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4 Sequences and series — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in 4, read straight through: 6 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 9MA0 section 4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

4.1 · Understand and use the binomial expansion of (a + bx)ⁿ for positive integer n; the notations n! and nCr; link to binomial probabilities; extend to any rational n, including for approximation, valid for |bx/a| < 1.

Definition. Specification 4.1: understand and use the binomial expansion of (a+bx)n(a+bx)^n for positive integer nn; the notations n!n! and (nr)\binom{n}{r}; link to binomial probabilities; extend to any rational nn, including its use for approximation; be aware that the expansion is valid for ∣bxa∣<1\left|\dfrac{bx}{a}\right|<1.

  • For a positive integer nn, (a+bx)n=∑(nr)an−r(bx)r(a+bx)^n=\sum\binom{n}{r}a^{n-r}(bx)^r, and the same coefficients give binomial probabilities (nr)pr(1−p)n−r\binom{n}{r}p^r(1-p)^{n-r}. For a rational or negative nn use (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots, an infinite series valid only for ∣x∣<1|x|<1.
  • Replace xx by the whole of bxbx in brackets, including its sign and coefficient, so (1−9x)1/2(1-9x)^{1/2} has second term 12(−9x)\tfrac12(-9x) and third 12⋅(−12)2(−9x)2\dfrac{\frac12\cdot(-\frac12)}{2}(-9x)^2.
  • When the first term is not 11, take it out first: 4+5x=2(1+54x)1/2\sqrt{4+5x}=2(1+\tfrac54x)^{1/2}, and the validity becomes ∣54x∣<1\left|\tfrac54x\right|<1.
  • State the range of validity, ∣x∣<19|x|<\dfrac{1}{9} or ∣9x∣<1|9x|<1, and check a substituted value against it before using the series for an approximation; a value outside the range gives no approximation at all.
  • Simplify each coefficient fully.

Worked example

1.

Find the first three terms, in ascending powers of xx, of the binomial expansion of (1−2x)−12(1-2x)^{-\frac12}, giving each coefficient in its simplest form. State the range of values of xx for which the expansion is valid.

(4)

(Total for Question 1 is 4 marks)

Working

  1. 1.Use (1+u)n=1+nu+n(n−1)2!u2+⋯(1+u)^n=1+nu+\dfrac{n(n-1)}{2!}u^2+\cdots with n=−12n=-\dfrac12 and u=−2xu=-2x.
  2. 2.1+(−12)(−2x)+(−12)(−32)2(−2x)2=1+x+38×4x21+\left(-\dfrac12\right)(-2x)+\dfrac{(-\frac12)(-\frac32)}{2}(-2x)^2=1+x+\dfrac{3}{8}\times4x^2.
  3. 3.So the expansion is 1+x+32x2+⋯1+x+\dfrac32x^2+\cdots, valid for ∣−2x∣<1|-2x|<1, that is ∣x∣<12|x|<\dfrac12.

Answer: 1+x+32x21+x+\dfrac32x^2; valid for ∣x∣<12|x|<\dfrac12

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1.Attempts the binomial expansion with index −12-\frac12 and the correct structure for the x2x^2 term, eg (−12)(−32)2!(±2x)2\dfrac{(-\frac12)(-\frac32)}{2!}(\pm2x)^2M1
Correct unsimplified expansion with brackets, 1+(−12)(−2x)+…1+(-\frac12)(-2x)+\ldotsA1
1+x+32x21+x+\dfrac32x^2A1
∣x∣<12|x|<\dfrac12 or −12<x<12-\dfrac12<x<\dfrac12B1
Notes
A1: cao
B1: oe

(4 marks)

Mark-scheme wording to use

  • “M1: For an attempt at the binomial expansion with n = 1/2 and obtains the correct structure for term 3 or term 4. … A1: Correct unsimplified expression as shown but the bracketing must be correct unless any missing brackets are implied by subsequent work. … OR allow this mark for at least 2 correct simplified terms from −(9/2)x, −(81/8)x² and −(729/16)x³ … A1: 1 − (9/2)x − (81/8)x² − (729/16)x³ or simplified equivalent.” — Pearson 9MA0, June 2024 Paper 1, Q2(a)
  • “B1: Expansion is valid for |x| < 1/9 or |x| ≤ 1/9 and x = −2/9 is outside this range. … Do not accept vague statements such as "it is too big", "it is outside the range" without any mention of what the range is.” — Pearson 9MA0, June 2024 Paper 1, Q2(b)
  • “B1 for extracting the factor 4^(1/2), equivalently writing 2(1 + 5x/4)^(1/2). M1 for applying the binomial expansion to at least two non-constant terms. A1ft for a correct consistent expansion through x^2. A1 for the fully correct result k = -25/64” — Pearson 9MA0, Specimen Paper 1, Q2(a)

What the examiner wants. Brackets round the whole term and a stated range. The June 2024 Paper 1 report says the binomial structure was generally very good but inconsistency with brackets and not simplifying cost accuracy marks, the most common error being 99 instead of −9-9; on validity, many could not state ∣x∣<19|x|<\frac19 and misread the modulus, writing x>19x>\frac19.

Common mistakes

  • Don't expand (a+bx)n(a+bx)^n without first factoring out aa, so the generalised binomial coefficients are applied to the wrong expression.
  • Don't use the finite binomial formula for a negative or fractional power and omit the convergence condition.

Exam tip. For a rational-power expansion, factor the constant first and state the validity condition from the transformed bracket.

4.2 · Work with sequences including those given by a formula for the nth term and those generated by a simple relation of the form xₙ₊₁ = f(xₙ); increasing sequences; decreasing sequences; periodic sequences.

  • An explicit rule gives xnx_n directly from nn, whereas a recurrence relation defines each term from one or more preceding terms and needs an initial value.
  • A sequence is increasing when xn+1>xnx_{n+1}>x_n and decreasing when xn+1<xnx_{n+1}<x_n throughout the stated range of nn.
  • A sequence is periodic if its terms repeat after a fixed positive number of steps; the least such number is its period.
  • When using a recurrence, retain sufficient accuracy between steps and check any invariant interval.
  • A common error is to apply ff repeatedly to the initial term instead of to the latest term.

Worked example

A sequence is defined by x1=1x_1=1 and xn+1=5−xnx_{n+1}=5-x_n. Find x2,x3,x4,x5x_2,x_3,x_4,x_5 and state its period.

  1. 1.Apply the recurrence to the most recent term: x2=5−1=4x_2=5-1=4, x3=5−4=1x_3=5-4=1, and the pair 1,41,4 then repeats.
  2. 2.Thus x4=4x_4=4, x5=1x_5=1, and the least period is 22.

Answer: x2=4x_2=4, x3=1x_3=1, x4=4x_4=4, x5=1x_5=1; Period 22

Common mistakes

  • Don't declare a sequence increasing because x2>x1x_2>x_1 without checking later terms or the recurrence.
  • Don't classify a recurrence from too few terms and miss a repeating cycle.

Exam tip. Generate enough consecutive terms to demonstrate the claimed monotonicity or period rather than guessing from the first change.

4.3 · Understand and use sigma notation for sums of series.

  • The notation ∑r=abf(r)\sum_{r=a}^{b}f(r) means add the values of f(r)f(r) for each integer rr from aa to bb inclusive.
  • To write a series in sigma notation, identify a formula for its general term together with the correct first and last index values.
  • Use linearity to split sums: for example, ∑r=1n(ar+b)=a∑r=1nr+b∑r=1n1\sum_{r=1}^{n}(ar+b)=a\sum_{r=1}^{n}r+b\sum_{r=1}^{n}1, with ∑r=1n1=n\sum_{r=1}^{n}1=n.
  • Check the endpoint terms after changing an index.
  • A common error is to treat the upper limit as the number of terms when the lower limit is not 11.

Worked example

Write the series 5+9+13+⋯+415+9+13+\cdots+41 using sigma notation.

  1. 1.The terms have common difference 44, and 4r+14r+1 gives 55 when r=1r=1.
  2. 2.Solving 4r+1=414r+1=41 gives r=10r=10, so the series is ∑r=110(4r+1)\sum_{r=1}^{10}(4r+1).

Answer: ∑r=110(4r+1)\sum_{r=1}^{10}(4r+1)

Common mistakes

  • Don't shift the index but leave the summand unchanged, so the re-indexed sum represents different terms.
  • Don't write a sigma expression whose lower and upper limits produce the wrong first or last term.

Exam tip. Check a sigma answer by substituting both endpoint indices and confirming the number of generated terms.

4.4 · Understand and work with arithmetic sequences and series, including the formulae for nth term and the sum to n terms.

  • An arithmetic sequence has constant difference dd and nth term un=a+(n−1)du_n=a+(n-1)d, where aa is the first term.
  • The sum of its first nn terms is Sn=n[2a+(n−1)d]/2S_n=n[2a+(n-1)d]/2, equivalently Sn=n(a+l)/2S_n=n(a+l)/2 when the last term ll is known.
  • Pairing the first and last terms, then the second and penultimate terms, gives nn equal pairs of total a+la+l across two copies of the series, proving Sn=n(a+l)/2S_n=n(a+l)/2.
  • The term number nn must be a positive integer.
  • A common error is to use a+nda+nd for the nth term, shifting every term by one difference.

Worked example

An arithmetic series has first term 77, last term 151151 and common difference 88. Find the number of terms and the sum of the series.

  1. 1.Use 151=7+(n−1)8151=7+(n-1)8, giving 144=8(n−1)144=8(n-1) and n=19n=19.
  2. 2.Then S19=19(7+151)/2=19×79=1501S_{19}=19(7+151)/2=19\times79=1501.

Answer: 1919 terms; 15011501

Common mistakes

  • Don't solve un=Lu_n=L and round a non-integer term number instead of checking whether LL is actually a term.
  • Don't use the last term as the number of terms in the arithmetic-series formula.

Exam tip. Find the integer term count from the nth-term formula before applying the arithmetic sum formula.

4.5 · Understand and work with geometric sequences and series, including the formulae for the nth term and the sum of a finite geometric series; the sum to infinity of a convergent geometric series, including the use of |r| < 1; modulus notation.

  • A geometric sequence has constant ratio rr and nth term un=arn−1u_n=ar^{n-1}, where aa is the first term.
  • For r≠1r\ne1, subtracting rSnrS_n from SnS_n proves Sn=a(1−rn)/(1−r)S_n=a(1-r^n)/(1-r); an equivalent form may be more convenient when r>1r>1.
  • A geometric series has a finite sum to infinity only when ∣r∣<1|r|<1, in which case S∞=a/(1−r)S_\infty=a/(1-r); for a finite-sum threshold, isolate rnr^n and use logarithms.
  • A negative ratio makes term signs alternate, but convergence still depends on its modulus.
  • A common error is to use r<1r<1 instead of ∣r∣<1|r|<1, which would wrongly accept ratios below −1-1.

Worked example

Find the exact sum to infinity of 12−3+0.75−⋯12-3+0.75-\cdots.

  1. 1.The common ratio is r=−3/12=−1/4r=-3/12=-1/4, and ∣r∣=1/4<1|r|=1/4<1, so the series converges.
  2. 2.Its sum is S∞=12/[1−(−1/4)]=12/(5/4)=48/5S_\infty=12/[1-(-1/4)]=12/(5/4)=48/5.

Answer: 485\dfrac{48}{5}

Common mistakes

  • Don't use arnar^n rather than arn−1ar^{n-1} for the nth term of a geometric sequence.
  • Don't apply the sum-to-infinity formula without checking that the common ratio has modulus less than one.

Exam tip. State the common ratio and verify convergence before calculating an infinite geometric sum.

4.6 · Use sequences and series in modelling.

  • A constant additive change suggests an arithmetic model, while a constant multiplier or percentage change suggests a geometric model.
  • State what the term number represents and whether the initial value is u1u_1 or u0u_0 before forming a term or sum.
  • Repeated deposits, withdrawals or other fixed adjustments can be represented by a recurrence and often rewritten using a finite geometric sum.
  • Interpret results within the context and identify assumptions such as fixed rates or indefinite continuation.
  • A common error is to use an infinite sum when the process has not converged or has a finite stopping point.

Worked example

A ball is dropped from a height of 1212 m. After each impact it rebounds to 65%65\% of its previous maximum height. Find the total vertical distance travelled before it comes to rest, giving your answer to 33 significant figures.

  1. 1.The initial downward distance is 1212 m.
  2. 2.The rebound heights form a geometric series with first term 12(0.65)=7.812(0.65)=7.8 and ratio 0.650.65.
  3. 3.Each rebound height is travelled once upwards and once downwards, so the total is 12+2[7.8/(1−0.65)]=56.5714…12+2[7.8/(1-0.65)]=56.5714\ldots m.
  4. 4.To 33 significant figures this is 56.656.6 m.

Answer: 56.656.6 m

Common mistakes

  • Don't model a fixed percentage change with an additive arithmetic sequence instead of a multiplicative ratio.
  • Don't count the initial drop twice or omit either the upward or downward part of each rebound.

Exam tip. Build the distance model by separating the first drop from the two-way geometric rebound distances.