Method of Differences: A-Level Further Maths

A long sum can look like a test of endurance. In reality, the method of differences A-Level Further Maths technique is often testing whether you can spot cancellation.

Imagine a sixth-form student faced with a sum containing $n$ complicated terms. Expanding every term feels impossible. Then the first few terms are written out, almost everything disappears, and the problem becomes short. That is the central idea: rewrite the general term as a difference so that consecutive terms cancel.

By the end of this lesson, you should be able to:

• express a term in the form $f(r)-f(r+1)$;
• expand the beginning and end of a finite sum;
• identify the surviving terms;
• find and simplify a formula for the sum;
• handle shifted differences and prove results;
• recognise common traps in exam questions.

What is the method of differences?

The method of differences evaluates certain series by writing each term as the difference between two related expressions. The standard form is

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

Now consider the finite sum

$$\sum _{r=1}^n{\nu }_r=\sum _{r=1}^n\left(f(r)-f(r+1)\right).$$

Writing out the terms gives

$$\left(f(1)-f(2)\right)+\left(f(2)-f(3)\right)+\cdots +\left(f(n)-f(n+1)\right).$$

Every middle expression appears once positively and once negatively. They cancel, leaving

$$\sum _{r=1}^n\left(f(r)-f(r+1)\right)=f(1)-f(n+1).$$

This is called a telescoping sum. Like a collapsing telescope, a long structure contracts into something much shorter.

The key is not memorising one answer. It is learning to see three things: the difference form, the cancellation pattern and the terms at the boundaries.

A reliable method for exam questions

Use this process whenever a question asks you to sum a term by the method of differences.

1. Rewrite the summand as a difference, usually using partial fractions or a supplied identity.
2. Substitute several small values of $r$.
3. Write the first three terms explicitly.
4. Use dots only after the cancellation pattern is clear.
5. Write the final one or two terms using the upper limit.
6. Cancel matching expressions carefully.
7. Simplify the surviving boundary terms.
8. Check the answer with a small value such as $n=1$.

Once this process makes sense, use the Maths Genie A-Level revision lessons and Question Bank to practise identifying the same structure under exam pressure.

Method of differences worked examples

Worked example: consecutive linear factors

Find

$$S_n=\sum _{r=1}^n\frac{1}{r(r+1)}.$$

Begin with partial fractions:

$$\frac{1}{r(r+1)}=\frac{1}{r}-\frac{1}{r+1}.$$

Therefore,

$$S_n=\sum _{r=1}^n\left(\frac{1}{r}-\frac{1}{r+1}\right).$$

Write the first few and final terms:

$$S_n=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{n}-\frac{1}{n+1}\right).$$

The middle fractions cancel:

$$S_n=1-\frac{1}{n+1}.$$

Hence,

$$\boxed{S_n=\frac{n}{n+1}}.$$

A quick check with $n=1$ gives

$$S_1=\frac{1}{1\times 2}=\frac{1}{2},$$

while the formula gives

$$\frac{1}{1+1}=\frac{1}{2}.$$

The check works. It does not prove the formula, but it can expose an indexing error.

Worked example: cancellation across two places

Evaluate

$$T_n=\sum _{r=1}^n\frac{1}{r(r+2)}.$$

Partial fractions give

$$\frac{1}{r(r+2)}=\frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+2}\right).$$

So

$$T_n=\frac{1}{2}\sum _{r=1}^n\left(\frac{1}{r}-\frac{1}{r+2}\right).$$

Expand enough terms to see both ends:

$$T_n=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{n-1}-\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{1}{n+2}\right)\right].$$

This time two terms survive at each end:

$$T_n=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right).$$

Therefore,

$$\boxed{T_n=\frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}}.$$

The important lesson is that the shift controls the number of uncancelled terms. A shift of $1$ usually leaves one term at each boundary. A shift of $2$ usually leaves two.

Worked example: proving a finite sum

Prove that

$$\sum _{r=1}^n\frac{2}{(2r-1)(2r+1)}=1-\frac{1}{2n+1}.$$

First observe that

$$\frac{2}{(2r-1)(2r+1)}=\frac{1}{2r-1}-\frac{1}{2r+1}.$$

The sum becomes

$$\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).$$

All intermediate terms cancel, so

$$\sum _{r=1}^n\frac{2}{(2r-1)(2r+1)}=1-\frac{1}{2n+1},$$

as required.

For a proof question, the phrase “as required” matters less than the logic before it. Show the identity, display the cancellation and state the surviving terms.

How to approach method of differences exam questions

Exam questions often hide the telescoping structure. You may be asked to establish a partial-fraction identity first and then use it to find a sum. Treat that first part as a signpost.

If an identity is supplied, verify its coefficient and shift before using it. Then pay close attention to the limits. For

$$\sum _{r=a}^n\left(f(r)-f(r+1)\right),$$

the answer is

$$f(a)-f(n+1),$$

not automatically $f(1)-f(n+1)$.

For exam-style practice, try this without looking at the outline:

$$\sum _{r=1}^n\frac{1}{(r+1)(r+2)}.$$

Use

$$\frac{1}{(r+1)(r+2)}=\frac{1}{r+1}-\frac{1}{r+2}.$$

Your final result should be

$$\boxed{\frac{1}{2}-\frac{1}{n+2}}.$$

Then change the lower limit from $1$ to $3$ and repeat the process. This small alteration tests whether you understand the boundaries rather than merely remember a formula. You can build fluency through A-Level past papers, checking each solution against the relevant mark scheme.

Common mistakes

Cancelling before expanding

It is tempting to jump from sigma notation to an answer. Write enough terms to make the pattern visible. This is especially important when the shift is greater than $1$.

Losing a partial-fraction coefficient

For example,

$$\frac{1}{r(r+2)}\ne \frac{1}{r}-\frac{1}{r+2}.$$

The correct identity includes a factor of $\frac{1}{2}$. Recombine the fractions if you are uncertain.

Using the wrong final index

If $r=n$, then $r+1=n+1$ and $r+2=n+2$. Substitute the upper limit directly rather than guessing the last denominator.

Cancelling terms that do not match exactly

Only identical expressions with opposite signs cancel. A term such as $\frac{1}{n}$ cannot cancel with $\frac{1}{n+1}$.

Ignoring the lower limit

A sum beginning at $r=3$ has a different first surviving term from one beginning at $r=1$. Circle both limits before expanding.

From spotting the pattern to earning the marks

Method of differences becomes manageable when you stop seeing a long series and start seeing its edges. Rewrite the term, expose the cancellation, and protect the boundary terms. That is the whole architecture of the method.

Knowing the pattern is one step; recognising it quickly in unfamiliar questions is the next. Start the matching Maths Genie lesson and Question Bank practice on the A-Level Maths revision page, then work through method of differences questions until writing the boundary terms feels automatic.