Series: method of differences Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Series: method of differences questions. See exactly how to solve problems on method-of-differences, telescoping, partial-fractions, sum-to-infinity.

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.

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 1(r+1)(r+2)≡1r+1−1r+2\frac{1}{\left(r+1\right)\left(r+2\right)}\equiv\frac{1}{r+1}-\frac{1}{r+2}, find ∑r=1n1(r+1)(r+2)\sum_{r=1}^{n}\frac{1}{\left(r+1\right)\left(r+2\right)}, giving your answer as a single fraction in terms of nn.

Worked solution

  1. Write the general term in partial fractions

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

    This is the form required for the method of differences.

  3. Add the terms and cancel the interior terms

    ∑r=1n1(r+1)(r+2)=(12−13)+(13−14)+⋯+(1n−1n+1)+(1n+1−1n+2)\sum_{r=1}^{n}\frac{1}{\left(r+1\right)\left(r+2\right)}=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)+\left(\frac{1}{n+1}-\frac{1}{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=1n1(r+1)(r+2)=n2(n+2)\sum_{r=1}^{n}\frac{1}{\left(r+1\right)\left(r+2\right)}=\frac{n}{2\left(n+2\right)}

    Collecting over a common denominator gives a single fraction.

Answer
n2(n+2)\frac{n}{2\left(n+2\right)}
Question 3
2 markseasy
Given that 2(2r−1)(2r+1)≡12r−1−12r+1\frac{2}{\left(2r-1\right)\left(2r+1\right)}\equiv\frac{1}{2r-1}-\frac{1}{2r+1}, find ∑r=1n2(2r−1)(2r+1)\sum_{r=1}^{n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}, giving your answer as a single fraction in terms of nn.

Worked solution

  1. Write the general term in partial fractions

    ur=2(2r−1)(2r+1)≡A2r−1+B2r+1u_{r}=\frac{2}{\left(2r-1\right)\left(2r+1\right)}\equiv\frac{A}{2r-1}+\frac{B}{2r+1}

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

  2. State the general term as a difference

    ur=2(2r−1)(2r+1)=f(r)−f(r+1)withf(r)=12r−1u_{r}=\frac{2}{\left(2r-1\right)\left(2r+1\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{2r-1}

    This is the form required for the method of differences.

  3. Add the terms and cancel the interior terms

    ∑r=1n2(2r−1)(2r+1)=(1−13)+(13−15)+⋯+(12n−3−12n−1)+(12n−1−12n+1)\sum_{r=1}^{n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{2n-3}-\frac{1}{2n-1}\right)+\left(\frac{1}{2n-1}-\frac{1}{2n+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.

  4. Simplify to obtain the closed form

    ∑r=1n2(2r−1)(2r+1)=2n2n+1\sum_{r=1}^{n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}=\frac{2n}{2n+1}

    Collecting over a common denominator gives a single fraction.

Answer
2n2n+1\frac{2n}{2n+1}
Question 4
2 markseasy
Given that 1(r+2)(r+3)≡1r+2−1r+3\frac{1}{\left(r+2\right)\left(r+3\right)}\equiv\frac{1}{r+2}-\frac{1}{r+3}, find ∑r=1n1(r+2)(r+3)\sum_{r=1}^{n}\frac{1}{\left(r+2\right)\left(r+3\right)}, giving your answer as a single fraction in terms of nn.

Worked solution

  1. State the general term as a difference

    ur=1(r+2)(r+3)=f(r)−f(r+1)withf(r)=1r+2u_{r}=\frac{1}{\left(r+2\right)\left(r+3\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{r+2}

    This is the form required for the method of differences.

  2. Add the terms and cancel the interior terms

    ∑r=1n1(r+2)(r+3)=(13−14)+(14−15)+⋯+(1n+1−1n+2)+(1n+2−1n+3)\sum_{r=1}^{n}\frac{1}{\left(r+2\right)\left(r+3\right)}=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+\left(\frac{1}{n+2}-\frac{1}{n+3}\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=1n1(r+2)(r+3)=n3(n+3)\sum_{r=1}^{n}\frac{1}{\left(r+2\right)\left(r+3\right)}=\frac{n}{3\left(n+3\right)}

    Collecting over a common denominator gives a single fraction.

Answer
n3(n+3)\frac{n}{3\left(n+3\right)}
Question 5
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}, show that the series converges and find ∑r=1∞1r(r+1)\sum_{r=1}^{\infty}\frac{1}{r\left(r+1\right)}.

Worked solution

  1. Write the general term in partial fractions

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

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

  2. 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.

  3. 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.

  4. State the sum to infinity

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

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

Answer
11

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