Skip to main content
EnglandMathsSyllabus dot point

How do you describe, sum and expand sequences and series?

Arithmetic and geometric sequences and series, sigma notation, the conditions for convergence of a geometric series, the binomial expansion for positive integer and rational powers, and recurrence relations.

A focused answer to the AQA A-Level Mathematics sequences and series content, covering arithmetic and geometric progressions, sigma notation, the sum to infinity of a convergent geometric series, the binomial expansion for positive integer and rational powers, and recurrence relations.

Generated by Claude Opus 4.810 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Arithmetic sequences and series
  3. Geometric sequences and series
  4. Sigma notation
  5. The binomial expansion
  6. Recurrence relations
  7. Notation, periodicity and modelling

What this dot point is asking

AQA wants you to handle arithmetic and geometric sequences and series, use sigma notation, know when a geometric series converges and find its sum to infinity, perform binomial expansions for positive integer powers and for rational and negative powers, and use recurrence relations. Geometric-series problems that combine two given facts into a quadratic, and binomial expansions for rational powers, are perennial exam favourites.

Arithmetic sequences and series

The nn-th term is un=a+(n−1)du_n = a + (n - 1)d, where aa is the first term and dd the common difference. The sum of the first nn terms is Sn=n2(2a+(n−1)d)S_n = \dfrac{n}{2}(2a + (n - 1)d), equivalently n2(a+l)\dfrac{n}{2}(a + l) where ll is the last term. Many problems give you two pieces of information (for instance a particular term and a particular sum) and ask you to set up and solve simultaneous equations in aa and dd.

Geometric sequences and series

The nn-th term is un=arn−1u_n = ar^{n-1}, with common ratio rr. The sum of the first nn terms is Sn=a(1−rn)1−rS_n = \dfrac{a(1 - r^n)}{1 - r} (equivalently a(rn−1)r−1\dfrac{a(r^n - 1)}{r - 1}, more convenient when r>1r > 1).

A common structure combines ar=ar = (a given term) with a1−r=\frac{a}{1 - r} = (a given sum). Eliminating aa produces a quadratic in rr, as in the worked exam question above, so expect to factorise or use the formula.

Sigma notation

The symbol ∑\sum compactly represents a sum. For example ∑r=14(2r+1)=3+5+7+9=24\displaystyle\sum_{r=1}^{4}(2r + 1) = 3 + 5 + 7 + 9 = 24. Standard results, such as ∑r=1n1=n\sum_{r=1}^{n} 1 = n and ∑r=1nc=cn\sum_{r=1}^{n} c = cn for a constant cc, let you split and simplify sigma expressions before evaluating.

The binomial expansion

For a positive integer nn, (a+b)n=∑r=0n(nr)an−rbr(a + b)^n = \displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r, where (nr)=n!r! (n−r)!\binom{n}{r} = \dfrac{n!}{r!\,(n - r)!}. This is a finite expansion with n+1n + 1 terms.

Recurrence relations

A recurrence relation defines each term from the previous one, for example un+1=2un+3u_{n+1} = 2u_n + 3 with u1=1u_1 = 1. You generate terms by repeated substitution, and some sequences settle to a limit LL found by solving L=2L+3L = 2L + 3 (where a stable limit exists). Recurrence questions may ask for a specific term, the sum of several terms, or a limit.

Notation, periodicity and modelling

A sequence may be defined by a position-to-term rule (a formula for unu_n directly in terms of nn) or by a term-to-term rule (a recurrence). You should be able to move between them where possible, and to recognise an arithmetic or geometric sequence from either form. Some recurrences produce a periodic sequence that repeats with a fixed period; for these, the sum of a large number of terms is found by summing one full period and multiplying, then adding the leftover terms. Sigma sums of periodic sequences are a recurring exam idea.

Series model real situations: arithmetic series describe quantities that change by a fixed amount each step (such as regular savings increasing by a constant), while geometric series describe repeated proportional change (such as compound interest or a bouncing ball losing a fixed fraction of height). The sum to infinity of a convergent geometric series gives the total distance a ball travels or the eventual total of an infinite repayment, but only when ∣r∣<1|r| < 1. Setting up the right type of series, with the correct first term and common difference or ratio, is the modelling skill these questions test.

Exam-style practice questions

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

AQA 20197 marksPaper 1, Section B. A geometric series has first term aa and common ratio rr. The second term is 66 and the sum to infinity is 3232. (a) Show that 32r2−32r+6=032r^2 - 32r + 6 = 0. (b) Find the two possible values of rr, and the corresponding values of aa.
Show worked answer →

The second term is ar=6ar = 6, and the sum to infinity is a1−r=32\frac{a}{1 - r} = 32 (valid for ∣r∣<1|r| < 1). From the first, a=6ra = \frac{6}{r}. Substitute: 6/r1−r=32\frac{6/r}{1 - r} = 32, so 6=32r(1−r)6 = 32r(1 - r), giving 6=32r−32r26 = 32r - 32r^2, that is 32r2−32r+6=032r^2 - 32r + 6 = 0 as required. For (b), divide by 22: 16r2−16r+3=016r^2 - 16r + 3 = 0, which factorises as (4r−1)(4r−3)=0(4r - 1)(4r - 3) = 0, so r=14r = \frac{1}{4} or r=34r = \frac{3}{4}. Then a=6ra = \frac{6}{r} gives a=24a = 24 or a=8a = 8. Markers reward forming both equations, eliminating aa, and solving the quadratic; both roots satisfy ∣r∣<1|r| < 1.

AQA 20216 marksPaper 1, Section A. (a) Find the binomial expansion of (1+4x)1/2(1 + 4x)^{1/2} up to and including the term in x2x^2. (b) State the range of values of xx for which the expansion is valid. (c) Use the expansion with a suitable value of xx to estimate 1.4\sqrt{1.4}.
Show worked answer →

Using (1+y)n=1+ny+n(n−1)2!y2+⋯(1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \cdots with n=12n = \frac{1}{2} and y=4xy = 4x: the linear term is 12(4x)=2x\frac{1}{2}(4x) = 2x; the quadratic term is 12(−12)2(4x)2=−18×16x2=−2x2\frac{\frac{1}{2}(-\frac{1}{2})}{2}(4x)^2 = -\frac{1}{8}\times 16x^2 = -2x^2. So (1+4x)1/2≈1+2x−2x2(1 + 4x)^{1/2} \approx 1 + 2x - 2x^2. For (b), it is valid when ∣4x∣<1|4x| < 1, that is ∣x∣<14|x| < \frac{1}{4}. For (c), set 1+4x=1.41 + 4x = 1.4, so x=0.1x = 0.1 (within range): 1.4≈1+2(0.1)−2(0.1)2=1+0.2−0.02=1.18\sqrt{1.4} \approx 1 + 2(0.1) - 2(0.1)^2 = 1 + 0.2 - 0.02 = 1.18. Markers reward the general binomial formula for a rational power, the validity condition, and a sensible substitution.

Related dot points

Sources & how we know this