EnglandMathsSyllabus dot point

How do you describe and sum patterns of numbers, and expand binomials for any power?

Sequences and series including arithmetic and geometric sequences, sigma notation, sums to infinity, recurrence relations, and the binomial expansion for any rational power.

A focused answer to the Edexcel A-Level Mathematics sequences and series content, covering arithmetic and geometric sequences and their sums, sigma notation, the condition for convergence and sum to infinity, recurrence relations, and the binomial expansion for any rational power.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Edexcel wants you to handle arithmetic and geometric sequences and series, use sigma notation, apply the sum formulae, know when a geometric series converges and find its sum to infinity, work with recurrence relations and periodic sequences, and expand $(1 + x)^n$ for any rational $n$ with the binomial series.

The answer

Arithmetic sequences

Each term increases by a fixed common difference $d$ .

Geometric sequences and sum to infinity

Each term is multiplied by a fixed common ratio $r$ .

Sigma notation and recurrence

Sigma notation is shorthand for a sum. A recurrence relation defines each term from previous ones.

A sequence can also be periodic, repeating after a fixed number of terms; for instance $u_{n+1} = \dfrac{1}{1 - u_n}$ cycles with period $3$ . Recognising periodicity lets you find a far-off term such as $u_{100}$ without listing every term: divide the index by the period and use the remainder.

Binomial expansion for any rational power

For non-integer or negative powers the expansion is an infinite series, valid for $|x| < 1$ .

Examples in context

A geometric series has first term

a = 24

and common ratio

r = \dfrac{2}{3}

. Find the sum to infinity and the sum of the first

4

terms

step 1 Check convergence and find $S_\infty$

$|r| = \dfrac{2}{3} < 1$ , so $S_\infty = \dfrac{a}{1 - r} = \dfrac{24}{1 - \frac{2}{3}} = \dfrac{24}{\frac{1}{3}} = 72$ .

step 2 Apply the finite-sum formula

$S_4 = \dfrac{a(1 - r^4)}{1 - r} = \dfrac{24\left(1 - \left(\frac{2}{3}\right)^4\right)}{\frac{1}{3}}$ .

step 3 Evaluate

$\left(\dfrac{2}{3}\right)^4 = \dfrac{16}{81}$ , so $S_4 = 72\left(1 - \dfrac{16}{81}\right) = 72 \times \dfrac{65}{81} \approx 57.8$ .

Try this

Q1. An arithmetic sequence has first term $3$ and common difference $4$ . Find the 10th term. [2 marks]

Cue. $u_{10} = 3 + 9 \times 4 = 39$ .

Q2. Find the sum to infinity of a geometric series with $a = 20$ and $r = 0.1$ . [2 marks]

Cue. $S_\infty = \dfrac{20}{1 - 0.1} = \dfrac{20}{0.9} \approx 22.2$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksAn arithmetic series has first term

7

and common difference

5

. Find the number of terms needed for the sum to first exceed

1000

Show worked answer →

Use $S_n = \dfrac{n}{2}(2a + (n - 1)d)$ with $a = 7$ , $d = 5$ (M1): $S_n = \dfrac{n}{2}(14 + 5(n - 1)) = \dfrac{n}{2}(5n + 9)$ (A1).

Set $S_n > 1000$ : $\dfrac{n}{2}(5n + 9) > 1000$ , so $5n^2 + 9n - 2000 > 0$ (M1).

Solve the quadratic $5n^2 + 9n - 2000 = 0$ : $n = \dfrac{-9 + \sqrt{81 + 40000}}{10} = \dfrac{-9 + \sqrt{40081}}{10} \approx \dfrac{-9 + 200.2}{10} \approx 19.1$ (M1, A1).

Since $n$ must be a positive integer and the sum increases, take $n = 20$ (A1).

Markers reward the sum formula, forming the quadratic inequality, solving it, and rounding up to the next integer.

Edexcel 20214 marksFind the first three terms, in ascending powers of

x

, of the binomial expansion of

(1 - 2x)^{1/2}

, and state the values of

x

for which the expansion is valid.

Show worked answer →

Use $(1 + X)^n = 1 + nX + \dfrac{n(n-1)}{2!}X^2 + \dots$ with $n = \dfrac{1}{2}$ and $X = -2x$ (M1).

Term in $x$ : $nX = \dfrac{1}{2}(-2x) = -x$ (A1).

Term in $x^2$ : $\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(-2x)^2 = \dfrac{-\frac{1}{4}}{2}(4x^2) = -\dfrac{1}{2}x^2$ (A1).

So $(1 - 2x)^{1/2} \approx 1 - x - \dfrac{1}{2}x^2$ , valid for $|{-2x}| < 1$ , that is $|x| < \dfrac{1}{2}$ (A1).

Markers reward the substitution, the first two terms, the $x^2$ term, and the validity condition.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)