 10 comma space 5

How do I expand harder binomials?

  • To expand harder binomials like open parentheses 1 plus 2 x close parentheses to the power of 5, put brackets around the terms

    • then put the powers outside the brackets

      • e.g. open parentheses 2 x close parentheses cubed

    • and use index laws

      • e.g. open parentheses 2 x close parentheses cubed equals 8 x cubed

  • To expand negative terms like open parentheses 1 minus 3 y close parentheses to the power of n, use brackets

    • e.g. open parentheses negative 3 y close parentheses squared equals 9 y squared

How do I find a particular term in a binomial expansion?

  • To find a particular term

    • either expand the whole binomial

    • or use the general term formula table row blank blank cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n a to the power of n minus r end exponent b to the power of r end cell end table

      • and form an equation to find r

Examiner Tips and Tricks

The formula for a general term in a binomial expansion is shown within the binomial theorem formula itself.

  • For example, find the term in x to the power of 14 in the expansion of open parentheses 2 x plus x cubed close parentheses to the power of 8

    • first substitute n equals 8, a equals open parentheses 2 x close parentheses and b equals open parentheses x cubed close parentheses into table row blank blank cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n a to the power of n minus r end exponent b to the power of r end cell end table

      • table row blank blank cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript 8 open parentheses 2 x close parentheses to the power of 8 minus r end exponent open parentheses x cubed close parentheses to the power of r end cell end table

    • then just look at the power of x only

      • x to the power of 8 minus r end exponent cross times x to the power of 3 r end exponent equals x to the power of 8 minus r plus 3 r end exponent equals x to the power of 8 plus 2 r end exponent

    • For this to equal x to the power of 14, you need 8 plus 2 r equals 14

      • which solves to give r equals 3

    • Substitute r equals 3 back into table row blank blank cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript 8 open parentheses 2 x close parentheses to the power of 8 minus r end exponent open parentheses x cubed close parentheses to the power of r end cell end table to get the whole term

      • table row cell scriptbase straight C subscript 3 end scriptbase presubscript blank presuperscript 8 open parentheses 2 x close parentheses to the power of 5 open parentheses x cubed close parentheses cubed end cell equals cell scriptbase straight C subscript 3 end scriptbase presubscript blank presuperscript 8 2 to the power of 5 x to the power of 5 x to the power of 9 equals scriptbase straight C subscript 3 end scriptbase presubscript blank presuperscript 8 32 x to the power of 14 end cell end table

      • and C presubscript blank presuperscript 8 subscript 3 equals fraction numerator 8 factorial over denominator 3 factorial open parentheses 8 minus 3 close parentheses factorial end fraction equals fraction numerator 8 factorial over denominator 3 factorial 5 factorial end fraction equals 56

      • giving table row blank blank cell 56 cross times 32 x to the power of 14 end cell end table

    • The answer is 1792 x to the power of 14

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.

1-5-1-binomial-theorem-we-solution-1

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    Author
    Reviewer
    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.

    Mark Curtis

    Reviewer: Mark Curtis

    Expertise: Maths Content Creator

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.