Further Maths/Lessons/Series (Year 2)/Method of Differences
Core Pure · Series · Year 2

The Method of Differences

Split each term into a difference and watch almost the whole sum cancel — only the first and last few terms survive.

13 min read 3 worked examples

The idea behind the method of differences

Method 2.1 — summing a series that telescopes

If the general term $u_r$ of a series can be written as a difference (or similar):

$u_r=f(r)-f(r+1)$

then summing from $r=1$ to $r=n$ causes almost every term to cancel with a neighbouring term, leaving only a handful of terms from the very start and very end of the sum:

$\displaystyle\sum_{r=1}^{n}\big(f(r)-f(r+1)\big)$

$=\big(f(1)-f(2)\big)+\big(f(2)-f(3)\big)+\cdots+\big(f(n)-f(n+1)\big)$

$=f(1)-f(n+1)$

This cancellation pattern is called telescoping. The whole method hinges on finding the right way to split $u_r$ into such a difference — usually done using partial fractions.

Example 1 · A telescoping sum via partial fractions
Show that $\dfrac1{r(r+1)}=\dfrac1r-\dfrac1{r+1}$, and hence find $\displaystyle\sum_{r=1}^{n}\dfrac1{r(r+1)}$.

$\dfrac1r-\dfrac1{r+1}=\dfrac{(r+1)-r}{r(r+1)}=\dfrac1{r(r+1)}$, confirming the identity.

So $\displaystyle\sum_{r=1}^n\dfrac1{r(r+1)}=\sum_{r=1}^n\left(\dfrac1r-\dfrac1{r+1}\right)$. Substitute $r=1,2,3,\ldots,n$ into $\dfrac1r-\dfrac1{r+1}$ and list the results:

$$\begin{aligned} r=1:&\quad \dfrac11-\cancel{\dfrac12}\\ r=2:&\quad \cancel{\dfrac12}-\cancel{\dfrac13}\\ r=3:&\quad \cancel{\dfrac13}-\cancel{\dfrac14}\\ r=n:&\quad \cancel{\dfrac1n}-\dfrac1{n+1} \end{aligned}$$

Adding all $n$ lines, every term cancels with the term diagonally next to it, except the $\dfrac11$ at the very start and the $-\dfrac1{n+1}$ at the very end. So $\displaystyle\sum_{r=1}^n\dfrac1{r(r+1)}=1-\dfrac1{n+1}=\dfrac{n}{n+1}$.

Watch out — check what actually cancels

Always write out the first three or four terms and the last two or three terms explicitly before cancelling — it's easy to misjudge which terms survive, especially when the difference has a coefficient or a shift larger than $1$ (e.g. $f(r)-f(r+2)$ leaves two terms at each end, not one).

Differences with a wider gap

Example 2 · A gap of 2 between terms
Given that $\dfrac1{(2r-1)(2r+1)}=\dfrac12\left(\dfrac1{2r-1}-\dfrac1{2r+1}\right)$, find $\displaystyle\sum_{r=1}^{n}\dfrac1{(2r-1)(2r+1)}$, and state the value of the sum to infinity.

$\displaystyle\sum_{r=1}^n\dfrac1{(2r-1)(2r+1)}=\dfrac12\sum_{r=1}^n\left(\dfrac1{2r-1}-\dfrac1{2r+1}\right)$. Substitute $r=1,2,3,\ldots,n$ into $\dfrac1{2r-1}-\dfrac1{2r+1}$ and list the results:

$$\begin{aligned} r=1:&\quad \dfrac11-\cancel{\dfrac13}\\ r=2:&\quad \cancel{\dfrac13}-\cancel{\dfrac15}\\ r=3:&\quad \cancel{\dfrac15}-\cancel{\dfrac17}\\ r=n:&\quad \cancel{\dfrac1{2n-1}}-\dfrac1{2n+1} \end{aligned}$$

Adding all $n$ lines, only the $\dfrac11$ at the very start and the $-\dfrac1{2n+1}$ at the very end survive: $\displaystyle\sum_{r=1}^n\dfrac1{(2r-1)(2r+1)}=\dfrac12\left(1-\dfrac1{2n+1}\right)=\dfrac12\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}$.

As $n\to\infty$, $\dfrac{n}{2n+1}\to\dfrac12$, so the sum to infinity is $\dfrac12$.

Series that need algebraic manipulation first

Method — when the difference isn't given

If a question doesn't hand you the partial-fraction split, find it yourself using the standard partial fractions technique (from Pure Year 2), then apply the method of differences exactly as above. Sometimes the series won't telescope in its given form at all, and needs an algebraic rearrangement (such as multiplying by a well-chosen factor, or using a trigonometric identity) before a telescoping structure appears — this is common when the terms involve $\cot$, $\csc$, or similar functions.

Example 3 · Finding the partial fractions first
Find $\displaystyle\sum_{r=1}^{n}\dfrac{4}{(2r+1)(2r+3)}$.

Split into partial fractions: $\dfrac4{(2r+1)(2r+3)}=\dfrac A{2r+1}+\dfrac B{2r+3}$. Multiplying through: $4=A(2r+3)+B(2r+1)$. Setting $r=-\dfrac32$: $4=B(-2)\Rightarrow B=-2$. Setting $r=-\dfrac12$: $4=A(2)\Rightarrow A=2$.

So $\dfrac4{(2r+1)(2r+3)}=\dfrac2{2r+1}-\dfrac2{2r+3}$. Substitute $r=1,2,3,\ldots,n$ into $\dfrac2{2r+1}-\dfrac2{2r+3}$ and list the results:

$$\begin{aligned} r=1:&\quad \dfrac23-\cancel{\dfrac25}\\ r=2:&\quad \cancel{\dfrac25}-\cancel{\dfrac27}\\ r=3:&\quad \cancel{\dfrac27}-\cancel{\dfrac29}\\ r=n:&\quad \cancel{\dfrac2{2n+1}}-\dfrac2{2n+3} \end{aligned}$$

Adding all $n$ lines, only the $\dfrac23$ at the very start and the $-\dfrac2{2n+3}$ at the very end survive: $\displaystyle\sum_{r=1}^n\dfrac4{(2r+1)(2r+3)}=\dfrac23-\dfrac2{2n+3}$, which simplifies to $\dfrac{2(2n+3)-6}{3(2n+3)}=\dfrac{4n}{3(2n+3)}$.

Exam-style questions

Question 1 6 marks

Given that $\dfrac1{r(r+2)}=\dfrac12\left(\dfrac1r-\dfrac1{r+2}\right)$, use the method of differences to show that $\displaystyle\sum_{r=1}^{n}\dfrac1{r(r+2)}=\dfrac34-\dfrac{2n+3}{2(n+1)(n+2)}$.

$\displaystyle\sum_{r=1}^n\dfrac1{r(r+2)}=\dfrac12\sum_{r=1}^n\left(\dfrac1r-\dfrac1{r+2}\right)$: since the gap is $2$, two terms survive at each end. (M1)

Substitute $r=1,2,3,\ldots,n$ into $\dfrac1r-\dfrac1{r+2}$ and list the results: (M1)

$$\begin{aligned} r=1:&\quad 1-\cancel{\dfrac13}\\ r=2:&\quad \dfrac12-\cancel{\dfrac14}\\ r=3:&\quad \cancel{\dfrac13}-\cancel{\dfrac15}\\ r=4:&\quad \cancel{\dfrac14}-\cancel{\dfrac16}\\ r=n-1:&\quad \cancel{\dfrac1{n-1}}-\dfrac1{n+1}\\ r=n:&\quad \cancel{\dfrac1n}-\dfrac1{n+2} \end{aligned}$$

Surviving terms: $1,\dfrac12$ from the start and $-\dfrac1{n+1},-\dfrac1{n+2}$ from the end: $\displaystyle\sum_{r=1}^n\dfrac1{r(r+2)}=\dfrac12\left(1+\dfrac12-\dfrac1{n+1}-\dfrac1{n+2}\right)=\dfrac34-\dfrac12\left(\dfrac1{n+1}+\dfrac1{n+2}\right)$. (M1, A1)

$\dfrac12\left(\dfrac1{n+1}+\dfrac1{n+2}\right)=\dfrac12\cdot\dfrac{(n+2)+(n+1)}{(n+1)(n+2)}=\dfrac{2n+3}{2(n+1)(n+2)}$, giving the required result. (M1, A1)

Question 2 7 marks

Show that $\displaystyle\sum_{r=1}^{n}\big(r^3-(r-1)^3\big)=n^3$, and use this result to derive the formula $\displaystyle\sum_{r=1}^{n}(3r^2-3r+1)=n^3$.

Substitute $r=1,2,3,\ldots,n$ into $r^3-(r-1)^3$ and list the results: (M1)

$$\begin{aligned} r=1:&\quad \cancel{1^3}-0^3\\ r=2:&\quad \cancel{2^3}-\cancel{1^3}\\ r=3:&\quad \cancel{3^3}-\cancel{2^3}\\ r=n:&\quad n^3-\cancel{(n-1)^3} \end{aligned}$$

Adding all $n$ lines, only the $0^3$ from the first line and the $n^3$ from the last line survive: $\displaystyle\sum_{r=1}^n\big(r^3-(r-1)^3\big)=n^3-0^3=n^3$. (M1, A1)

Expanding $r^3-(r-1)^3=r^3-(r^3-3r^2+3r-1)=3r^2-3r+1$. (M1, A1)

So $\displaystyle\sum_{r=1}^n(3r^2-3r+1)=\sum_{r=1}^n\big(r^3-(r-1)^3\big)=n^3$, as required. (M1, A1)

Question 3 6 marks

Find $\displaystyle\sum_{r=1}^{n}\dfrac{6}{(3r-2)(3r+1)}$, giving your answer as a single fraction in terms of $n$.

Partial fractions: $\dfrac6{(3r-2)(3r+1)}=\dfrac2{3r-2}-\dfrac2{3r+1}$ (solving $6=A(3r+1)+B(3r-2)$ gives $A=2,B=-2$). (M1, A1)

Substitute $r=1,2,3,\ldots,n$ into $\dfrac2{3r-2}-\dfrac2{3r+1}$ and list the results: (M1)

$$\begin{aligned} r=1:&\quad \dfrac21-\cancel{\dfrac24}\\ r=2:&\quad \cancel{\dfrac24}-\cancel{\dfrac27}\\ r=3:&\quad \cancel{\dfrac27}-\cancel{\dfrac2{10}}\\ r=n:&\quad \cancel{\dfrac2{3n-2}}-\dfrac2{3n+1} \end{aligned}$$

Adding all $n$ lines, only the $\dfrac21$ at the very start and the $-\dfrac2{3n+1}$ at the very end survive: $\displaystyle\sum_{r=1}^n\dfrac6{(3r-2)(3r+1)}=\dfrac21-\dfrac2{3n+1}$. (M1, A1)

$=\dfrac{2(3n+1)-2}{3n+1}=\dfrac{6n}{3n+1}$. (A1)

← Previous
Solving Geometric Problems