Method of Differences (Edexcel A Level Further Maths) : Revision Note

Jamie Wood

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Method of Differences

What is the Method of Differences?

  • The Method of Differences is a way of turning longer and more complicated sums into shorter and simpler ones

  • Sometimes when summing series, you will notice that many of the terms (or parts of the terms) simply “cancel out” or eliminate each other

    • This can turn a very long series summation, into a much simpler shorter one

  • In the case of sum from r equals 1 to n of left parenthesis f left parenthesis r right parenthesis minus f left parenthesis r plus 1 right parenthesis right parenthesis

    • f left parenthesis 1 right parenthesis minus f left parenthesis 2 right parenthesis

    • f left parenthesis 2 right parenthesis minus f left parenthesis 3 right parenthesis

    • f left parenthesis 3 right parenthesis minus f left parenthesis 4 right parenthesis

    • f left parenthesis 4 right parenthesis minus f left parenthesis 5 right parenthesis

    • (and so on, until…)

    • f left parenthesis n minus 1 right parenthesis minus f left parenthesis n right parenthesis

      • This is the penultimate term

    • f left parenthesis n right parenthesis minus f left parenthesis n plus 1 right parenthesis

      • This is the last term

  • You can see that when these are summed, most of the terms will cancel out

    • This leaves just f left parenthesis 1 right parenthesis minus f left parenthesis n plus 1 right parenthesis

    • We can say that sum from r equals 1 to n of left parenthesis f left parenthesis r right parenthesis minus f left parenthesis r plus 1 right parenthesis right parenthesis equals f left parenthesis 1 right parenthesis minus f left parenthesis n plus 1 right parenthesis

How can I use partial fractions along with the method of differences?

  • You will often need to use partial fractions to change the general term into a sum of two or three terms, rather than a single fraction

    • For example, fraction numerator 6 over denominator left parenthesis r plus 1 right parenthesis left parenthesis r plus 3 right parenthesis end fraction can be rewritten as fraction numerator 3 over denominator r plus 1 end fraction minus fraction numerator 3 over denominator r plus 3 end fraction 

  • This may lead to a more interesting pattern of cancellations than was seen for f left parenthesis r right parenthesis minus f left parenthesis r plus 1 right parenthesis

  • For example, sum from r equals 1 to n of fraction numerator 3 over denominator r plus 1 end fraction minus fraction numerator 3 over denominator r plus 3 end fraction can be written as sum from r equals 1 to n of f left parenthesis r right parenthesis minus f left parenthesis r plus 2 right parenthesis, where f left parenthesis r right parenthesis equals fraction numerator 3 over denominator r plus 1 end fraction, and the terms can then be listed as:   

    • f left parenthesis 1 right parenthesis minus f left parenthesis 3 right parenthesis

      • f left parenthesis 2 right parenthesis minus f left parenthesis 4 right parenthesis

      • f left parenthesis 3 right parenthesis minus f left parenthesis 5 right parenthesis

      • f left parenthesis 4 right parenthesis minus f left parenthesis 6 right parenthesis

      • f left parenthesis 5 right parenthesis minus f left parenthesis 7 right parenthesis

      • (and so on, until…)

      • f left parenthesis n minus 1 right parenthesis minus f left parenthesis n plus 1 right parenthesis

      • f left parenthesis n right parenthesis minus f left parenthesis n plus 2 right parenthesis

    • When these are summed, it will just leave f left parenthesis 1 right parenthesis plus f left parenthesis 2 right parenthesis minus f left parenthesis n plus 1 right parenthesis minus f left parenthesis n plus 2 right parenthesis

    • You can then evaluate this expression with f left parenthesis r right parenthesis equals fraction numerator 3 over denominator r plus 1 end fraction to get to your final answer

  • It is helpful to use f left parenthesis r right parenthesis notation to spot the pattern, rather than substituting r equals 1 comma space r equals 2 comma space... into the expression every time, especially with more complicated expressions

    • You need to consider carefully which term to make f left parenthesis r right parenthesis and then how the other terms in the expression relate to it

    • If this is difficult for a particular expression, it may be more straightforward to substitute r equals 1 comma space r equals 2 comma space... into each term in the series and spot any patterns that way 

    • This is essentially writing the series out in full until you spot which terms will cancel

    • In your working, however, you should still write out the last two or three terms in terms of n comma space n minus 1 comma space n minus 2, etc.

How can I use the method of differences for series with expressions containing more than two terms?

  • The general term of the series may have more than two terms, which can sometimes make spotting which terms will cancel more challenging

    • For example, sum from r equals 1 to n of 2 over r minus fraction numerator 3 over denominator r plus 1 end fraction plus fraction numerator 1 over denominator r plus 2 end fraction 

    • This can be written as 2 f left parenthesis r right parenthesis minus 3 f left parenthesis r plus 1 right parenthesis plus f left parenthesis r plus 2 right parenthesis where f left parenthesis r right parenthesis equals 1 over r

    • Writing out the first five terms and the last three terms we get:

      • 2 f left parenthesis 1 right parenthesis minus 3 f left parenthesis 2 right parenthesis bold plus bold italic f bold left parenthesis bold 3 bold right parenthesis

      • 2 f left parenthesis 2 right parenthesis bold minus bold 3 bold italic f bold left parenthesis bold 3 bold right parenthesis plus f left parenthesis 4 right parenthesis

      • bold 2 bold italic f bold left parenthesis bold 3 bold right parenthesis minus 3 f left parenthesis 4 right parenthesis plus f left parenthesis 5 right parenthesis

      • 2 f left parenthesis 4 right parenthesis minus 3 f left parenthesis 5 right parenthesis plus f left parenthesis 6 right parenthesis

      • 2 f left parenthesis 5 right parenthesis minus 3 f left parenthesis 6 right parenthesis plus f left parenthesis 7 right parenthesis

      • (and so on until…)

      • 2 f left parenthesis n minus 2 right parenthesis minus 3 f left parenthesis n minus 1 right parenthesis plus f left parenthesis n right parenthesis

      • 2 f left parenthesis n minus 1 right parenthesis minus 3 left parenthesis f n right parenthesis plus f left parenthesis n plus 1 right parenthesis

      • 2 f left parenthesis n right parenthesis minus 3 f left parenthesis n plus 1 right parenthesis plus f left parenthesis n plus 2 right parenthesis

    • In this case, look at the diagonals starting at the top right

      • We have f left parenthesis 3 right parenthesis comma space minus 3 f left parenthesis 3 right parenthesis and 2 f left parenthesis 3 right parenthesis which sum to 0

      • This pattern repeats for the other diagonals

    • We will eventually be left with only 2 f left parenthesis 1 right parenthesis minus 3 f left parenthesis 2 right parenthesis plus 2 f left parenthesis 2 right parenthesis plus f left parenthesis n plus 1 right parenthesis minus 3 f left parenthesis n plus 1 right parenthesis plus f left parenthesis n plus 2 right parenthesis equals 2 f left parenthesis 1 right parenthesis minus f left parenthesis 2 right parenthesis minus 2 f left parenthesis n plus 1 right parenthesis plus f left parenthesis n plus 2 right parenthesis

    • Evaluating this with f left parenthesis r right parenthesis equals 1 over r results in an answer of 3 over 2 minus fraction numerator 2 over denominator n plus 1 end fraction plus fraction numerator 1 over denominator n plus 2 end fraction 

What other uses are there for the method of differences?

  • Method of differences can also be used to prove the formulae for the sum of space r comma space r squared and r cubed

    • For example, this can be used to prove that the result for the sum of squares is indeed 1 over 6 n left parenthesis n plus 1 right parenthesis left parenthesis 2 n plus 1 right parenthesis

      • By expanding brackets it can be shown that left parenthesis 2 r plus 1 right parenthesis cubed minus left parenthesis 2 r minus 1 right parenthesis cubed equals 24 r to the power of 2 end exponent plus 2,  and then the sum of both sides of that equation from r equals 1 to r equals n can be considered

      • sum from r equals 1 to n of left parenthesis left parenthesis 2 r plus 1 right parenthesis cubed minus left parenthesis 2 r minus 1 right parenthesis cubed right parenthesis equals stack sum left parenthesis 24 with r equals 1 below and n on top r to the power of 2 end exponent plus 2 right parenthesis

      • The left-hand side can be found in terms of n using method of differences, and the right-hand side can be rearranged to give 2 n plus 24 stack sum r squared with r equals 1 below and n on top

      • That equation can then be rearranged to give an expression for stack sum r squared with r equals 1 below and n on top

      • When these proofs have appeared previously in exams, they have tended to be structured to help you work through the steps

  • You may have to use your algebraic method of differences result to find a numerical answer, usually in the last part of a question

    • The question will often ask you to evaluate the sum starting from r equals 20 (or some other arbitrary value) rather than from r equals 1

    • To help with this, remember that: sum from r equals p to q of u subscript r equals sum from r equals 1 to q of u subscript r minus sum from r equals 1 to p minus 1 of u subscript r

    • You may also find it helpful to recall that for constants a and bsum from r equals p to q of left parenthesis a u subscript r plus b right parenthesis equals stack a sum with r equals p below and q on top u subscript r plus sum from r equals p to q of b

Examiner Tips and Tricks

  • Mark schemes often specify how many terms from the start and end of the series should be written down – it is usually two or three, so always write down the first three and last three terms

  • Don’t be afraid to write out more terms than this to make sure you spot the pattern, and can easily decide which terms will cancel and which will not

  • Check your algebraic answer by substituting in numbers to make sure it works; you can use your calculator to find summations in sigma notation

Worked Example

(a) Express fraction numerator 2 over denominator r left parenthesis r plus 2 right parenthesis end fraction in partial fractions.

3-2-2-edx-a-fm-we1a-soltn

(b) Hence show that sum from r equals 1 to n of fraction numerator 2 over denominator r left parenthesis r plus 2 right parenthesis end fraction equals fraction numerator 3 n squared plus 5 n over denominator 2 left parenthesis n plus 1 right parenthesis left parenthesis n plus 2 right parenthesis end fraction using the method of differences.

3-2-2-edx-a-fm-we1b-soltn
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Jamie Wood

Author: Jamie Wood

Expertise: Maths Content Creator

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

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