Further Maths Series: method of differences Practice Questions

Free Further Maths Series: method of differences practice questions with full step-by-step worked solutions. Covers method-of-differences, telescoping, partial-fractions, sum-to-infinity. Practise exam-style problems and check your method.

method-of-differencestelescopingpartial-fractionssum-to-infinitylimitsnumerical-sum
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Given that 1r(r+1)≡1r−1r+1\frac{1}{r\left(r+1\right)}\equiv\frac{1}{r}-\frac{1}{r+1}, find ∑r=1n1r(r+1)\sum_{r=1}^{n}\frac{1}{r\left(r+1\right)}, giving your answer as a single fraction in terms of nn.
Show worked solution

Worked solution

  1. State the general term as a difference

    ur=1r(r+1)=f(r)−f(r+1)withf(r)=1ru_{r}=\frac{1}{r\left(r+1\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{r}

    This is the form required for the method of differences.

  2. Add the terms and cancel the interior terms

    ∑r=1n1r(r+1)=(1−12)+(12−13)+⋯+(1n−1−1n)+(1n−1n+1)\sum_{r=1}^{n}\frac{1}{r\left(r+1\right)}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  3. Simplify to obtain the closed form

    ∑r=1n1r(r+1)=nn+1\sum_{r=1}^{n}\frac{1}{r\left(r+1\right)}=\frac{n}{n+1}

    Collecting over a common denominator gives a single fraction.

Answer
nn+1\frac{n}{n+1}
Question 2
2 markseasy
Given that 5(r+1)(r+2)≡5r+1−5r+2\frac{5}{\left(r+1\right)\left(r+2\right)}\equiv\frac{5}{r+1}-\frac{5}{r+2}, find ∑r=1n5(r+1)(r+2)\sum_{r=1}^{n}\frac{5}{\left(r+1\right)\left(r+2\right)}, giving your answer as a single fraction in terms of nn.
Show worked solution

Worked solution

  1. Write the general term in partial fractions

    ur=5(r+1)(r+2)≡Ar+1+Br+2u_{r}=\frac{5}{\left(r+1\right)\left(r+2\right)}\equiv\frac{A}{r+1}+\frac{B}{r+2}

    The denominator is a product of two linear factors, so a two-term decomposition exists.

  2. State the general term as a difference

    ur=5(r+1)(r+2)=f(r)−f(r+1)withf(r)=5r+1u_{r}=\frac{5}{\left(r+1\right)\left(r+2\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{5}{r+1}

    This is the form required for the method of differences.

  3. Add the terms and cancel the interior terms

    ∑r=1n5(r+1)(r+2)=(52−53)+(53−54)+⋯+(5n−5n+1)+(5n+1−5n+2)\sum_{r=1}^{n}\frac{5}{\left(r+1\right)\left(r+2\right)}=\left(\frac{5}{2}-\frac{5}{3}\right)+\left(\frac{5}{3}-\frac{5}{4}\right)+\cdots+\left(\frac{5}{n}-\frac{5}{n+1}\right)+\left(\frac{5}{n+1}-\frac{5}{n+2}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  4. Simplify to obtain the closed form

    ∑r=1n5(r+1)(r+2)=5n2(n+2)\sum_{r=1}^{n}\frac{5}{\left(r+1\right)\left(r+2\right)}=\frac{5n}{2\left(n+2\right)}

    Collecting over a common denominator gives a single fraction.

Answer
5n2(n+2)\frac{5n}{2\left(n+2\right)}
Question 3
4 marksintermediate
The general term of a series is ur=5(5r+1)(5r+6)u_{r}=\frac{5}{\left(5r+1\right)\left(5r+6\right)}. Express uru_{r} in partial fractions and hence show that the series converges and find ∑r=1∞5(5r+1)(5r+6)\sum_{r=1}^{\infty}\frac{5}{\left(5r+1\right)\left(5r+6\right)}.
Show worked solution

Worked solution

  1. Write the general term in partial fractions

    ur=5(5r+1)(5r+6)≡A5r+1+B5r+6u_{r}=\frac{5}{\left(5r+1\right)\left(5r+6\right)}\equiv\frac{A}{5r+1}+\frac{B}{5r+6}

    The denominator is a product of two linear factors, so a two-term decomposition exists.

  2. Substitute r=−15r=-\frac{1}{5} to find AA

    5=A(5)⇒A=15=A\left(5\right)\Rightarrow A=1

    This root of the first factor kills the BB term.

  3. Substitute r=−65r=-\frac{6}{5} to find BB

    5=B(−5)⇒B=−15=B\left(-5\right)\Rightarrow B=-1

    This root of the second factor kills the AA term.

  4. State the general term as a difference

    ur=5(5r+1)(5r+6)=f(r)−f(r+1)withf(r)=15r+1u_{r}=\frac{5}{\left(5r+1\right)\left(5r+6\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{5r+1}

    This is the form required for the method of differences.

  5. Write the term when r=nr=n

    f(n)−f(n+1)=15n+1−15n+6f\left(n\right)-f\left(n+1\right)=\frac{1}{5n+1}-\frac{1}{5n+6}

    The last term of the sum; note the second part involves f(n+1)f\left(n+1\right).

  6. Add the terms and cancel the interior terms

    ∑r=1n5(5r+1)(5r+6)=(16−111)+(111−116)+⋯+(15n−4−15n+1)+(15n+1−15n+6)\sum_{r=1}^{n}\frac{5}{\left(5r+1\right)\left(5r+6\right)}=\left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{16}\right)+\cdots+\left(\frac{1}{5n-4}-\frac{1}{5n+1}\right)+\left(\frac{1}{5n+1}-\frac{1}{5n+6}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  7. State the sum to infinity

    ∑r=1∞5(5r+1)(5r+6)=16\sum_{r=1}^{\infty}\frac{5}{\left(5r+1\right)\left(5r+6\right)}=\frac{1}{6}

    The limit exists, so the series converges to this value.

Answer
16\frac{1}{6}
Question 4
6 markshard
The function ff is defined by f(r)=1r(r+1)f(r)=\frac{1}{r\left(r+1\right)}, and f(r)−f(r+1)=2r(r+1)(r+2)f(r)-f(r+1)=\frac{2}{r\left(r+1\right)\left(r+2\right)}. Which of the following is the value of ∑r=182r(r+1)(r+2)\sum_{r=1}^{8}\frac{2}{r\left(r+1\right)\left(r+2\right)}?
Show worked solution

Worked solution

  1. State the function to be used

    f(r)=1r(r+1)f(r)=\frac{1}{r\left(r+1\right)}

    The general term will be written as a difference of values of ff.

  2. Form the difference f(r)−f(r+1)f(r)-f(r+1)

    f(r)−f(r+1)=1r(r+1)−1(r+1)(r+2)f(r)-f(r+1)=\frac{1}{r\left(r+1\right)}-\frac{1}{\left(r+1\right)\left(r+2\right)}

    This is the difference that will telescope.

  3. Put the difference over a common denominator

    f(r)−f(r+1)=2r(r+1)(r+2)f(r)-f(r+1)=\frac{2}{r\left(r+1\right)\left(r+2\right)}

    The two terms combine to give exactly the general term of the series.

  4. State the general term as a difference

    ur=2r(r+1)(r+2)=f(r)−f(r+1)withf(r)=1r(r+1)u_{r}=\frac{2}{r\left(r+1\right)\left(r+2\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{r\left(r+1\right)}

    This is the form required for the method of differences.

  5. Write the term when r=1r=1

    f(1)−f(2)=12−16f\left(1\right)-f\left(2\right)=\frac{1}{2}-\frac{1}{6}

    Each term is the difference of two values of ff.

  6. Write the term when r=2r=2

    f(2)−f(3)=16−112f\left(2\right)-f\left(3\right)=\frac{1}{6}-\frac{1}{12}

    Each term is the difference of two values of ff.

  7. Write the term when r=nr=n

    f(n)−f(n+1)=1n(n+1)−1(n+1)(n+2)f\left(n\right)-f\left(n+1\right)=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}

    The last term of the sum; note the second part involves f(n+1)f\left(n+1\right).

  8. Add the terms and cancel the interior terms

    ∑r=1n2r(r+1)(r+2)=(12−16)+(16−112)+⋯+(1n(n−1)−1n(n+1))+(1n(n+1)−1(n+1)(n+2))\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{12}\right)+\cdots+\left(\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\right)+\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  9. Identify the surviving terms

    ∑r=1n2r(r+1)(r+2)=12−1(n+1)(n+2)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}

    Exactly 1 term survives at each end of the sum.

  10. Simplify to obtain the closed form

    ∑r=1n2r(r+1)(r+2)=n(n+3)2(n+1)(n+2)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}

    Collecting over a common denominator gives a single fraction.

  11. Verify by adding the 8 terms directly

    13+112+⋯+1360=2245\frac{1}{3}+\frac{1}{12}+\cdots+\frac{1}{360}=\frac{22}{45}

    Direct addition of every term reproduces the same value.

  12. Select the option equal to this value

    ∑r=182r(r+1)(r+2)=2245\sum_{r=1}^{8}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{22}{45}

    This is the value of the sum.

Answer
2245\frac{22}{45}
Question 5
9 markschallenging
The function ff is defined by f(r)=1r(r+1)f(r)=\frac{1}{r\left(r+1\right)}. Show that f(r)−f(r+1)=2r(r+1)(r+2)f(r)-f(r+1)=\frac{2}{r\left(r+1\right)\left(r+2\right)} and hence find the value of nn for which ∑r=1n2r(r+1)(r+2)=65132\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{65}{132}.
Show worked solution

Worked solution

  1. State the function to be used

    f(r)=1r(r+1)f(r)=\frac{1}{r\left(r+1\right)}

    The general term will be written as a difference of values of ff.

  2. Form the difference f(r)−f(r+1)f(r)-f(r+1)

    f(r)−f(r+1)=1r(r+1)−1(r+1)(r+2)f(r)-f(r+1)=\frac{1}{r\left(r+1\right)}-\frac{1}{\left(r+1\right)\left(r+2\right)}

    This is the difference that will telescope.

  3. Put the difference over a common denominator

    f(r)−f(r+1)=2r(r+1)(r+2)f(r)-f(r+1)=\frac{2}{r\left(r+1\right)\left(r+2\right)}

    The two terms combine to give exactly the general term of the series.

  4. State the general term as a difference

    ur=2r(r+1)(r+2)=f(r)−f(r+1)withf(r)=1r(r+1)u_{r}=\frac{2}{r\left(r+1\right)\left(r+2\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{r\left(r+1\right)}

    This is the form required for the method of differences.

  5. Write the sum as a difference of two shifted sums

    ∑r=1n2r(r+1)(r+2)=∑r=1nf(r)−∑r=2n+1f(r)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\sum_{r=1}^{n}f(r)-\sum_{r=2}^{n+1}f(r)

    Re-indexing the second sum shows which terms are common to both.

  6. Write the term when r=1r=1

    f(1)−f(2)=12−16f\left(1\right)-f\left(2\right)=\frac{1}{2}-\frac{1}{6}

    Each term is the difference of two values of ff.

  7. Write the term when r=2r=2

    f(2)−f(3)=16−112f\left(2\right)-f\left(3\right)=\frac{1}{6}-\frac{1}{12}

    Each term is the difference of two values of ff.

  8. Write the term when r=n−1r=n-1

    f(n−1)−f(n)=1n(n−1)−1n(n+1)f\left(n-1\right)-f\left(n\right)=\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}

    The penultimate term of the sum.

  9. Write the term when r=nr=n

    f(n)−f(n+1)=1n(n+1)−1(n+1)(n+2)f\left(n\right)-f\left(n+1\right)=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}

    The last term of the sum; note the second part involves f(n+1)f\left(n+1\right).

  10. Add the terms and cancel the interior terms

    ∑r=1n2r(r+1)(r+2)=(12−16)+(16−112)+⋯+(1n(n−1)−1n(n+1))+(1n(n+1)−1(n+1)(n+2))\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{12}\right)+\cdots+\left(\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\right)+\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  11. Identify the surviving terms

    ∑r=1n2r(r+1)(r+2)=12−1(n+1)(n+2)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}

    Exactly 1 term survives at each end of the sum.

  12. Simplify to obtain the closed form

    ∑r=1n2r(r+1)(r+2)=n(n+3)2(n+1)(n+2)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}

    Collecting over a common denominator gives a single fraction.

  13. Set the closed form equal to the given total

    n(n+3)2(n+1)(n+2)=65132\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}=\frac{65}{132}

    This turns the summation into an equation in nn.

  14. Cross-multiply and collect all terms on one side

    n2+3n−130=0n^{2}+3n-130=0

    Clearing the denominator leaves a polynomial equation in nn.

  15. Factorise the polynomial

    (n−10)(n+13)=0\left(n-10\right)\left(n+13\right)=0

    Factorising exposes the roots of the equation.

  16. Check by summing the 10 terms directly

    ∑r=1102r(r+1)(r+2)=65132\sum_{r=1}^{10}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{65}{132}

    Direct addition of the terms reproduces the given total.

  17. State the value of nn

    n=10n=10

    This is the required upper limit.

Answer
n=10n=10

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