Method of Differences

What is the Method of Differences?

The Method of Differences is a way of turning longer and more complicated sums into shorter and simpler ones
Sometimes when summing series, you will notice that many of the terms (or parts of the terms) simply “cancel out” or eliminate each other

This can turn a very long series summation, into a much simpler shorter one

In the case of $\sum_{r = 1}^{n} (f (r) - f (r + 1))$

$f (1) - f (2)$
$f (2) - f (3)$
$f (3) - f (4)$
$f (4) - f (5)$
(and so on, until…)
$f (n - 1) - f (n)$

This is the penultimate term

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

This is the last term

You can see that when these are summed, most of the terms will cancel out

This leaves just $f (1) - f (n + 1)$
We can say that $\sum_{r = 1}^{n} (f (r) - f (r + 1)) = f (1) - f (n + 1)$

How can I use partial fractions along with the method of differences?

You will often need to use partial fractions to change the general term into a sum of two or three terms, rather than a single fraction
- For example, $\frac{6}{(r + 1) (r + 3)}$ can be rewritten as $\frac{3}{r + 1} - \frac{3}{r + 3}$
This may lead to a more interesting pattern of cancellations than was seen for $f (r) - f (r + 1)$
For example, $\sum_{r = 1}^{n} \frac{3}{r + 1} - \frac{3}{r + 3}$ can be written as $\sum_{r = 1}^{n} f (r) - f (r + 2)$ , where $f (r) = \frac{3}{r + 1}$ , and the terms can then be listed as:
- - $f (1) - f (3)$
  - $f (2) - f (4)$
  - $f (3) - f (5)$
  - $f (4) - f (6)$
  - $f (5) - f (7)$
  - (and so on, until…)
  - $f (n - 1) - f (n + 1)$
  - $f (n) - f (n + 2)$
- When these are summed, it will just leave $f (1) + f (2) - f (n + 1) - f (n + 2)$
- You can then evaluate this expression with $f (r) = \frac{3}{r + 1}$ to get to your final answer
It is helpful to use $f (r)$ notation to spot the pattern, rather than substituting $r = 1, r = 2, . . .$ into the expression every time, especially with more complicated expressions
- You need to consider carefully which term to make $f (r)$ and then how the other terms in the expression relate to it
- If this is difficult for a particular expression, it may be more straightforward to substitute $r = 1, r = 2, . . .$ into each term in the series and spot any patterns that way
- This is essentially writing the series out in full until you spot which terms will cancel
- In your working, however, you should still write out the last two or three terms in terms of $n, n - 1, n - 2$ , etc.

How can I use the method of differences for series with expressions containing more than two terms?

The general term of the series may have more than two terms, which can sometimes make spotting which terms will cancel more challenging

For example, $\sum_{r = 1}^{n} \frac{2}{r} - \frac{3}{r + 1} + \frac{1}{r + 2}$
This can be written as $2 f (r) - 3 f (r + 1) + f (r + 2)$ where $f (r) = \frac{1}{r}$
Writing out the first five terms and the last three terms we get:

$2 f (1) - 3 f (2) + f (3)$
$2 f (2) - 3 f (3) + f (4)$
$2 f (3) - 3 f (4) + f (5)$
$2 f (4) - 3 f (5) + f (6)$
$2 f (5) - 3 f (6) + f (7)$
(and so on until…)
$2 f (n - 2) - 3 f (n - 1) + f (n)$
$2 f (n - 1) - 3 (f n) + f (n + 1)$
$2 f (n) - 3 f (n + 1) + f (n + 2)$

In this case, look at the diagonals starting at the top right

We have $f (3), - 3 f (3)$ and $2 f (3)$ which sum to $0$
This pattern repeats for the other diagonals

We will eventually be left with only
$2 f (1) - 3 f (2) + 2 f (2) + f (n + 1) - 3 f (n + 1) + f (n + 2) = 2 f (1) - f (2) - 2 f (n + 1) + f (n + 2)$
Evaluating this with $f (r) = \frac{1}{r}$ results in an answer of $\frac{3}{2} - \frac{2}{n + 1} + \frac{1}{n + 2}$

What other uses are there for the method of differences?

Method of differences can also be used to prove the formulae for the sum of $r, r^{2}$ and $r^{3}$

For example, this can be used to prove that the result for the sum of squares is indeed $\frac{1}{6} n (n + 1) (2 n + 1)$

By expanding brackets it can be shown that ${(2 r + 1)}^{3} - {(2 r - 1)}^{3} = 24 r^{2} + 2$ , and then the sum of both sides of that equation from $r = 1$ to $r = n$ can be considered
$\sum_{r = 1}^{n} ({(2 r + 1)}^{3} - {(2 r - 1)}^{3}) = {\sum (24}_{r = 1}^{n} r^{2} + 2)$
The left-hand side can be found in terms of $n$ using method of differences, and the right-hand side can be rearranged to give $2 n + 24 {\sum r^{2}}_{r = 1}^{n}$
That equation can then be rearranged to give an expression for ${\sum r^{2}}_{r = 1}^{n}$
When these proofs have appeared previously in exams, they have tended to be structured to help you work through the steps

You may have to use your algebraic method of differences result to find a numerical answer, usually in the last part of a question

The question will often ask you to evaluate the sum starting from $r = 20$ (or some other arbitrary value) rather than from $r = 1$
To help with this, remember that: $\sum_{r = p}^{q} u_{r} = \sum_{r = 1}^{q} u_{r} - \sum_{r = 1}^{p - 1} u_{r}$
You may also find it helpful to recall that for constants $a$ and $b$ : $\sum_{r = p}^{q} (a u_{r} + b) = {a \sum}_{r = p}^{q} u_{r} + \sum_{r = p}^{q} b$

Examiner Tip

Mark schemes often specify how many terms from the start and end of the series should be written down – it is usually two or three, so always write down the first three and last three terms
Don’t be afraid to write out more terms than this to make sure you spot the pattern, and can easily decide which terms will cancel and which will not
Check your algebraic answer by substituting in numbers to make sure it works; you can use your calculator to find summations in sigma notation

Worked example

(a)

Express

\frac{2}{r (r + 2)}

in partial fractions.

3-2-2-edx-a-fm-we1a-soltn

(b)

Hence show that

\sum_{r = 1}^{n} \frac{2}{r (r + 2)} = \frac{3 n^{2} + 5 n}{2 (n + 1) (n + 2)}

using the method of differences.

3-2-2-edx-a-fm-we1b-soltn

Method of Differences (Edexcel A Level Further Maths: Core Pure)

Revision Note

Method of Differences

What is the Method of Differences?

How can I use partial fractions along with the method of differences?

How can I use the method of differences for series with expressions containing more than two terms?

What other uses are there for the method of differences?

Examiner Tip

Worked example

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Author: Jamie W