Why is "Writing $\sum_{r=1}^{n} c = c$ instead of $cn$" a common mistake in Method of differences?

Why it happens: Treating the constant as a single number rather than a term added once for each of the n values of r. How to avoid it: Remember a constant is summed n times: \sum_r=1^n 1 = n, \sum_r=1^n 3 = 3n. Source: 9231 Paper 1 Examiner Reports — summation questions

Why is "Jumping straight to $f(1) - f(n+1)$ without writing out the terms" a common mistake in Method of differences?

Why it happens: Over-confidence; memorising the result instead of demonstrating the cancellation. How to avoid it: Always write the first two-three and last two-three terms explicitly — examiners credit this and it prevents miscounting survivors. Source: 9231 Paper 1 Examiner Reports

Why is "On a gap-2 telescope, keeping only one term at each end" a common mistake in Method of differences?

Why it happens: Applying the gap-1 result f(1) - f(n+1) when the term is actually f(r) - f(r+2). How to avoid it: For f(r) - f(r+2), TWO terms survive at each end: f(1) + f(2) - f(n+1) - f(n+2). Write out four terms to confirm. Source: 9231 Paper 1 mark schemes

Why is "Sign error in the partial-fraction constants, so nothing cancels" a common mistake in Method of differences?

Why it happens: Cover-up arithmetic slips, especially with a negative substitution value like r = -1. How to avoid it: Check the split by recombining over the common denominator; if the telescope fails to cancel, the partial fractions are wrong. Source: 9231 Paper 1 Examiner Reports

Why is "Leaving the answer as $1 - \frac{1}{n+1}$ instead of $\frac{n}{n+1}$" a common mistake in Method of differences?

Why it happens: Stopping at the telescoped form rather than combining into the simplest single fraction the mark scheme wants. How to avoid it: When the question says 'simplest form' or 'a single fraction', always combine over a common denominator as the last step. Source: 9231 Paper 1 mark schemes

Cambridge International A Levels Further Mathematics (9231)

Method of differences

Q: Why is "Trying standard results on a fraction, or differences on a polynomial" a common mistake in Method of differences?

Why it happens: Not reading the structure of the general term before starting. How to avoid it: Polynomial in r → standard results; fraction over a product of factors → method of differences. A 'partial fractions, hence find ' question is always a telescope. Source: 9231 Paper 1 Examiner Reports

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Detailed Study Notes

Detailed notes on Summation of Series for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Summation of Series — Method of Differences (Cambridge International AS & A Level Further Mathematics 9231 Paper 1, 2026-2027 syllabus)

Two tools for summing series. First, the standard results $\sum r$ , $\sum r^2$ , $\sum r^3$ (in MF19) for polynomials in $r$ . Second, the method of differences (not in MF19): rewrite the general term as $f(r) - f(r+1)$ so almost everything cancels and the sum telescopes to a few surviving end terms.

At a glance

Standard results $\sum r = \tfrac{1}{2}n(n+1)$ , $\sum r^2 = \tfrac{1}{6}n(n+1)(2n+1)$ , $\sum r^3 = \tfrac{1}{4}n^2(n+1)^2$ are in MF19 — split any polynomial in $r$ and apply them.
Method of differences: write the term as $u_r = f(r) - f(r+1)$ so consecutive terms cancel.
Telescoping leaves only the first and last surviving pieces: $\sum_{r=1}^{n} u_r = f(1) - f(n+1)$ .
Get the $f(r)$ from partial fractions, e.g. $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ .
Always write out the first 2-3 and last 2-3 terms to see exactly what survives — this is the examiner-credited method.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.3a — Use the standard results for $\sum r$ , $\sum r^2$ and $\sum r^3$ to find related sums (e.g. sums of polynomials in $r$ ).
1.3b — Use the method of differences to obtain the sum of a finite series, e.g. by expressing the general term in partial fractions.
1.3c — Recognise from the structure of the general term whether to use standard results or the method of differences.

Standard results — summing polynomials in r

Σr, Σr² and Σr³ are in MF19. Split a polynomial term-by-term and apply them.

The three results you can quote (all printed in the MF19 formula booklet, so no need to derive them):

$\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$ $\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$ $\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2 = \left[\frac{1}{2}n(n+1)\right]^2$

Two facts that make these usable on any polynomial:

$\sum (a u_r + b v_r) = a\sum u_r + b\sum v_r$ — the sum is linear, so split it.
$\sum_{r=1}^{n} c = cn$ — a constant term summed $n$ times.

Worked idea. To find $\displaystyle\sum_{r=1}^{n}(2r+1)$ , split it: $\sum_{r=1}^{n}(2r+1) = 2\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1 = 2\cdot\tfrac{1}{2}n(n+1) + n = n(n+1) + n = n(n+2).$

Stacking 1, 2, 3, … unit squares builds half a rectangle of area n(n+1), giving the closed form ½n(n+1).

Cambridge tip. Leave the answer fully factorised where possible — examiners often demand the result "in the form $n(\ldots)$ ". A common final move is to take out the common factor and tidy the bracket.

$\sum r$ , $\sum r^2$ , $\sum r^3$ are in MF19 — quote, don't derive.
Sums are linear: split $\sum(2r+1)$ into $2\sum r + \sum 1$ .
$\sum_{r=1}^{n} c = cn$ for a constant term.

See the full worked example for method of differences →

The method of differences — how telescoping works

Write the term as f(r) − f(r+1); consecutive pieces cancel, leaving the ends.

The method of differences (also called telescoping) is not in MF19 — you must know the technique. It works whenever the general term can be written as the difference of consecutive values of some function $f$ :

$u_r = f(r) - f(r+1).$

Why it collapses. Add up $u_1, u_2, \ldots, u_n$ and stack the differences: $\sum_{r=1}^{n} u_r = \big[f(1)-f(2)\big] + \big[f(2)-f(3)\big] + \big[f(3)-f(4)\big] + \cdots + \big[f(n)-f(n+1)\big].$

Each $-f(2)$ kills the next $+f(2)$ , each $-f(3)$ kills the next $+f(3)$ , and so on. Everything in the middle cancels, leaving only the very first and very last pieces: $\sum_{r=1}^{n} u_r = f(1) - f(n+1).$

Inner terms cancel in pairs; only the first and last survive. Here, summing four terms leaves f(1) − f(5).

Some terms split two apart. A few series telescope as $u_r = f(r) - f(r+2)$ . Then two terms survive at each end: $\sum_{r=1}^{n} \big[f(r) - f(r+2)\big] = f(1) + f(2) - f(n+1) - f(n+2).$

Cambridge tip. Whatever the gap, write out the first two or three terms and the last two or three terms in full before cancelling. Examiners credit this explicit display, and it stops you mis-counting which terms survive.

Telescoping needs $u_r = f(r) - f(r+1)$ .
Simple gap-1 case: $\sum_{r=1}^n u_r = f(1) - f(n+1)$ .
Gap-2 case ( $f(r)-f(r+2)$ ): two terms survive at each end.

See the full worked example for method of differences →

Getting f(r) from partial fractions

Split a product-of-factors fraction into partial fractions, then telescope.

Most exam terms are fractions like $\dfrac{1}{r(r+1)}$ or $\dfrac{1}{r(r+2)}$ . Partial fractions turn them into a difference $f(r) - f(r+\text{gap})$ ready to telescope.

The classic gap-1 split. $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}.$ So $f(r) = \dfrac{1}{r}$ and $\sum_{r=1}^{n}\frac{1}{r(r+1)} = f(1) - f(n+1) = 1 - \frac{1}{n+1} = \frac{n}{n+1}.$

A gap-2 split. $\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right),$ which telescopes two-apart, leaving the first two and last two pieces.

A three-factor (gap-2 in disguise) split. For $\dfrac{1}{r(r+1)(r+2)}$ the slick route is to spot $\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left[\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right],$ so $f(r) = \dfrac{1}{r(r+1)}$ and the sum is $\dfrac{1}{2}\big[f(1) - f(n+1)\big]$ .

Routine for any fraction:

Factorise the denominator.
Set up partial fractions and find the constants (cover-up method is fastest).
Identify $f(r)$ and the gap.
Write out the first/last terms, cancel, and read off the closed form.
Simplify into a single fraction — the mark scheme answer is almost always one tidy fraction.

Cambridge tip. If the partial-fraction constants are e.g. $\tfrac{1}{2}, -\tfrac{1}{2}$ , keep the common $\tfrac{1}{2}$ outside the sum so the telescoping inside is clean, then multiply back at the end.

$\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ (gap 1).
$\dfrac{1}{r(r+2)} = \tfrac{1}{2}\big(\dfrac{1}{r} - \dfrac{1}{r+2}\big)$ (gap 2).
Pull constant factors outside before telescoping; combine into one fraction at the end.

See the full worked example for method of differences →

Which method? — reading the general term

Polynomial in r → standard results. Fraction over a product of factors → method of differences.

The fastest mark-saver is deciding correctly which tool to reach for before you start.

Use the standard results when the general term is a polynomial in $r$ (or expands to one):

$\sum (3r^2 - 2r + 1)$ , $\sum r(r+1)$ , $\sum (2r-1)^2$ — multiply out, then quote $\sum r$ , $\sum r^2$ , $\sum r^3$ .

Use the method of differences when the term is a fraction whose denominator is a product of factors that differ by a fixed gap, or is otherwise a difference of consecutive things:

$\dfrac{1}{r(r+1)}$ , $\dfrac{1}{(2r-1)(2r+1)}$ , $\dfrac{r}{(r+1)!}$ — split and telescope.

The tell-tale signs of a "differences" question:

A denominator that factorises into a product.
The question first asks you to "express in partial fractions" — a strong hint the next part telescopes.
The phrase "hence find $\sum \ldots$ " linking the two parts.

Cambridge tip. If you ever set up a telescoping sum and nothing cancels, you have the wrong split — re-check your partial fractions before grinding on.

Polynomial in $r$ → standard results $\sum r, \sum r^2, \sum r^3$ .
Fraction over a product of factors → method of differences.
"Express in partial fractions, hence find $\sum$ " almost always means telescope.

See the full worked example for method of differences →

Quick recap

$\sum r = \tfrac{1}{2}n(n+1)$ , $\sum r^2 = \tfrac{1}{6}n(n+1)(2n+1)$ , $\sum r^3 = \tfrac{1}{4}n^2(n+1)^2$ — all in MF19.
Sums are linear: split a polynomial term-by-term and apply the standard results.
Method of differences: $u_r = f(r) - f(r+1)$ telescopes to $f(1) - f(n+1)$ .
Get $f(r)$ from partial fractions, e.g. $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ .
Write out the first and last few terms explicitly to see what survives.
The method of differences itself is NOT in MF19 — know the technique cold.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\sum r = \tfrac{1}{2}n(n+1)$ , $\;\sum r^2 = \tfrac{1}{6}n(n+1)(2n+1)$ , $\;\sum r^3 = \tfrac{1}{4}n^2(n+1)^2$ (in MF19).
Telescoping result: $\sum_{r=1}^{n}\big[f(r)-f(r+1)\big] = f(1) - f(n+1)$ .
Gap-2 result: $\sum_{r=1}^{n}\big[f(r)-f(r+2)\big] = f(1)+f(2)-f(n+1)-f(n+2)$ .
$\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ and $\dfrac{1}{r(r+2)} = \tfrac{1}{2}\big(\dfrac{1}{r}-\dfrac{1}{r+2}\big)$ .
Display the first/last few terms before cancelling — examiner-credited working.

How it’s examined

Summation of series appears in most Paper 1 papers (5-9 marks). A standard-results question asks for $\sum$ of a polynomial in $r$ and demands a fully factorised closed form. A method-of-differences question usually has a part (a) 'express in partial fractions' followed by 'hence find $\sum_{r=1}^{n}\ldots$ ' — and often a follow-on linking to the sum to infinity (subtopic 3.2). Marks are lost through (i) dropping or mis-counting the surviving end terms, (ii) sign errors in the partial fractions, and (iii) not simplifying to a single fraction. Showing the first/last terms written out is the safest route to full method marks.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11, 9231/12, 9231/13 Further Pure Mathematics 1 — past papers and mark schemes; MF19 List of Formulae. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Method of differences

Step-by-step solutions to past-paper-style questions on method of differences, written exactly the way a tutor would explain them at the board.

Question type:

1Summing a linear term with the standard results
Getting startedDirect calculation• standard-results, polynomial
Question
Find $\displaystyle\sum_{r=1}^{n}(2r+1)$ , giving your answer in a fully factorised form.
Step-by-step solution
1. Step 1
  Split the sum using linearity into a multiple of $\sum r$ plus a sum of a constant.
  $\sum_{r=1}^{n}(2r+1) = 2\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1$
2. Step 2
  Quote the standard result $\sum r = \tfrac{1}{2}n(n+1)$ (in MF19) and $\sum_{r=1}^{n}1 = n$ .
  $= 2\cdot\tfrac{1}{2}n(n+1) + n = n(n+1) + n$
3. Step 3
  Factorise the result.
  $= n\big[(n+1)+1\big] = n(n+2)$
Answer
$\displaystyle\sum_{r=1}^{n}(2r+1) = n(n+2)$ .
Examiner tip
Linearity ( $\sum(2r+1) = 2\sum r + \sum 1$ ) and remembering $\sum 1 = n$ (not $1$ ) are the two things being tested. Leave the answer factorised.
2Summing a quadratic term
Getting startedDirect calculation• standard-results, polynomial
Question
Find $\displaystyle\sum_{r=1}^{n} r(r+1)$ , simplifying as far as possible.
Step-by-step solution
1. Step 1
  Multiply out the term into powers of $r$ .
  $r(r+1) = r^2 + r$
2. Step 2
  Apply the standard results for $\sum r^2$ and $\sum r$ .
  $\sum_{r=1}^{n} (r^2 + r) = \tfrac{1}{6}n(n+1)(2n+1) + \tfrac{1}{2}n(n+1)$
3. Step 3
  Take out the common factor $\tfrac{1}{6}n(n+1)$ and tidy the bracket.
  $= \tfrac{1}{6}n(n+1)\big[(2n+1) + 3\big] = \tfrac{1}{6}n(n+1)(2n+4)$
4. Step 4
  Simplify the even bracket.
  $= \tfrac{1}{3}n(n+1)(n+2)$
Answer
$\displaystyle\sum_{r=1}^{n} r(r+1) = \tfrac{1}{3}n(n+1)(n+2)$ .
Examiner tip
Common factor $\tfrac{1}{6}n(n+1)$ comes straight out of both standard results — pulling it out first avoids messy expansion. $2n+4 = 2(n+2)$ cancels the 2 against the 6.
3A first telescoping sum
Building confidenceMulti-step problem• method-of-differences, telescoping
Question
Given that $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ , find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)}$ .
Step-by-step solution
1. Step 1
  Identify $f(r) = \dfrac{1}{r}$ , so each term is $f(r) - f(r+1)$ . Write out the first three and last two terms.
  $\Big(\tfrac{1}{1}-\tfrac{1}{2}\Big)+\Big(\tfrac{1}{2}-\tfrac{1}{3}\Big)+\Big(\tfrac{1}{3}-\tfrac{1}{4}\Big)+\cdots+\Big(\tfrac{1}{n}-\tfrac{1}{n+1}\Big)$
2. Step 2
  Cancel the inner terms ( $-\tfrac12$ with $+\tfrac12$ , etc.). Only $f(1)$ and $-f(n+1)$ survive.
  $= 1 - \frac{1}{n+1}$
3. Step 3
  Combine into a single fraction.
  $= \frac{(n+1) - 1}{n+1} = \frac{n}{n+1}$
Answer
$\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)} = \frac{n}{n+1}$ .
Examiner tip
Showing the written-out terms before cancelling is the credited method. End with one tidy fraction, not $1 - \tfrac{1}{n+1}$ .
4Express in partial fractions, then telescope
Building confidenceMulti-step problem• method-of-differences, partial-fractions
Question
(a) Express $\dfrac{2}{(2r-1)(2r+1)}$ in partial fractions. (b) Hence find $\displaystyle\sum_{r=1}^{n}\frac{2}{(2r-1)(2r+1)}$ .
Step-by-step solution
1. Step 1
  Set up partial fractions: $\dfrac{2}{(2r-1)(2r+1)} = \dfrac{A}{2r-1} + \dfrac{B}{2r+1}$ . Cover-up gives $A = 1$ , $B = -1$ .
  $\frac{2}{(2r-1)(2r+1)} = \frac{1}{2r-1} - \frac{1}{2r+1}$
2. Step 2
  This is $f(r) - f(r+1)$ with $f(r) = \dfrac{1}{2r-1}$ (since $2(r+1)-1 = 2r+1$ ). Write out the terms.
  $\Big(\tfrac{1}{1}-\tfrac{1}{3}\Big)+\Big(\tfrac{1}{3}-\tfrac{1}{5}\Big)+\cdots+\Big(\tfrac{1}{2n-1}-\tfrac{1}{2n+1}\Big)$
3. Step 3
  Telescope — only the first and last survive.
  $= 1 - \frac{1}{2n+1} = \frac{2n}{2n+1}$
Answer
(a) $\dfrac{1}{2r-1} - \dfrac{1}{2r+1}$ . (b) $\displaystyle\sum_{r=1}^{n}\frac{2}{(2r-1)(2r+1)} = \frac{2n}{2n+1}$ .
Examiner tip
Recognising $f(r) = \tfrac{1}{2r-1}$ and that $f(r+1) = \tfrac{1}{2r+1}$ is the key step. A sign slip in $B$ ( $+1$ instead of $-1$ ) destroys the cancellation.
5A three-factor denominator (gap-2 telescoping)
StretchMulti-step problem• method-of-differences, partial-fractions, three-factor
Question
Show that $\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2}\left[\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right]$ , and hence find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}$ .
Step-by-step solution
1. Step 1
  Verify the split. Combine the right-hand side over the common denominator $r(r+1)(r+2)$ . The numerator is $\tfrac12[(r+2) - r] = \tfrac12\cdot 2 = 1$ .
  $\tfrac{1}{2}\cdot\frac{(r+2) - r}{r(r+1)(r+2)} = \frac{1}{r(r+1)(r+2)}$
2. Step 2
  Identify $f(r) = \dfrac{1}{r(r+1)}$ , so the term is $\tfrac12\big[f(r) - f(r+1)\big]$ . Pull the $\tfrac12$ out front and telescope $f(1) - f(n+1)$ .
  $\sum_{r=1}^{n} = \tfrac{1}{2}\Big[f(1) - f(n+1)\Big] = \tfrac{1}{2}\left[\frac{1}{1\cdot 2} - \frac{1}{(n+1)(n+2)}\right]$
3. Step 3
  Simplify to a single fraction.
  $= \tfrac{1}{2}\left[\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right] = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$
4. Step 4
  Combine over a common denominator.
  $= \frac{(n+1)(n+2) - 2}{4(n+1)(n+2)} = \frac{n^2 + 3n}{4(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$
Answer
$\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$ .
Examiner tip
Treating $f(r) = \tfrac{1}{r(r+1)}$ as a single object keeps the telescoping clean. The final factorisation $(n+1)(n+2) - 2 = n^2+3n = n(n+3)$ is a frequent stumbling block — expand carefully.
6A two-apart cancellation
StretchMulti-step problem• method-of-differences, gap-2
Question
Given that $\dfrac{1}{r(r+2)} = \dfrac{1}{2}\left(\dfrac{1}{r} - \dfrac{1}{r+2}\right)$ , find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)}$ .
Step-by-step solution
1. Step 1
  $f(r) = \dfrac{1}{r}$ and the gap is 2 — each term is $\tfrac12\big[f(r) - f(r+2)\big]$ . Write out terms to see the gap-2 pattern.
  $\tfrac{1}{2}\Big[\Big(\tfrac{1}{1}-\tfrac{1}{3}\Big)+\Big(\tfrac{1}{2}-\tfrac{1}{4}\Big)+\Big(\tfrac{1}{3}-\tfrac{1}{5}\Big)+\Big(\tfrac{1}{4}-\tfrac{1}{6}\Big)+\cdots\Big]$
2. Step 2
  Cancel terms two apart. The $-\tfrac13$ kills the later $+\tfrac13$ , etc. The first two ( $\tfrac11, \tfrac12$ ) and last two ( $-\tfrac{1}{n+1}, -\tfrac{1}{n+2}$ ) survive.
  $= \tfrac{1}{2}\left[1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right]$
3. Step 3
  Tidy the constant part.
  $= \tfrac{1}{2}\left[\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right] = \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}$
Answer
$\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)} = \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}$ .
Examiner tip
With a gap-2 split, TWO terms survive at each end — forgetting the second one ( $\tfrac12$ at the start, or one of the two end terms) is the most common error. Writing out four terms makes the pattern unmistakable.

Model Answers — Method of differences

High-scoring sample answers for method of differences on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
Find $\displaystyle\sum_{r=1}^{n}(r^3 + 2r)$ , giving your answer in a fully factorised form.
Model answer
Split using linearity and quote the standard results (both in MF19): $\sum_{r=1}^{n}(r^3 + 2r) = \sum_{r=1}^{n} r^3 + 2\sum_{r=1}^{n} r = \tfrac{1}{4}n^2(n+1)^2 + 2\cdot\tfrac{1}{2}n(n+1)$

$= \tfrac{1}{4}n^2(n+1)^2 + n(n+1).$

Take out the common factor $\tfrac{1}{4}n(n+1)$ : $= \tfrac{1}{4}n(n+1)\big[n(n+1) + 4\big] = \tfrac{1}{4}n(n+1)(n^2 + n + 4).$
Why this scores
M1 for splitting and quoting both standard results, A1 for the unsimplified sum, M1 for taking out the common factor, A1 for the factorised form. The bracket $n^2+n+4$ does not factorise further — leaving it is fine.
Question 2
4 marks
Find $\displaystyle\sum_{r=1}^{n}(2r-1)^2$ , simplifying your answer fully.
Model answer
Expand the general term: $(2r-1)^2 = 4r^2 - 4r + 1.$

Apply the standard results: $\sum_{r=1}^{n}(4r^2 - 4r + 1) = 4\cdot\tfrac{1}{6}n(n+1)(2n+1) - 4\cdot\tfrac{1}{2}n(n+1) + n.$

$= \tfrac{2}{3}n(n+1)(2n+1) - 2n(n+1) + n.$

Take out $n$ and simplify the bracket: $= n\left[\tfrac{2}{3}(n+1)(2n+1) - 2(n+1) + 1\right] = n\left[\tfrac{2}{3}(2n^2 + 3n + 1) - 2n - 1\right]$ $= n\left[\tfrac{4}{3}n^2 + 2n + \tfrac{2}{3} - 2n - 1\right] = n\left[\tfrac{4}{3}n^2 - \tfrac{1}{3}\right] = \tfrac{1}{3}n(4n^2 - 1) = \tfrac{1}{3}n(2n-1)(2n+1).$
Why this scores
M1 for expansion, M1 for the three standard results substituted, M1 for factoring out $n$ , A1 for $\tfrac{1}{3}n(2n-1)(2n+1)$ . Recognising $4n^2-1$ as a difference of squares secures the neat final form.
Question 3
6 marks
(a) Express $\dfrac{1}{r(r+1)}$ in partial fractions. (b) Hence use the method of differences to find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)}$ , showing your working clearly.
Model answer
(a) Write $\dfrac{1}{r(r+1)} = \dfrac{A}{r} + \dfrac{B}{r+1}$ . Then $1 = A(r+1) + Br$ .

$r = 0$ : $1 = A$ , so $A = 1$ .

$r = -1$ : $1 = -B$ , so $B = -1$ .

$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}.$

(b) With $f(r) = \dfrac{1}{r}$ , the term is $f(r) - f(r+1)$ . Writing out the sum: $\sum_{r=1}^{n}\frac{1}{r(r+1)} = \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right).$

All interior terms cancel, leaving the first and last: $= 1 - \frac{1}{n+1} = \frac{n}{n+1}.$
Why this scores
Part (a): M1 for the partial-fraction setup, A1 for $A=1, B=-1$ . Part (b): M1 for writing the terms out, M1 for the cancellation leaving $1 - \tfrac{1}{n+1}$ , A1 for the single fraction $\tfrac{n}{n+1}$ . 'Showing your working clearly' means the written-out terms must appear.
Question 4
6 marks
(a) Show that $\dfrac{4}{(2r-1)(2r+3)} = \dfrac{1}{2r-1} - \dfrac{1}{2r+3}$ . (b) Hence find $\displaystyle\sum_{r=1}^{n}\frac{4}{(2r-1)(2r+3)}$ .
Model answer
(a) Combining the right-hand side over a common denominator: $\frac{1}{2r-1} - \frac{1}{2r+3} = \frac{(2r+3) - (2r-1)}{(2r-1)(2r+3)} = \frac{4}{(2r-1)(2r+3)}. \;\checkmark$

(b) Here $f(r) = \dfrac{1}{2r-1}$ , and $f(r+2) = \dfrac{1}{2(r+2)-1} = \dfrac{1}{2r+3}$ — this is a gap-2 telescope. Writing out terms: $\sum_{r=1}^{n} = \left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\cdots+\left(\frac{1}{2n-1}-\frac{1}{2n+3}\right).$

The $-\tfrac15$ cancels the later $+\tfrac15$ , the $-\tfrac17$ cancels the later $+\tfrac17$ , and so on. Two terms survive at each end — the first two positives $\tfrac11, \tfrac13$ and the last two negatives $-\tfrac{1}{2n+1}, -\tfrac{1}{2n+3}$ : $= 1 + \frac{1}{3} - \frac{1}{2n+1} - \frac{1}{2n+3} = \frac{4}{3} - \frac{1}{2n+1} - \frac{1}{2n+3}.$
Why this scores
Part (a): M1 A1 for the common-denominator verification (a 'Show that', so the numerator $4$ must be displayed). Part (b): M1 for spotting the gap-2 structure, M1 for the cancellation keeping all four surviving terms, A1 for $\tfrac43 - \tfrac{1}{2n+1} - \tfrac{1}{2n+3}$ . Losing one of the four surviving end terms is the standard error.
Question 5
8 marks
(a) Express $\dfrac{1}{r(r+1)(r+2)}$ in partial fractions. (b) Hence find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}$ as a single fraction in its simplest form.
Model answer
(a) Write $\dfrac{1}{r(r+1)(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+1} + \dfrac{C}{r+2}$ , so $1 = A(r+1)(r+2) + Br(r+2) + Cr(r+1)$ .

$r=0$ : $1 = A(1)(2) = 2A \Rightarrow A = \tfrac12$ .

$r=-1$ : $1 = B(-1)(1) = -B \Rightarrow B = -1$ .

$r=-2$ : $1 = C(-2)(-1) = 2C \Rightarrow C = \tfrac12$ .

$\frac{1}{r(r+1)(r+2)} = \frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2}.$

(b) To telescope, regroup the partial fractions as a difference of two-factor pieces: $\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left[\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right].$ (Check: $\tfrac12\cdot\dfrac{(r+2)-r}{r(r+1)(r+2)} = \dfrac{1}{r(r+1)(r+2)}$ . ✓)

With $f(r) = \dfrac{1}{r(r+1)}$ , the term is $\tfrac12\big[f(r) - f(r+1)\big]$ , which telescopes: $\sum_{r=1}^{n} = \frac{1}{2}\Big[f(1) - f(n+1)\Big] = \frac{1}{2}\left[\frac{1}{1\cdot 2} - \frac{1}{(n+1)(n+2)}\right] = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}.$

Combine into a single fraction: $= \frac{(n+1)(n+2) - 2}{4(n+1)(n+2)} = \frac{n^2 + 3n}{4(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$
Why this scores
Part (a): M1 setup, A1 A1 for $A=\tfrac12, B=-1, C=\tfrac12$ . Part (b): M1 for the two-factor regrouping, M1 for the telescope $f(1)-f(n+1)$ , M1 for combining over a common denominator, A1 for $\dfrac{n(n+3)}{4(n+1)(n+2)}$ . The single-fraction-simplest-form instruction means the unsimplified $\tfrac14 - \tfrac{1}{2(n+1)(n+2)}$ would lose the final A1.
Question 6
8 marks
The $r$ th term of a series is $u_r = \dfrac{2r+1}{r^2(r+1)^2}$ . (a) Show that $u_r = \dfrac{1}{r^2} - \dfrac{1}{(r+1)^2}$ . (b) Hence find $\displaystyle\sum_{r=1}^{n} u_r$ and write down the smallest $n$ for which the sum exceeds $0.99$ .
Model answer
(a) Combine the proposed difference over a common denominator $r^2(r+1)^2$ : $\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{(r^2 + 2r + 1) - r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2} = u_r. \;\checkmark$

(b) With $f(r) = \dfrac{1}{r^2}$ , the term is $f(r) - f(r+1)$ , a standard gap-1 telescope: $\sum_{r=1}^{n} u_r = \left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\cdots+\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right) = 1 - \frac{1}{(n+1)^2}.$

Smallest $n$ with sum $> 0.99$ : require $1 - \frac{1}{(n+1)^2} > 0.99 \;\Rightarrow\; \frac{1}{(n+1)^2} < 0.01 \;\Rightarrow\; (n+1)^2 > 100 \;\Rightarrow\; n+1 > 10 \;\Rightarrow\; n > 9.$

So the smallest such integer is $n = 10$ .
Why this scores
Part (a): M1 for the common denominator, A1 for showing the numerator equals $2r+1$ (a 'Show that' — every step shown). Part (b): M1 for $f(r)=\tfrac{1}{r^2}$ and telescoping, A1 for $1 - \tfrac{1}{(n+1)^2}$ , M1 for the inequality, A1 for $n=10$ . Many candidates write $n=9$ — check the strict inequality carefully.

Key Formulae — Method of differences

The formulae you need to memorise for method of differences on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Sum of the first n natural numbers
$\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$
$n$
the upper limit (number of terms)
$r$
the summation index
When to use
Linear terms in $r$ , or as a building block. In MF19 — quote freely.
Sum of squares
$\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$
$n$
the upper limit
When to use
Quadratic terms in $r$ (after expanding). In MF19.
Sum of cubes
$\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2 = \left[\tfrac{1}{2}n(n+1)\right]^2$
$n$
the upper limit
When to use
Cubic terms in $r$ . In MF19. Note it is the square of $\sum r$ .
Method of differences (gap 1)
$\sum_{r=1}^{n}\big[f(r) - f(r+1)\big] = f(1) - f(n+1)$
$f(r)$
the function whose consecutive difference equals the term
When to use
After writing the term as a difference of consecutive values (often via partial fractions). NOT in MF19 — memorise the technique.
Method of differences (gap 2)
$\sum_{r=1}^{n}\big[f(r) - f(r+2)\big] = f(1) + f(2) - f(n+1) - f(n+2)$
$f(r)$
the function whose two-apart difference equals the term
When to use
When the partial-fraction split jumps two terms (e.g. $\tfrac{1}{r(r+2)}$ ). Two terms survive at each end.
Classic partial-fraction splits
$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}, \qquad \frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$
$r$
the summation index
When to use
The two most common starting points for a telescoping sum. Recognise them on sight.

Key Definitions and Keywords — Method of differences

Definitions to memorise and the exact keywords mark schemes credit for method of differences answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Method of differences
Examiner keyword
A technique for summing a finite series by writing the general term as $u_r = f(r) - f(r+k)$ , so that consecutive terms cancel and the sum collapses ('telescopes') to a few surviving end terms. NOT in MF19.
Telescoping series
Examiner keyword
A series in which most terms cancel in pairs when summed, because each term is a difference of consecutive values of some function. The result depends only on the first and last surviving terms.
General term $u_r$
Examiner keyword
The expression for the $r$ th term of a series, as a function of $r$ . Determining its structure (polynomial vs product-fraction) decides whether to use standard results or differences.
Partial fractions
Examiner keyword
Splitting a single fraction with a factorised denominator into a sum of simpler fractions, e.g. $\tfrac{1}{r(r+1)} = \tfrac{1}{r} - \tfrac{1}{r+1}$ . The standard route to setting up a method-of-differences sum.
Standard results for $\sum$
Examiner keyword
The closed forms $\sum r = \tfrac12 n(n+1)$ , $\sum r^2 = \tfrac16 n(n+1)(2n+1)$ , $\sum r^3 = \tfrac14 n^2(n+1)^2$ , all given in MF19 and quotable without proof.
Linearity of summation
$\sum(a u_r + b v_r) = a\sum u_r + b\sum v_r$ , with $\sum_{r=1}^{n} c = cn$ for a constant. Lets you split a polynomial term and apply the standard results piece by piece.

Common Mistakes and Misconceptions — Method of differences

The traps other students keep falling into on method of differences questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Writing $\sum_{r=1}^{n} c = c$ instead of $cn$
9231 Paper 1 Examiner Reports — summation questions
Why it happens
Treating the constant as a single number rather than a term added once for each of the $n$ values of $r$ .
How to avoid it
Remember a constant is summed $n$ times: $\sum_{r=1}^{n} 1 = n$ , $\sum_{r=1}^{n} 3 = 3n$ .
✕Jumping straight to $f(1) - f(n+1)$ without writing out the terms
9231 Paper 1 Examiner Reports
Why it happens
Over-confidence; memorising the result instead of demonstrating the cancellation.
How to avoid it
Always write the first two-three and last two-three terms explicitly — examiners credit this and it prevents miscounting survivors.
✕On a gap-2 telescope, keeping only one term at each end
9231 Paper 1 mark schemes
Why it happens
Applying the gap-1 result $f(1) - f(n+1)$ when the term is actually $f(r) - f(r+2)$ .
How to avoid it
For $f(r) - f(r+2)$ , TWO terms survive at each end: $f(1) + f(2) - f(n+1) - f(n+2)$ . Write out four terms to confirm.
✕Sign error in the partial-fraction constants, so nothing cancels
9231 Paper 1 Examiner Reports
Why it happens
Cover-up arithmetic slips, especially with a negative substitution value like $r = -1$ .
How to avoid it
Check the split by recombining over the common denominator; if the telescope fails to cancel, the partial fractions are wrong.
✕Leaving the answer as $1 - \frac{1}{n+1}$ instead of $\frac{n}{n+1}$
9231 Paper 1 mark schemes
Why it happens
Stopping at the telescoped form rather than combining into the simplest single fraction the mark scheme wants.
How to avoid it
When the question says 'simplest form' or 'a single fraction', always combine over a common denominator as the last step.
✕Trying standard results on a fraction, or differences on a polynomial
9231 Paper 1 Examiner Reports
Why it happens
Not reading the structure of the general term before starting.
How to avoid it
Polynomial in $r$ → standard results; fraction over a product of factors → method of differences. A 'partial fractions, hence find $\sum$ ' question is always a telescope.

Method of differences

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Summing a linear term with the standard results

2Summing a quadratic term

3A first telescoping sum

4Express in partial fractions, then telescope

5A three-factor denominator (gap-2 telescoping)

6A two-apart cancellation

Sum of the first n natural numbers

Sum of squares

Sum of cubes

Method of differences (gap 1)

Method of differences (gap 2)

Classic partial-fraction splits

Method of differences

Telescoping series

General term uru_rur​

Partial fractions

Standard results for ∑\sum∑

Linearity of summation

✕Writing ∑r=1nc=c\sum_{r=1}^{n} c = c∑r=1n​c=c instead of cncncn

✕Jumping straight to f(1)−f(n+1)f(1) - f(n+1)f(1)−f(n+1) without writing out the terms

✕On a gap-2 telescope, keeping only one term at each end

✕Sign error in the partial-fraction constants, so nothing cancels

✕Leaving the answer as 1−1n+11 - \frac{1}{n+1}1−n+11​ instead of nn+1\frac{n}{n+1}n+1n​

✕Trying standard results on a fraction, or differences on a polynomial

General term $u_r$

Standard results for $\sum$

✕Writing $\sum_{r=1}^{n} c = c$ instead of $cn$

✕Jumping straight to $f(1) - f(n+1)$ without writing out the terms

✕Leaving the answer as $1 - \frac{1}{n+1}$ instead of $\frac{n}{n+1}$