Notes · MathematicsUK · A-Levels

Sequences and series

Sequences and series cover arithmetic and geometric progressions and their sums, sigma notation, the binomial expansion for positive-integer and general rational indices, and recurrence relations, including increasing, decreasing and periodic sequences. The sum formulae and the binomial theorem are among the most frequently used results in the pure papers.

4 sections~9 min reading time3 competenciesLevel Standard 3 · Advanced 1

T·0555 / 20

Exam profile

AO1 · Use arithmetic, geometric and binomial series formulae accuratelyAO2 · Justify convergence conditions and reason with recurrence relationsAO3 · Model growth, decay and finance with series

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basic level

AS-Level covers arithmetic and geometric series, sigma notation and the binomial expansion for positive integer $n$ .

higher level

The full A-Level adds the binomial expansion for any rational $n$ (with its validity condition) and recurrence relations including periodic sequences.

Depth

Text

Contents · 4 sections

Sequences and series

Sequences and recurrence relations#

StandardDfE GCE Mathematics D1AQA 7357 D

A convergent recurrence

Fig. 1

u_{n+1} = 0.5u_n + 3

with

u_1 = 10

converges to the limit

L = 6

A sequence is an ordered list of terms, and it can be specified in two ways: by a formula for the

n

th term, such as

u_n = 3n - 1

, or by a recurrence relation that gives each term from previous ones, such as

u_{n+1} = 0.5u_n + 3

with a stated first term. The

n

th-term rule lets you jump straight to any term; the recurrence generates the sequence step by step and is the natural model for iterative processes.

Sequences are classified by their long-term behaviour. A sequence is increasing if

u_{n+1} > u_n

for all

n

, decreasing if

u_{n+1} < u_n

for all

n

, and periodic if the terms repeat in a fixed cycle, so

u_{n+k} = u_n

for some fixed period

k

. For example

u_{n+1} = -u_n

with

u_1 = 2

gives

2, -2, 2, -2, \ldots

, which is periodic with period 2.

A sequence converges to a limit

L

if the terms approach

L

n

increases without bound. For a recurrence, any limit must satisfy the equation obtained by replacing both

u_{n+1}

and

u_n

L

: from

u_{n+1} = 0.5u_n + 3

the limit satisfies

L = 0.5L + 3

, giving

L = 6

. This 'fixed-point' idea reappears in the numerical-methods topic.

Recurrence relations underlie the numerical iteration methods and many modelling situations (savings with regular deposits, drug concentration after repeated doses). Being able to list the first few terms accurately, to describe the behaviour in words, and to find a limit where one exists, are all examinable skills.

u_{n+1} = 0.5\,u_n + 3, \quad u_1 = 10 \ \Rightarrow\ L = 6

Limit of a convergent recurrence

Set $u_{n+1} = u_n = L$ and solve $L = 0.5L + 3$ to find the limit.

Worked example

Terms and limit of a recurrence

For $u_{n+1} = 0.2u_n + 8$ with $u_1 = 3$ , find $u_2$ , $u_3$ and the limit.

Compute terms
$u_2 = 0.2(3) + 8 = 8.6$ ; $u_3 = 0.2(8.6) + 8 = 9.72$ .
Set up the limit equation
A limit $L$ satisfies $L = 0.2L + 8$ .
Solve
$0.8L = 8 \Rightarrow L = 10$ .

Result: $u_2 = 8.6$ , $u_3 = 9.72$ ; the sequence increases towards the limit $L = 10$ .

Typical mistakes

Confusing the position-to-term rule with a recurrence and substituting into the wrong one.
Assuming every recurrence converges; oscillating or diverging sequences have no limit.

Active revision

A sequence is defined by $u_{n+1} = 3 - \dfrac{1}{u_n}$ with $u_1 = 2$ . Write down the first four terms and, assuming it converges, find its limit.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Arithmetic sequences and series#

StandardDfE GCE Mathematics D2AQA 7357 D

An arithmetic sequence

Fig. 2The terms of

u_n = 5 + 3(n-1)

increase by the constant common difference 3.

An arithmetic sequence has a constant difference

d

between consecutive terms, so the

n

th term is

u_n = a + (n - 1)d

, where

a

is the first term. This linear form means arithmetic terms sit on a straight line when plotted against

n

; the common difference is the gradient. Recognising this constant-difference structure is the first step in any arithmetic problem.

The sum of the first

n

terms,

S_n

, has a memorable derivation. Write the sum forwards and backwards and add: each of the

n

paired terms totals

a + \ell

(first plus last), so

2S_n = n(a + \ell)

and

S_n = \dfrac{n}{2}(a + \ell)

. Substituting

\ell = a + (n - 1)d

gives the alternative form

S_n = \dfrac{n}{2}\big(2a + (n - 1)d\big)

. Being able to reproduce this pairing argument, attributed to the young Gauss, is worth knowing.

The two forms of

S_n

are used according to what is given: use

\dfrac{n}{2}(a + \ell)

when you know the last term, and

\dfrac{n}{2}(2a + (n-1)d)

when you know

a

and

d

. Many problems give two facts (for instance two particular terms, or a term and a sum) leading to simultaneous equations in

a

and

d

Sigma notation writes such sums compactly:

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

means the sum of

u_r

r

runs from 1 to

n

. Splitting and shifting sums, and recognising an arithmetic sum inside sigma notation, let you evaluate expressions such as

\sum_{r=1}^{20}(3r + 2)

quickly using the arithmetic formula.

u_n = a + (n - 1)d

nth term of an arithmetic sequence

$a$ is the first term and $d$ the common difference.

S_n = \frac{n}{2}\big(2a + (n - 1)d\big) = \frac{n}{2}(a + \ell)

Sum of an arithmetic series

Derived by adding the series to its reverse; $\ell$ is the last term.

Worked example

Sum of an arithmetic series

An arithmetic series has first term 7 and common difference 4. Find the sum of the first 25 terms.

Identify a, d, n
$a = 7$ , $d = 4$ , $n = 25$ .
Apply the sum formula
$S_{25} = \dfrac{25}{2}\big(2(7) + 24(4)\big) = \dfrac{25}{2}(14 + 96)$ .
$S_{25} = \frac{25}{2}(14 + 96)$
Evaluate
$S_{25} = \dfrac{25}{2}(110) = 25 \times 55 = 1375$ .

Result: $S_{25} = 1375$ .

Typical mistakes

Using $n$ instead of $n - 1$ in $u_n = a + (n - 1)d$ .
Confusing the $n$ th term with the sum of $n$ terms.

Active revision

The 5th term of an arithmetic sequence is 17 and the 12th term is 45. Find $a$ , $d$ and the sum of the first 20 terms.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Geometric sequences and series#

StandardDfE GCE Mathematics D3AQA 7357 D

Geometric decay

Fig. 3

u_n = 16\left(\tfrac{1}{2}\right)^{n-1}

: each term is half the last, so

|r| = \tfrac{1}{2} < 1

and the series converges.

A geometric sequence multiplies by a constant ratio

r

each step, so the

n

th term is

u_n = ar^{n-1}

. The ratio may be greater than 1 (growth), between 0 and 1 (decay) or negative (alternating signs). Geometric progressions model compound interest, radioactive decay and any process with a constant proportional change, which makes them a favourite modelling context.

The sum of the first

n

terms is derived by the 'multiply and subtract' trick: writing

S_n = a + ar + \cdots + ar^{n-1}

and

rS_n = ar + ar^2 + \cdots + ar^{n}

, subtracting gives

S_n(1 - r) = a(1 - r^n)

, so

S_n = \dfrac{a(1 - r^n)}{1 - r}

for

r \neq 1

. Almost every step of the geometric-series toolkit follows from this one identity.

When

|r| < 1

the term

r^n \to 0

n \to \infty

, so the partial sums approach a finite limit: the sum to infinity is

S_\infty = \dfrac{a}{1 - r}

. This convergence holds only for

|r| < 1

; outside that range the series diverges and has no sum. Stating and using this condition correctly is a common source of marks.

Geometric ideas link to logarithms when you must find how many terms are needed to exceed a target, since solving

ar^{n-1} > k

requires taking logs. Recognising whether a real-world scenario is arithmetic (constant addition) or geometric (constant multiplication) is itself a modelling decision the papers test.

u_n = ar^{n-1}, \qquad S_n = \frac{a(1 - r^n)}{1 - r}\ (r \neq 1)

nth term and sum of a geometric series

Derived by subtracting $rS_n$ from $S_n$ .

S_\infty = \frac{a}{1 - r}, \qquad |r| < 1

Sum to infinity

Valid only when $|r| < 1$ , so that $r^n \to 0$ .

Partial sums approaching a limit

Fig. 4The partial sums of

16 + 8 + 4 + \cdots

approach the sum to infinity

\dfrac{16}{1 - 0.5} = 32

Worked example

Sum to infinity

Find the sum to infinity of the geometric series $18 + 12 + 8 + \cdots$ .

Find the ratio
$r = \dfrac{12}{18} = \dfrac{2}{3}$ , and $\left|\tfrac{2}{3}\right| < 1$ , so the sum to infinity exists.
Apply the formula
$S_\infty = \dfrac{a}{1 - r} = \dfrac{18}{1 - \tfrac{2}{3}} = \dfrac{18}{\tfrac{1}{3}}$ .
$S_\infty = \frac{18}{1 - \tfrac{2}{3}} = \frac{18}{\tfrac{1}{3}}$
Evaluate
$S_\infty = 18 \times 3 = 54$ .

Result: $S_\infty = 54$ .

Typical mistakes

Applying $S_\infty = \dfrac{a}{1 - r}$ when $|r| \geq 1$ , where no sum exists.
Using $r^{n}$ instead of $r^{n-1}$ in the $n$ th term.

Active revision

A geometric series has first term 24 and common ratio $\tfrac{2}{3}$ . Find the sum to infinity and the least number of terms whose sum exceeds 70.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

The binomial expansion#

AdvancedDfE GCE Mathematics D4AQA 7357 D

Binomial coefficients are symmetric

Fig. 5The coefficients of

(a + b)^5

are

1, 5, 10, 10, 5, 1

— Pascal's row 5, symmetric about the centre.

The binomial theorem expands

(a + b)^n

for a positive integer

n

as a finite sum

\sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

, where the binomial coefficients

\binom{n}{r} = \dfrac{n!}{r!(n-r)!}

are the entries of Pascal's triangle. The coefficients are symmetric,

\binom{n}{r} = \binom{n}{n-r}

, which is why the row of coefficients reads the same forwards and backwards.

For a positive integer index the expansion terminates after

n + 1

terms, and each term is

\binom{n}{r} a^{n-r} b^r

. A common task is to find a particular term or coefficient without expanding fully — for instance the coefficient of

x^3

(2 + x)^7

\binom{7}{3} 2^4 = 35 \times 16 = 560

, obtained by picking the single relevant term.

For any rational index

n

(including negative and fractional),

(1 + x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots

as an infinite series, valid only for

|x| < 1

. This is the tool for approximating roots and reciprocals: for example

\sqrt{1 + x} = (1 + x)^{1/2} \approx 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2

for small

x

. Stating the validity condition is essential.

To expand

(a + bx)^n

for rational

n

you first take out

a^n

(a + bx)^n = a^n\left(1 + \dfrac{b}{a}x\right)^n

, so the series is valid for

\left|\dfrac{b}{a}x\right| < 1

, i.e.

|x| < \dfrac{a}{b}

. Partial fractions combine with this expansion to produce series for more complicated rational functions.

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}, \qquad \binom{n}{r} = \frac{n!}{r!\,(n-r)!}

Binomial theorem (positive integer n)

A finite sum of $n+1$ terms; coefficients are the rows of Pascal's triangle.

(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots, \quad |x| < 1

Expansion for any rational n

An infinite series, valid only for $|x| < 1$ ; used to approximate roots and reciprocals.

Worked example

A specific term of a binomial expansion

Find the coefficient of $x^2$ in the expansion of $(3 + 2x)^5$ .

Identify the term
The term in $x^2$ is $\binom{5}{2}(3)^{3}(2x)^{2}$ .
Evaluate the pieces
$\binom{5}{2} = 10$ , $3^3 = 27$ , $(2x)^2 = 4x^2$ .
$\binom{5}{2}(3)^3(2x)^2 = 10 \times 27 \times 4x^2$
Multiply
$10 \times 27 \times 4 = 1080$ .

Result: The coefficient of $x^2$ is $1080$ .

Typical mistakes

Omitting the validity condition for the general binomial series.
Forgetting to raise the numerical part to its power, e.g. dropping the $2^4$ in a coefficient of $(2 + x)^7$ .

Active revision

Find the first three terms of the expansion of $(1 - 2x)^{1/2}$ in ascending powers of $x$ , and state the values of $x$ for which the expansion is valid.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Sequences and series

Key points

Terms and limit of a recurrence

Key points

Sum of an arithmetic series

Key points

Sum to infinity

Key points

A specific term of a binomial expansion

Sources