All subjects · Further Mathematics · WFM02 Further Pure Mathematics 2

The method of differences

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

Some series collapse when you add them up: nearly every term cancels with a neighbour. Writing the general term as a difference f(r) - f(r + 1) makes this happen and gives a formula for the sum.

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The idea

If u_r = f(r) - f(r + 1), then

u_1 + u_2 + ... + u_n = [f(1) - f(2)] + [f(2) - f(3)] + ... + [f(n) - f(n + 1)] = f(1) - f(n + 1).

Everything in the middle cancels. This is called a telescoping sum.

Using partial fractions

Rational terms are split first: 1/(r(r + 1)) = 1/r - 1/(r + 1).

Then list the rows r = 1, 2, 3, ..., n - 1, n underneath each other, so you can see which terms cancel and which survive at the start and the end.

Layout and limits

Write the first three rows and the last two rows in full. The scheme of work stresses careful layout, because poor layout can lead you to miscopy your own work.

For a sum from r = k to n, use Σ from k to n = Σ from 1 to n minus Σ from 1 to k - 1, or start your rows at r = k.

When the answer is given ("show that"), show every step, including combining the surviving terms over a common denominator.

Using the identity in reverse

Differences also prove standard results. Since (r + 1)2 - r2 = 2r + 1, summing from 1 to n gives (n + 1)2 - 1 = 2Σr + n, so Σr = n(n + 1)/2.

Worked example

Find Σ from r = 1 to n of 1/(r(r + 1)).

  1. 1/(r(r + 1)) = 1/r - 1/(r + 1).
  2. r = 1: 1 - 1/2; r = 2: 1/2 - 1/3; r = 3: 1/3 - 1/4; ...; r = n: 1/n - 1/(n + 1).
  3. All terms cancel except the first and the last: 1 - 1/(n + 1).

Answer: n/(n + 1)

Key idea: Write the term as f(r) - f(r + 1), list rows to see the cancelling, and keep only the terms that survive at each end.

Check yourself

  1. Find the exact value of Σ from r = 1 to 20 of 1/(r(r + 1)). Give your answer as a fraction in its simplest form.
    Show answer

    20/21. Using n/(n + 1) with n = 20 gives 20/21.

  2. Use 1/((2r - 1)(2r + 1)) = ½[1/(2r - 1) - 1/(2r + 1)] to find Σ from r = 1 to 10 of 1/((2r - 1)(2r + 1)). Give your answer as a fraction in its simplest form.
    Show answer

    10/21. The sum telescopes to ½[1 - 1/(2n + 1)] = n/(2n + 1). With n = 10: 10/21.

  3. Find Σ from r = 11 to 30 of 1/(r(r + 1)). Give your answer as a fraction in its simplest form.
    Show answer

    20/341. The terms from r = 11 to 30 leave 1/11 - 1/31 = 20/341.

  4. Which expression equals Σ from r = 1 to n of [(r + 1)2 - r2]?
    • n2 - 1
    • (n + 1)2 - 1
    • (n + 1)2
    • n2
    Show answer

    (n + 1)2 - 1. It telescopes: the survivors are (n + 1)2 from the last row and -12 from the first row.

  5. Given 1/r2 - 1/(r + 1)2 = (2r + 1)/(r2(r + 1)2), find Σ from r = 1 to n of (2r + 1)/(r2(r + 1)2).
    • 1 - 1/n2
    • 1/(n + 1)2
    • 1 + 1/(n + 1)2
    • 1 - 1/(n + 1)2
    Show answer

    1 - 1/(n + 1)2. The difference telescopes to 1/12 - 1/(n + 1)2.

  6. Find the smallest integer n for which Σ from r = 1 to n of 1/(r(r + 1)) > 0.95.
    Show answer

    20. n/(n + 1) > 0.95 gives n > 0.95n + 0.95, so 0.05n > 0.95, n > 19. The inequality is strict, so n = 20.

  7. When Σ from r = 1 to n of [1/r - 1/(r + 1)] is listed row by row, which terms survive?
    • 1 and -1/n
    • 1/2 and -1/(n + 1)
    • Only 1
    • 1 from the first row and -1/(n + 1) from the last row
    Show answer

    1 from the first row and -1/(n + 1) from the last row. Each -1/(r + 1) cancels the +1/(r + 1) of the next row, except in the last row.

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Last updated 8 November 2026. Spotted a mistake? Tell us.