Revision NotesPast Papers
International ASFurther MathsEdexcelFurther Pure 1Revision NotesSeriesSeriesCombinations of Series

Combinations of Series (Edexcel International AS Further Maths) : Revision Note

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on

Combinations of Series

How do I find combinations of series?

  • Use the three formulae:

    • sum from r equals 1 to n of r equals 1 half n open parentheses n plus 1 close parentheses

    • sum from r equals 1 to n of r squared equals 1 over 6 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses

    • sum from r equals 1 to n of r cubed equals 1 fourth n squared open parentheses n plus 1 close parentheses squared

  • Coefficients of r, r squared and r cubed can come out of the sum

    • sum from r equals 1 to n of open parentheses a r cubed plus b r squared plus c r plus d close parentheses equals a sum from r equals 1 to n of r cubed space plus b sum from r equals 1 to n of r squared plus c sum from r equals 1 to n of r plus d n

      • Constants are multiplied by n

    • You may have to expand expressions in r

      • sum from r equals 1 to n of open parentheses r plus 3 close parentheses open parentheses r plus 4 close parentheses equals sum from r equals 1 to n of open parentheses r squared plus 7 r plus 12 close parentheses

  • It is often possible to factorise final answers, in particular taking out:

    • fractional coefficients

    • n

    • open parentheses n plus 1 close parentheses

  • You may be required to write series as the difference of two sums

    • sum from r equals 5 to 10 of r equals sum from r equals 1 to 10 of r minus sum from r equals 1 to 4 of r

Examiner Tips and Tricks

Exam questions often refer to these three formulae as the "standard summation formulae".

Worked Example

(a) Using standard summation formulae, show that

sum from r equals 1 to n of open parentheses r minus 1 close parentheses r open parentheses r plus 1 close parentheses equals 1 fourth open parentheses n minus 1 close parentheses n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses

Expand and simplify the left-hand side

table row cell open parentheses r minus 1 close parentheses r open parentheses r plus 1 close parentheses end cell equals cell r open parentheses r squared minus 1 close parentheses end cell row blank equals cell r cubed minus r end cell end table

Substitute in the formulae sum from r equals 1 to n of r cubed equals 1 fourth n squared open parentheses n plus 1 close parentheses squared and sum from r equals 1 to n of r equals 1 half n open parentheses n plus 1 close parentheses

table row cell sum from r equals 1 to n of open parentheses r cubed minus r close parentheses end cell equals cell sum from r equals 1 to n of r cubed minus sum from r equals 1 to n of r end cell row blank equals cell 1 fourth n squared open parentheses n plus 1 close parentheses squared minus 1 half n open parentheses n plus 1 close parentheses end cell end table

Factorise out the quarter, n and open parentheses n plus 1 close parentheses then simplify

table row cell sum from r equals 1 to n of open parentheses r cubed minus r close parentheses end cell equals cell 1 fourth n open parentheses n plus 1 close parentheses open square brackets n open parentheses n plus 1 close parentheses minus 2 close square brackets end cell row blank equals cell 1 fourth n open parentheses n plus 1 close parentheses open parentheses n squared plus n minus 2 close parentheses end cell end table

Factorise the quadratic on the right-hand side

table row cell sum from r equals 1 to n of open parentheses r cubed minus r close parentheses end cell equals cell equals 1 fourth n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses open parentheses n minus 1 close parentheses end cell row blank blank blank end table

Rearrange into the form asked for by the question

table row cell sum from r equals 1 to n of open parentheses r minus 1 close parentheses r open parentheses r plus 1 close parentheses end cell equals cell 1 fourth open parentheses n minus 1 close parentheses n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses end cell end table

(b) Hence find the value of

0 cross times 1 cross times 2 space plus space 1 cross times 2 cross times 3 space plus space 2 cross times 3 cross times 4 space plus... plus space 20 cross times 21 cross times 22

It helps to write out the first few terms in the series table row blank blank cell sum from r equals 1 to n of open parentheses r minus 1 close parentheses r open parentheses r plus 1 close parentheses end cell end table

table row blank blank cell open parentheses 1 minus 1 close parentheses cross times 1 cross times open parentheses 1 plus 1 close parentheses space space plus space space open parentheses 2 minus 1 close parentheses cross times 2 cross times open parentheses 2 plus 1 close parentheses space space plus space space open parentheses 3 minus 1 close parentheses cross times 3 cross times open parentheses 3 plus 1 close parentheses space space plus... end cell row blank equals cell 0 cross times 1 cross times 2 space space plus space space 1 cross times 2 cross times 3 space space plus space space 2 cross times 3 cross times 4 space space plus... end cell end table

The question stops at 20 x 21 x 22
This is actually the 21st term (not the 20th term)
To find the sum of the first 21 terms, substitute n equals 21 into part (a)

table row cell sum from r equals 1 to 21 of open parentheses r minus 1 close parentheses r open parentheses r plus 1 close parentheses end cell equals cell 1 fourth open parentheses 21 minus 1 close parentheses cross times 21 cross times open parentheses 21 plus 1 close parentheses open parentheses 21 plus 2 close parentheses end cell row blank equals cell 1 fourth cross times 20 cross times 21 cross times 22 cross times 23 end cell end table

53130

(c) If sum from r equals 1 to n of open parentheses r minus 1 close parentheses r open parentheses r plus 1 close parentheses equals 10 space sum from r equals 1 to n of r squared, find n.

Substitute the answer from part (a) into the left-hand side
Substitute the standard formula sum from r equals 1 to n of r squared equals 1 over 6 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses into the right-hand side

table row cell 1 fourth open parentheses n minus 1 close parentheses n open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses end cell equals cell 10 cross times 1 over 6 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses end cell end table

Cancel the terms n and open parentheses n plus 1 close parentheses from both sides
It also helps to multiply both sides by 12 to remove fractions

3 table row blank blank cell open parentheses n minus 1 close parentheses open parentheses n plus 2 close parentheses end cell end table table row blank equals blank end table table row blank blank 20 end table table row blank blank cell open parentheses 2 n plus 1 close parentheses end cell end table

Expand, then form and solve a quadratic in

table row cell 3 open parentheses n squared plus n minus 2 close parentheses end cell equals cell 40 n plus 20 end cell row cell 3 n squared minus 37 n minus 26 end cell equals 0 row cell open parentheses 3 n plus 2 close parentheses open parentheses n minus 13 close parentheses end cell equals 0 row n equals cell negative 2 over 3 space space or space space n equals 13 end cell end table

n must be a positive integer (as it is used in sigma notation, sum from r equals 1 to n of)

n equals 13

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:Standard SeriesNext:Proof by Induction
    Author
    Reviewer
    Mark Curtis

    Author: 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.

    Dan Finlay

    Reviewer: 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.