Home · A-Level Further Maths 9231 · Summation of series · The method of differences · Practice
AS & A Level · Practice questions · The method of differences

Split, list, cancel.

Six original Cambridge-style questions on the method of differences: using a given difference, partial fractions for the general term, two survivors at each end, limits that do not start at 1, and deriving a standard result.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Split the term, write the rows, collect the survivors.

01
Recall
[2 marks]

Using \(\dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r + 1}\), find the exact value of \(\displaystyle\sum_{r=1}^{20} \dfrac{1}{r(r + 1)}\).

  • The rows \(\left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \dots + \left(\tfrac{1}{20} - \tfrac{1}{21}\right)\) cancel to \(1 - \dfrac{1}{21}\). ✓
  • \(= \dfrac{20}{21}\). ✓
02
Calculation
[4 marks]

(a) Express \(\dfrac{1}{(3r - 1)(3r + 2)}\) in partial fractions.

(b) Hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{(3r - 1)(3r + 2)}\), giving your answer as a single fraction in terms of \(n\).

  • (a) \(1 = A(3r + 2) + B(3r - 1)\). \(r = \tfrac{1}{3}\): \(1 = 3A\), \(A = \tfrac{1}{3}\). \(r = -\tfrac{2}{3}\): \(1 = -3B\), \(B = -\tfrac{1}{3}\). ✓
  • So \(\dfrac{1}{(3r - 1)(3r + 2)} = \dfrac{1}{3}\left(\dfrac{1}{3r - 1} - \dfrac{1}{3r + 2}\right)\); note \(3(r + 1) - 1 = 3r + 2\), so this is \(f(r) - f(r + 1)\) with \(f(r) = \dfrac{1}{3(3r - 1)}\). ✓
  • (b) Rows: \(\dfrac{1}{3}\left[\left(\tfrac{1}{2} - \tfrac{1}{5}\right) + \left(\tfrac{1}{5} - \tfrac{1}{8}\right) + \dots + \left(\tfrac{1}{3n - 1} - \tfrac{1}{3n + 2}\right)\right]\). ✓
  • Survivors: \(\dfrac{1}{3}\left(\dfrac{1}{2} - \dfrac{1}{3n + 2}\right) = \dfrac{1}{3} \times \dfrac{3n}{2(3n + 2)} = \dfrac{n}{2(3n + 2)}\). ✓
03
Calculation
[5 marks]

Find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r^{2} + 4r + 3}\), giving your answer in the form \(\dfrac{5}{12} - \dfrac{an + b}{2(n + 2)(n + 3)}\), where \(a\) and \(b\) are integers.

  • \(r^{2} + 4r + 3 = (r + 1)(r + 3)\), and \(\dfrac{1}{(r + 1)(r + 3)} = \dfrac{1}{2}\left(\dfrac{1}{r + 1} - \dfrac{1}{r + 3}\right)\). ✓
  • Rows: \(\dfrac{1}{2}\left[\left(\tfrac{1}{2} - \tfrac{1}{4}\right) + \left(\tfrac{1}{3} - \tfrac{1}{5}\right) + \left(\tfrac{1}{4} - \tfrac{1}{6}\right) + \dots + \left(\tfrac{1}{n} - \tfrac{1}{n + 2}\right) + \left(\tfrac{1}{n + 1} - \tfrac{1}{n + 3}\right)\right]\). ✓
  • A gap of two: survivors \(\dfrac{1}{2}\left(\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{n + 2} - \dfrac{1}{n + 3}\right)\). ✓
  • \(= \dfrac{5}{12} - \dfrac{1}{2} \times \dfrac{(n + 3) + (n + 2)}{(n + 2)(n + 3)} = \dfrac{5}{12} - \dfrac{2n + 5}{2(n + 2)(n + 3)}\). ✓
  • So \(a = 2\), \(b = 5\). Check \(n = 1\): \(\dfrac{5}{12} - \dfrac{7}{24} = \dfrac{1}{8}\), and the first term is \(\dfrac{1}{1 + 4 + 3} = \dfrac{1}{8}\). ✓
04
Analysis
[5 marks]

(a) Express \(\dfrac{1}{r(r + 1)(r + 2)}\) in partial fractions.

(b) Show that \(\dfrac{1}{r(r + 1)(r + 2)} = f(r) - f(r + 1)\), where \(f(r) = \dfrac{1}{2r(r + 1)}\).

(c) Hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 1)(r + 2)}\).

  • (a) \(1 = A(r + 1)(r + 2) + Br(r + 2) + Cr(r + 1)\). \(r = 0\): \(A = \tfrac{1}{2}\); \(r = -1\): \(1 = -B\), \(B = -1\); \(r = -2\): \(1 = 2C\), \(C = \tfrac{1}{2}\). ✓
  • So \(\dfrac{1}{r(r + 1)(r + 2)} = \dfrac{1}{2r} - \dfrac{1}{r + 1} + \dfrac{1}{2(r + 2)}\). ✓
  • (b) \(f(r) - f(r + 1) = \dfrac{1}{2r(r + 1)} - \dfrac{1}{2(r + 1)(r + 2)} = \dfrac{(r + 2) - r}{2r(r + 1)(r + 2)} = \dfrac{1}{r(r + 1)(r + 2)}\). ✓
  • (c) The sum telescopes to \(f(1) - f(n + 1) = \dfrac{1}{4} - \dfrac{1}{2(n + 1)(n + 2)}\). ✓
  • Check \(n = 1\): \(\dfrac{1}{4} - \dfrac{1}{12} = \dfrac{1}{6} = \dfrac{1}{1 \times 2 \times 3}\). ✓
05
Analysis
[4 marks]

Show that \(\displaystyle\sum_{r=n+1}^{2n} \dfrac{1}{r(r + 1)} = \dfrac{n}{(n + 1)(2n + 1)}\), and deduce that this sum is less than \(\dfrac{1}{2}\) for every positive integer \(n\).

  • \(\dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r + 1}\). Rows from \(r = n + 1\) to \(r = 2n\): \(\left(\tfrac{1}{n + 1} - \tfrac{1}{n + 2}\right) + \dots + \left(\tfrac{1}{2n} - \tfrac{1}{2n + 1}\right)\). ✓
  • Survivors: \(\dfrac{1}{n + 1} - \dfrac{1}{2n + 1} = \dfrac{(2n + 1) - (n + 1)}{(n + 1)(2n + 1)} = \dfrac{n}{(n + 1)(2n + 1)}\). ✓
  • The denominators are positive, so the sum is less than \(\dfrac{1}{2}\) exactly when \(2n < (n + 1)(2n + 1) = 2n^{2} + 3n + 1\). ✓
  • That is \(2n^{2} + n + 1 > 0\), true for every positive \(n\), so the sum is always less than \(\dfrac{1}{2}\). ✓
06
Proof
[6 marks]

(a) Show that \((r + 1)^{4} - r^{4} = 4r^{3} + 6r^{2} + 4r + 1\).

(b) Using the method of differences and the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^{2}\), prove that \(\displaystyle\sum_{r=1}^{n} r^{3} = \tfrac{1}{4}n^{2}(n + 1)^{2}\).

  • (a) \((r + 1)^{4} = r^{4} + 4r^{3} + 6r^{2} + 4r + 1\) by the binomial expansion; subtract \(r^{4}\). ✓
  • (b) Sum the left side for \(r = 1\) to \(n\): \((2^{4} - 1^{4}) + (3^{4} - 2^{4}) + \dots + ((n + 1)^{4} - n^{4}) = (n + 1)^{4} - 1\). ✓
  • Sum the right side, with \(S = \displaystyle\sum_{r=1}^{n} r^{3}\): \(4S + n(n + 1)(2n + 1) + 2n(n + 1) + n\). ✓
  • So \(4S = (n + 1)^{4} - (n + 1) - n(n + 1)(2n + 1) - 2n(n + 1)\), using \(1 + n = n + 1\). ✓
  • Take out \((n + 1)\): \(4S = (n + 1)\left[(n + 1)^{3} - 1 - n(2n + 1) - 2n\right] = (n + 1)\left(n^{3} + 3n^{2} + 3n - 2n^{2} - 3n\right)\). ✓
  • \(= (n + 1)(n^{3} + n^{2}) = n^{2}(n + 1)^{2}\), so \(S = \tfrac{1}{4}n^{2}(n + 1)^{2}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.