All subjects · Further Mathematics · WFM02 Further Pure Mathematics 2

Method of differences: gaps, three factors and limits

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

Exam questions push the method further: terms that cancel two rows apart, three-factor denominators, sums to infinity, and finding the smallest n for which a sum passes a value.

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Cancelling two rows apart

1/(r(r + 2)) = ½[1/r - 1/(r + 2)]. Now each negative term cancels with the positive term two rows later.

So two terms survive at the start (1 and 1/2) and two at the end (-1/(n + 1) and -1/(n + 2)):

Σ from 1 to n of 1/(r(r + 2)) = ½[1 + 1/2 - 1/(n + 1) - 1/(n + 2)].

Three factors

1/(r(r + 1)(r + 2)) can be written as ½[1/(r(r + 1)) - 1/((r + 1)(r + 2))], which is f(r) - f(r + 1) with f(r) = 1/(2r(r + 1)). The sum is 1/4 - 1/(2(n + 1)(n + 2)).

If you use three partial fractions A/r + B/(r + 1) + C/(r + 2) instead, look carefully at which terms cancel diagonally: the June 2025 examiners' report notes that some students missed cancellations.

Sums to infinity

If the surviving end terms tend to 0 as n → ∞, the series converges and its sum is what is left. For 1/(r(r + 2)), the sum to infinity is ½(1 + 1/2) = 3/4.

Finding n

To find the smallest n with S_n > k, write the inequality using your formula and solve it. n must be a positive integer, and a strict inequality matters: in June 2025 a significant number of students did not recognise that n cannot be negative, and others lost the mark by ignoring the strict inequality (examiners' report).

Worked example

Find Σ from r = 1 to n of 1/(r(r + 2)) in the form 3/4 - (an + b)/(2(n + 1)(n + 2)).

  1. 1/(r(r + 2)) = ½[1/r - 1/(r + 2)].
  2. Rows: r = 1: 1 - 1/3; r = 2: 1/2 - 1/4; r = 3: 1/3 - 1/5; ...; r = n - 1: 1/(n - 1) - 1/(n + 1); r = n: 1/n - 1/(n + 2).
  3. Survivors: ½[1 + 1/2 - 1/(n + 1) - 1/(n + 2)].
  4. 1/(n + 1) + 1/(n + 2) = (2n + 3)/((n + 1)(n + 2)).

Answer: 3/4 - (2n + 3)/(2(n + 1)(n + 2))

Key idea: If the gap in the partial fractions is k, then k terms survive at each end. Write enough rows to see it.

Check yourself

  1. Find the exact value of Σ from r = 1 to 10 of 1/(r(r + 2)). Give your answer as a fraction in its simplest form.
    Show answer

    175/264. 3/4 - (2(10) + 3)/(2 × 11 × 12) = 3/4 - 23/264 = 175/264.

  2. Find the sum to infinity of 1/(r(r + 2)), from r = 1. Give your answer as a fraction in its simplest form.
    Show answer

    3/4. As n → ∞, 1/(n + 1) and 1/(n + 2) tend to 0, leaving ½(1 + 1/2) = 3/4.

  3. Find the exact value of Σ from r = 1 to 8 of 1/(r(r + 1)(r + 2)). Give your answer as a fraction in its simplest form.
    Show answer

    11/45. 1/4 - 1/(2 × 9 × 10) = 1/4 - 1/180 = 11/45.

  4. Find the exact value of Σ from r = 1 to 7 of 4/((r + 1)(r + 3)). Give your answer as a fraction in its simplest form.
    Show answer

    56/45. 4/((r + 1)(r + 3)) = 2[1/(r + 1) - 1/(r + 3)]. Survivors: 2[1/2 + 1/3 - 1/9 - 1/10] = 5/3 - 2/9 - 1/5 = 56/45.

  5. Which identity lets you prove the formula for Σ r3 by the method of differences?
    • (r + 1)3 - r3 = 3r2
    • (r + 1)2 - r2 = 2r
    • r2(r + 1)2 - (r - 1)2 r2 = 4r3
    • r4 - (r - 1)4 = 4r3
    Show answer

    r2(r + 1)2 - (r - 1)2 r2 = 4r3. Summing r2(r + 1)2 - (r - 1)2 r2 from 1 to n telescopes to n2(n + 1)2, so Σr3 = n2(n + 1)2/4.

  6. Σ from r = 1 to n of ln((r + 1)/r) = ln(n + 1). Evaluate the sum for n = 99, to 3 decimal places.
    Show answer

    4.605. ln((r + 1)/r) = ln(r + 1) - ln r, which telescopes to ln(n + 1) - ln 1 = ln 100.

  7. S_n = Σ from r = 1 to n of 1/(r(r + 1)(r + 2)) = 1/4 - 1/(2(n + 1)(n + 2)). Find the smallest n for which S_n > 0.249.
    Show answer

    21. 1/(2(n + 1)(n + 2)) < 0.001 gives (n + 1)(n + 2) > 500. n = 20 gives 462, n = 21 gives 506, so n = 21.

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