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Method of differencesEdexcel A-Level Further Maths: Revision notes

Section 1

The idea: telescoping sums

If each term can be written as a difference ur=f(r)−f(r+1)u_r=f(r)-f(r+1), then adding the terms makes the middle ones cancel in pairs: ∑r=1n[f(r)−f(r+1)]=f(1)−f(n+1).\sum_{r=1}^{n}\left[f(r)-f(r+1)\right]=f(1)-f(n+1). This is the method of differences. Example: (r+1)3−r3=3r2+3r+1(r+1)^3-r^3=3r^2+3r+1, so ∑r=1n(3r2+3r+1)=(n+1)3−1\sum_{r=1}^{n}(3r^2+3r+1)=(n+1)^3-1. The same idea gives sums over any range: ∑r=1120[(r+1)3−r3]=213−113\sum_{r=11}^{20}\left[(r+1)^3-r^3\right]=21^3-11^3.

Key termsmethod of differencestelescoping
Common mistake

Mixing up the surviving terms: the first term of the first bracket and the last term of the last bracket remain, i.e. f(1)−f(n+1)f(1)-f(n+1), not f(1)−f(n)f(1)-f(n).

Exam tip

Always write out the first two or three and last two or three differences. Cancelling visibly earns the method mark.

Section 2

Using partial fractions

Many fractions are not given as a difference, so split them first. For 1r(r+1)\frac1{r(r+1)} the partial fractions are 1r−1r+1\frac1r-\frac1{r+1}, giving ∑r=1n1r(r+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1=nn+1.\sum_{r=1}^{n}\frac1{r(r+1)}=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)=1-\frac1{n+1}=\frac{n}{n+1}. To find the constants for 1(r+1)(r+3)=Ar+1+Br+3\frac1{(r+1)(r+3)}=\frac A{r+1}+\frac B{r+3}, write 1=A(r+3)+B(r+1)1=A(r+3)+B(r+1) and substitute r=−1r=-1 and r=−3r=-3: A=12A=\frac12, B=−12B=-\frac12.

Key termspartial fractions
Common mistake

Forgetting the factor 12\frac12 in 1(r+1)(r+3)=12(1r+1−1r+3)\frac1{(r+1)(r+3)}=\frac12\left(\frac1{r+1}-\frac1{r+3}\right).

Section 3

When the gap is more than 1

For 12(1r+1−1r+3)\frac12\left(\frac1{r+1}-\frac1{r+3}\right) the terms cancel two places apart, so four terms survive: two from the start and two from the end. ∑r=1n1(r+1)(r+3)=12[12+13−1n+2−1n+3]=512−2n+52(n+2)(n+3).\sum_{r=1}^{n}\frac1{(r+1)(r+3)}=\frac12\left[\frac12+\frac13-\frac1{n+2}-\frac1{n+3}\right]=\frac5{12}-\frac{2n+5}{2(n+2)(n+3)}. Three-factor denominators work the same way. 1r(r+1)(r+2)=12[1r(r+1)−1(r+1)(r+2)]\frac1{r(r+1)(r+2)}=\frac12\left[\frac1{r(r+1)}-\frac1{(r+1)(r+2)}\right], so the sum is 12[12−1(n+1)(n+2)]=n(n+3)4(n+1)(n+2)\frac12\left[\frac12-\frac1{(n+1)(n+2)}\right]=\frac{n(n+3)}{4(n+1)(n+2)}.

Key termssurviving terms
Exam tip

Count the survivors: a gap of kk between the two fractions leaves kk terms at the start and kk at the end.

Section 4

Squares and sums to infinity

Other differences also telescope. Since 1r2−1(r+1)2=2r+1r2(r+1)2\frac1{r^2}-\frac1{(r+1)^2}=\frac{2r+1}{r^2(r+1)^2}, we get ∑r=1n2r+1r2(r+1)2=1−1(n+1)2\sum_{r=1}^{n}\frac{2r+1}{r^2(r+1)^2}=1-\frac1{(n+1)^2}. Once the sum is in closed form, you can let n→∞n\to\infty: the leftover fractions tend to 0, so ∑r=1∞1r(r+1)=1\sum_{r=1}^{\infty}\frac1{r(r+1)}=1 and ∑r=1∞2r+1r2(r+1)2=1\sum_{r=1}^{\infty}\frac{2r+1}{r^2(r+1)^2}=1. To sum from r=ar=a rather than 1, start the list at aa: ∑r=10191r(r+1)=110−120=120\sum_{r=10}^{19}\frac1{r(r+1)}=\frac1{10}-\frac1{20}=\frac1{20}.

Exam tip

For a 'show that' with a given answer, combine the surviving terms over a common denominator and factorise the numerator.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Method of differences

  1. Let ur=1r(r+1)u_r=\frac{1}{r(r+1)}.
    Find the value of ∑r=1019ur\sum_{r=10}^{19}u_r.2 marks
  2. For all values of rr, (r+1)3−r3=3r2+3r+1(r+1)^3-r^3=3r^2+3r+1.
    Hence find the value of ∑r=1120(3r2+3r+1)\sum_{r=11}^{20}\left(3r^2+3r+1\right).2 marks
  3. Let ur=1(r+1)(r+3)u_r=\frac{1}{(r+1)(r+3)}.
    Express uru_r in the form Ar+1+Br+3\frac{A}{r+1}+\frac{B}{r+3}, where AA and BB are constants to be found.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).