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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9231Paper 1 · Further Pure Mathematics 1Summation of seriesSummation of series
Summation of series · Subtopic 1 of 1

Summation of Series

∑r=12n(n+1)\sum r=\frac12n(n+1), ∑r2=16n(n+1)(2n+1)\sum r^2=\frac16n(n+1)(2n+1), ∑r3=14n2(n+1)2\sum r^3=\frac14n^2(n+1)^2. Combine them for polynomial sums. When each term is a difference f(r)−f(r+1)f(r)-f(r+1), almost everything cancels: the method of differences.

Cambridge 9231 Paper 1Edexcel YFM01 WFM01Edexcel YFM01 WFM02Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2Edexcel 9FM0 Option≈ 40 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Use the standard results for ∑r\sum r, ∑r2\sum r^2, ∑r3\sum r^3.
  • Find sums between two limits.
  • Use the method of differences, including with partial fractions.
  • Find sums to infinity of telescoping series.
Pause and think

(1−12)+(12−13)+⋯+(19−110)\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac19-\tfrac1{10}\right). What is the total?

Section 2 of 14

Core idea

Start hereGauss and a hundred numbers

A teacher once set a class to add 1+2+⋯+1001+2+\cdots+100, hoping for peace. Young Gauss answered in seconds: 50505050. He paired the ends. Further maths takes that idea much further, to sums of squares, cubes, and series that collapse almost entirely.

The big idea

Use ∑r\sum r, ∑r2\sum r^2, ∑r3\sum r^3 for polynomial terms. If each term is f(r)−f(r+1)f(r)-f(r+1), the sum telescopes to f(1)−f(n+1)f(1)-f(n+1).

1. Standard results

∑r2\sum r^2 by telescoping cubes
  1. (r+1)3−r3=3r2+3r+1(r+1)^3-r^3=3r^2+3r+1
    Expand.
  2. ∑r=1n[(r+1)3−r3]=(n+1)3−1\sum_{r=1}^n\left[(r+1)^3-r^3\right]=(n+1)^3-1
    Telescopes.
  3. 3∑r2+3⋅12n(n+1)+n=(n+1)3−13\sum r^2+3\cdot\tfrac12n(n+1)+n=(n+1)^3-1
    Sum the right side term by term.
  4. ∑r2=16n(n+1)(2n+1)\sum r^2=\tfrac16n(n+1)(2n+1)
    Rearrange and factorise.
∑r=1nr=12n(n+1),∑r=1nr2=16n(n+1)(2n+1),∑r=1nr3=14n2(n+1)2\sum_{r=1}^n r=\tfrac12n(n+1),\quad\sum_{r=1}^n r^2=\tfrac16n(n+1)(2n+1),\quad\sum_{r=1}^n r^3=\tfrac14n^2(n+1)^2
where
nn
number of terms.
Worked example∑r(r+2)\sum r(r+2)

Find ∑r=1nr(r+2)\sum_{r=1}^n r(r+2) in factorised form.

Goal
Closed form.
Strategy
Expand, use results, factorise early.
  1. ∑r2+2∑r\sum r^2+2\sum r
    Linearity.
  2. 16n(n+1)(2n+1)+n(n+1)\tfrac16n(n+1)(2n+1)+n(n+1)
    Standard results.
  3. =16n(n+1)(2n+7)=\tfrac16n(n+1)(2n+7)
    Factor 16n(n+1)\tfrac16n(n+1): 2n+1+62n+1+6.
Answer16n(n+1)(2n+7)\tfrac16n(n+1)(2n+7).
✓ Checkn=2n=2: 3+8=113+8=11 and 16(2)(3)(11)=11\tfrac16(2)(3)(11)=11 ✓.
Worked exampleBetween limits

Find ∑r=1120r\sum_{r=11}^{20}r.

Goal
Sum.
Strategy
∑120−∑110\sum_1^{20}-\sum_1^{10}.
  1. 210−55210-55
    Keep term 11.
  2. =155=155
    Subtract.
Answer155155.
✓ CheckTen terms averaging 15.515.5 ✓.

2. Method of differences

∑1r(r+1)\sum\frac1{r(r+1)}
  1. 1r(r+1)=1r−1r+1\frac1{r(r+1)}=\frac1r-\frac1{r+1}
    Partial fractions.
  2. (1−12)+(12−13)+⋯+(1n−1n+1)\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1n-\tfrac1{n+1}\right)
    Write the terms out.
  3. =1−1n+1=nn+1=1-\frac1{n+1}=\frac n{n+1}
    Middle terms cancel.
  4. →1 as n→∞\to1\text{ as }n\to\infty
    Sum to infinity.
Worked exampleGaps of two

Find ∑r=1∞2r(r+2)\sum_{r=1}^\infty\frac2{r(r+2)}.

Goal
Sum to infinity.
Strategy
1r−1r+2\frac1r-\frac1{r+2}: cancelling terms are two apart.
  1. (1−13)+(12−14)+(13−15)+⋯\left(1-\tfrac13\right)+\left(\tfrac12-\tfrac14\right)+\left(\tfrac13-\tfrac15\right)+\cdots
    List terms.
  2. 1+12−1n+1−1n+21+\tfrac12-\tfrac1{n+1}-\tfrac1{n+2}
    Two survivors each end.
  3. →32\to\tfrac32
    Limit.
Answer32\tfrac32.
✓ CheckFirst three terms already give 0.67+0.25+0.13=1.050.67+0.25+0.13=1.05, heading to 1.5 ✓.
Common misconceptionWrong lower limit ∑1120\sum_{11}^{20} subtracts ∑110\sum_1^{10}, not ∑111\sum_1^{11}.

Summary

  • Three standard results.
  • Factorise early.
  • Differences: list terms; survivors at both ends.
Key words
Telescoping
A sum in which almost all terms cancel.
HistoryGauss in school. Legend says the young Carl Friedrich Gauss added 11 to 100100 instantly by pairing 1+1001+100, 2+992+99, …: fifty pairs of 101101, giving 50505050.
Fun factSquares of triangles. ∑r3=(∑r)2\sum r^3=\left(\sum r\right)^2: the sum of the first nn cubes is the square of the nnth triangular number. 1+8+27=36=621+8+27=36=6^2.
Section 3 of 14

Key formulae

Standard results∑r=1nr=12n(n+1),∑r=1nr2=16n(n+1)(2n+1),∑r=1nr3=14n2(n+1)2\sum_{r=1}^n r=\tfrac12n(n+1),\quad\sum_{r=1}^n r^2=\tfrac16n(n+1)(2n+1),\quad\sum_{r=1}^n r^3=\tfrac14n^2(n+1)^2
Between limits∑r=mnf(r)=∑r=1nf(r)−∑r=1m−1f(r)\sum_{r=m}^n f(r)=\sum_{r=1}^n f(r)-\sum_{r=1}^{m-1}f(r)
Differences∑r=1n[f(r)−f(r+1)]=f(1)−f(n+1)\sum_{r=1}^n\bigl[f(r)-f(r+1)\bigr]=f(1)-f(n+1)
TrickWrite out the first and last few terms. With differences, list r=1,2,3r=1,2,3 and r=n−1,nr=n-1,n in columns. You'll see exactly which terms survive at the start and end.
Section 4 of 14

Worked examples

Example 1 · Combining results0 / 2 steps
Find ∑r=1nr(r+2)\sum_{r=1}^n r(r+2) in factorised form.
    Example 2 · Differences0 / 3 steps
    Find ∑r=1n1r(r+1)\sum_{r=1}^n\dfrac1{r(r+1)} and its sum to infinity.
      Section 5 of 14

      Explore the graph

      Partial sums approach a limit
      Loading the interactive graph…
      Try this
      1. The point shows ∑r=1n1r(r+1)\sum_{r=1}^n\frac1{r(r+1)}. How close to 11 is it when n=20n=20?

      The partial sums nn+1\frac n{n+1} climb towards 11 but never reach it.

      In the real worldComputer running times. A program that loops rr times inside a loop of nn steps does ∑r=12n(n+1)\sum r=\frac12n(n+1) operations: that's why such algorithms are called "quadratic time".
      Section 6 of 14

      Common mistakes

      Incorrect∑r=1n3=3\sum_{r=1}^n 3=3
      Correct3n3n
      You add 33 a total of nn times.
      Incorrect∑r=1120=∑120−∑111\sum_{r=11}^{20}=\sum_1^{20}-\sum_1^{11}
      CorrectSubtract ∑110\sum_1^{10}
      Term 1111 must stay in.
      IncorrectCancelling only one term at each end
      CorrectCheck how far apart the cancelling terms are
      1r−1r+2\frac1r-\frac1{r+2} leaves two terms at each end.

      Me: adding 100 fractions one by one

      Method of differences: everything cancels except the ends. 🧹

      Section 7 of 14

      Exam tip

      Exam tipFactorise as you go. Take out 16n(n+1)\tfrac16n(n+1) (or similar) before expanding. "Show that" answers are usually in factorised form, and it saves lots of algebra.
      Section 8 of 14

      Guided practice

      SA-FP-SUM-G012 marksL1 FoundationNon-calculatorStandard sums
      Find ∑r=110(2r+1)\displaystyle\sum_{r=1}^{10}(2r+1).

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-SUM-0012 marksL1 FoundationCalculatorStandard sums
      Find ∑r=120r2\displaystyle\sum_{r=1}^{20}r^2.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-SUM-0022 marksL2 StandardNon-calculatorStandard sums
      Find ∑r=1120r\displaystyle\sum_{r=11}^{20}r.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-SUM-0032 marksL2 StandardNon-calculatorDifferences
      Using 1r(r+1)=1r−1r+1\dfrac1{r(r+1)}=\dfrac1r-\dfrac1{r+1}, find ∑r=191r(r+1)\displaystyle\sum_{r=1}^{9}\dfrac1{r(r+1)}.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-SUM-0041 markL2 StandardNon-calculatorDifferences
      State the sum to infinity of ∑r=1∞1r(r+1)\displaystyle\sum_{r=1}^{\infty}\dfrac1{r(r+1)}.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-SUM-0054 marksL4 ChallengingNon-calculatorDifferences
      Using 2r(r+2)=1r−1r+2\dfrac2{r(r+2)}=\dfrac1r-\dfrac1{r+2}, find ∑r=1∞2r(r+2)\displaystyle\sum_{r=1}^{\infty}\dfrac2{r(r+2)}.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Three standard results.
      • Between limits: subtract the sum up to m−1m-1.
      • Differences: f(1)−f(n+1)f(1)-f(n+1).
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-SUM-Q011 markL1 FoundationNon-calculatorStandard sums
      ∑r=1nr3\displaystyle\sum_{r=1}^{n}r^3 equals
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      9231 AS & A Level Further Mathematics
      Where
      Paper 1 · 1.3 Summation of series
      Expected
      Standard results for Σr, Σr², Σr³; the method of differences; convergence and sums to infinity.
      Calculator
      Permitted on all papers; exact answers are expected where the question asks.
      Typical style
      Common pattern: "Show that Σ … = ⅙n(n+1)(an+b)", then use the result to evaluate a sum between two limits.
      Open in
      9231 · Paper 1
      Section 14 of 14

      Past paper questions

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      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).