Summing Series and the Method of Differences Revision Notes for AQA A-Level Further Maths

Summing Series and the Method of Differences

Introduction to sigma notation

Sigma notation provides a concise way to write the sum of a sequence of terms. The Greek letter Σ (sigma) represents summation.

When writing the sum of a series containing n terms such as $u_1, u_2, \ldots, u_r, \ldots, u_{n-1}, u_n$ , we use the notation:

$\sum_{r=1}^{n} u_r$

This means "find the sum of all the terms from $u_1$ to $u_n$ ."

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The expression $u_r$ is called the general term, where r is the variable. If all terms are defined algebraically, $u_r$ is a function of r.

For example, if $u_r = r^2$ , then:

$\sum_{r=1}^{n} r^2 = 1^2 + 2^2 + 3^2 + \ldots + n^2$

Key notation:

$S_n$ means "the sum of these n terms"
The lower number (1) indicates where to start
The upper number (n) indicates where to finish
The expression after Σ tells you what to sum

Standard summation formulae

For A-Level Further Mathematics, you must memorise three essential formulae. These will not be provided in the exam and you must be able to quote them without proof.

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Memorization Required: These three formulae are NOT provided in the exam formula booklet. You must be able to recall and apply them accurately under exam conditions.

Sum of first n integers

The sum of the first n positive integers is given by:

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

In words: the sum equals n multiplied by (n+1), divided by 2.

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Memory aid: Think of this as "n lots of (n+1), then halve it"

Sum of first n squares

The sum of the squares of the first n positive integers is:

$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$

This formula can be derived using the method of differences, which we will explore later in this note.

Sum of first n cubes

The sum of the cubes of the first n positive integers is:

$\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4}$

Notice that this can also be written as:

$\sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2 = \left(\sum_{r=1}^{n} r\right)^2$

This shows an elegant relationship: the sum of the cubes equals the square of the sum of the integers.

Deriving the sum of integers formula

Understanding how to derive the formula for $\sum r$ helps build intuition for series manipulation. This elegant method demonstrates the power of algebraic symmetry.

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Derivation: Sum of First n Integers

Step 1: Write out the sum from 1 to n:

$\sum_{r=1}^{n} r = 1 + 2 + 3 + \ldots + (n-2) + (n-1) + n$

Step 2: Write the same sequence in reverse order:

$\sum_{r=1}^{n} r = n + (n-1) + (n-2) + \ldots + 3 + 2 + 1$

Step 3: Add the two expressions together term by term:

$2\sum_{r=1}^{n} r = (n+1) + (n+1) + (n+1) + \ldots + (n+1) + (n+1) + (n+1)$

Step 4: Notice that every term equals (n+1). Since there are n terms in total:

$2\sum_{r=1}^{n} r = n(n+1)$

Step 5: Divide both sides by 2:

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

This elegant proof demonstrates why the formula works and should help you remember it.

The method of differences

The method of differences is a technique used to sum certain types of series where the general term can be expressed as the difference of two consecutive function values.

When to use the method of differences

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There is no single universal method for summing all series. Some series cannot be expressed as simple algebraic expressions.

However, when the general term of a series can be written in the form:

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

we can find the sum using the method of differences, also known as the telescoping series method.

How the method works

The key insight is that when we write out successive differences vertically, most terms cancel out, leaving only the first and last terms.

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Step-by-step procedure for Method of Differences:

Express the general term as a difference: $u_r = f(r+1) - f(r)$
Write differences vertically for successive values of r:
- For $r = 1$ : $f(2) - f(1)$
- For $r = 2$ : $f(3) - f(2)$
- For $r = 3$ : $f(4) - f(3)$
- ...
- For $r = n$ : $f(n+1) - f(n)$
Eliminate terms wherever possible. Notice that $f(2)$ appears as +f(2) and -f(2), so cancels out. Similarly, $f(3), f(4), \ldots, f(n)$ all cancel.
Collect remaining terms: Only $f(n+1)$ and $-f(1)$ remain.

Therefore: $\sum_{r=1}^{n} u_r = f(n+1) - f(1)$

Worked examples

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Example 1: Sum of Even Numbers

Find the value of $2 + 4 + 6 + 8 + \ldots + 40$

Solution:

First, rewrite as a sum of integers by factoring out 2:

$2 + 4 + 6 + 8 + \ldots + 40 = 2(1 + 2 + 3 + 4 + \ldots + 20)$

This can be written using sigma notation:

$= 2\sum_{r=1}^{20} r$

Now substitute $n = 20$ into the standard formula $\sum r = \frac{n(n+1)}{2}$ :

$= 2 \times \frac{20(20+1)}{2} = 2 \times \frac{20 \times 21}{2}$

$= 2 \times \frac{420}{2} = 2 \times 210 = 420$

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Example 2: Sum Involving Products

Evaluate $\sum_{r=5}^{n} r(r+4)$

Solution:

Write out the first few terms to understand the pattern:

$\sum_{r=5}^{n} r(r+4) = 5(9) + 6(10) + \ldots + n(n+4)$

Expand the general term:

$= \sum_{r=5}^{n} (r^2 + 4r)$

Split into two separate sums:

$= \sum_{r=5}^{n} r^2 + 4\sum_{r=5}^{n} r$

To use standard formulae, we need sums starting from 1. Subtract the missing terms:

$= \sum_{r=1}^{n} r^2 - \sum_{r=1}^{4} r^2 + 4\left(\sum_{r=1}^{n} r - \sum_{r=1}^{4} r\right)$

Calculate the sums from 1 to 4:

$\sum_{r=1}^{4} r^2 = 1 + 4 + 9 + 16 = 30$

$\sum_{r=1}^{4} r = 1 + 2 + 3 + 4 = 10$

Substitute the standard formulae:

$= \frac{n(n+1)(2n+1)}{6} - 30 + 4\left(\frac{n(n+1)}{2} - 10\right)$

$= \frac{n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2} - 30 - 40$

$= \frac{n(n+1)(2n+1)}{6} + 2n(n+1) - 70$

Combine over a common denominator:

$= \frac{n(n+1)(2n+1) + 12n(n+1) - 420}{6}$

$= \frac{n(n+1)(2n+1+12) - 420}{6}$

$= \frac{n(n+1)(2n+13) - 420}{6}$

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Example 3: Basic Telescoping Series

Find the sum to n terms of the series $\frac{1}{r} - \frac{1}{r+1}$

Solution:

The general term is already in the form $f(r) - f(r+1)$ where $f(r) = \frac{1}{r}$ .

Write the differences vertically:

\begin{aligned} \sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) &= \left(\frac{1}{1} - \frac{1}{2}\right) \\ &\quad + \left(\frac{1}{2} - \frac{1}{3}\right) \\ &\quad + \left(\frac{1}{3} - \frac{1}{4}\right) \\ &\quad + \ldots \\ &\quad + \left(\frac{1}{n} - \frac{1}{n+1}\right) \end{aligned}

Eliminate terms wherever possible. Notice that $-\frac{1}{2}$ cancels with $+\frac{1}{2}$ , $-\frac{1}{3}$ cancels with $+\frac{1}{3}$ , and so on.

Collect up the remaining terms:

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

Therefore: The sum equals $\sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) = \frac{n}{n+1}$

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Example 4: Partial Fractions with Method of Differences

The expression $\frac{18}{(3r-2)(3r+1)(3r+4)}$ can be written as $\frac{1}{(3r-2)} - \frac{2}{(3r+1)} + \frac{1}{(3r+4)}$

Find the sum to n terms of the series $\frac{18}{(3r-2)(3r+1)(3r+4)}$

Solution:

Using the given partial fraction decomposition:

$\sum_{r=1}^{n} \frac{18}{(3r-2)(3r+1)(3r+4)} = \sum_{r=1}^{n} \left[\frac{1}{(3r-2)} - \frac{2}{(3r+1)} + \frac{1}{(3r+4)}\right]$

Write out the terms vertically to see the telescoping pattern:

For $r=1$ : $\frac{1}{1} - \frac{2}{4} + \frac{1}{7}$
For $r=2$ : $\frac{1}{4} - \frac{2}{7} + \frac{1}{10}$
For $r=3$ : $\frac{1}{7} - \frac{2}{10} + \frac{1}{13}$
For $r=4$ : $\frac{1}{10} - \frac{2}{13} + \frac{1}{16}$

Continue this pattern and observe the cancellations. After eliminating middle terms:

$= \frac{1}{1} - \frac{2}{4} + \frac{1}{4} + \frac{1}{3n+1} - \frac{2}{3n+1} + \frac{1}{3n+4}$

Simplify:

$= 1 - \frac{1}{4} - \frac{1}{3n+1} + \frac{1}{3n+4}$

$= \frac{3}{4} - \frac{1}{3n+1} + \frac{1}{3n+4}$

Therefore: $\sum_{r=1}^{n} \frac{18}{(3r-2)(3r+1)(3r+4)} = \frac{3}{4} - \frac{1}{3n+1} + \frac{1}{3n+4}$

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Example 5: Proving Standard Formulae Using Method of Differences

Use the method of differences to show that $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$

Solution:

Consider the identity $(2r+1)^3 - (2r-1)^3$ . Expanding both cubes:

(2r+1)^3 = 8r^3 + 12r^2 + 6r + 1 \\ (2r-1)^3 = 8r^3 - 12r^2 + 6r - 1

Subtracting:

$(2r+1)^3 - (2r-1)^3 = 24r^2 + 2$

Therefore: $r^2 = \frac{1}{24}[(2r+1)^3 - (2r-1)^3 - 2]$

Now sum from $r=1$ to $n$ and write differences vertically:

For $r=1$ : $\frac{1}{24}[3^3 - 1^3 - 2]$
For $r=2$ : $\frac{1}{24}[5^3 - 3^3 - 2]$
For $r=3$ : $\frac{1}{24}[7^3 - 5^3 - 2]$
...
For $r=n$ : $\frac{1}{24}[(2n+1)^3 - (2n-1)^3 - 2]$

The middle cubic terms cancel telescopically, leaving:

$\sum_{r=1}^{n} r^2 = \frac{1}{24}[(2n+1)^3 - 1^3 - 2n]$

$= \frac{1}{24}[(2n+1)^3 - 1 - 2n]$

Expanding $(2n+1)^3 = 8n^3 + 12n^2 + 6n + 1$ :

$= \frac{1}{24}[8n^3 + 12n^2 + 6n + 1 - 1 - 2n]$

$= \frac{1}{24}[8n^3 + 12n^2 + 4n]$

$= \frac{n(2n^2 + 3n + 1)}{6}$

$= \frac{n(2n+1)(n+1)}{6} = \frac{n(n+1)(2n+1)}{6}$

This proves the formula.

Exam tips and common pitfalls

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Memory requirement: You must memorise all three standard formulae. They will not be given in the formula booklet. Practice writing them out regularly until they become automatic.

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Common mistakes to avoid:

Forgetting to adjust limits when splitting sums (e.g., $\sum_{r=5}^{n}$ vs $\sum_{r=1}^{n}$ )
Not writing differences vertically in method of differences questions
Algebraic errors when combining fractions over common denominators
Missing the telescoping pattern in series

Exam strategy:

For questions involving sums, first try to express in terms of standard formulae
If the term looks like a difference, consider the method of differences
Always write differences vertically to clearly see cancellations
Check your final answer by substituting small values of n (e.g., n=1, n=2)

When using method of differences:

Rearrange the expression to identify the difference form $f(r+1) - f(r)$
Use partial fractions if necessary to create differences
Write at least 4-5 terms vertically to see the pattern clearly
Circle or cross out terms that cancel
Collect only the remaining terms

bookmarkSummary

Key Points to Remember:

Sigma notation ( $\Sigma$ ) provides a compact way to write sums of sequences
Three essential formulae must be memorised:
- $\sum r = \frac{n(n+1)}{2}$
- $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$
- $\sum r^3 = \frac{n^2(n+1)^2}{4}$
The method of differences applies when the general term can be written as $f(r+1) - f(r)$
Always write differences vertically to identify which terms cancel in telescoping series
The sum of cubes equals the square of the sum of integers: $(\sum r)^2 = \sum r^3$

Summing Series and the Method of Differences (AQA A-Level Further Maths): Revision Notes