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Edexcel A-level Maths revision notes

Sequences and series

Section 4
Both years
Both years: this holds AS subject content and content the exam board adds beyond it for the full A-level.
6 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section 4

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.

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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.

Notes
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A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a positive integer nn, (a+bx)n=r=0n(nr)anr(bx)r(a+bx)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}(bx)^r, where (nr)=n!/[r!(nr)!]\binom{n}{r}=n!/[r!(n-r)!], n!=n(n1)1n!=n(n-1)\cdots1 and 0!=10!=1; Pascal's relation is (nr)=(n1r1)+(n1r)\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r} for 1rn11\le r\le n-1. The same coefficients appear in binomial probabilities: P(X=r)=(nr)pr(1p)nrP(X=r)=\binom{n}{r}p^r(1-p)^{n-r} for XB(n,p)X\sim B(n,p).
  • For rational nn, write the expression as an(1+u)na^n(1+u)^n and use 1+nu+n(n1)u2/2!+1+nu+n(n-1)u^2/2!+\cdots, valid for u<1|u|<1.
  • For an approximation, choose a nearby convenient value and retain the requested number of terms.
  • A common error is to omit powers of the coefficient bb from terms involving (bx)r(bx)^r.
  • Use (1+u)n=1+nu+n(n1)u2/2+(1+u)^n=1+nu+n(n-1)u^2/2+\cdots with n=1/2n=-1/2 and u=2xu=-2x.
Worked example

For (12x)1/2(1-2x)^{-1/2}, obtain the constant, xx and x2x^2 terms. Also give the interval of xx on which this series is valid.

  1. 1.Use (1+u)n=1+nu+n(n1)u2/2+(1+u)^n=1+nu+n(n-1)u^2/2+\cdots with n=1/2n=-1/2 and u=2xu=-2x.
  2. 2.This gives 1+(1/2)(2x)+[(1/2)(3/2)/2](2x)2=1+x+3x2/21+(-1/2)(-2x)+[(-1/2)(-3/2)/2](-2x)^2=1+x+3x^2/2.
  3. 3.Validity requires 2x<1|-2x|<1, so x<1/2|x|<1/2.

Answer: 1+x+32x21+x+\dfrac{3}{2}x^2; x<12|x|<\dfrac{1}{2}

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.

Tier 1 · Easy

ORIGINAL

1.

Find the coefficient of x2x^2 in the expansion of (1+4x)5(1+4x)^5.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the coefficient of x3x^3 in the expansion of (2x)7(2-x)^7.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Use the first three terms of a binomial expansion of (64+x)1/3(64+x)^{1/3} to estimate 653\sqrt[3]{65}. Give the estimate to 55 decimal places and justify that the expansion is valid at the value of xx used.

(5)

(Total for Question 1 is 5 marks)

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

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.

Notes
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Your confidence:

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Explanation

  • 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=5xnx_{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=51=4x_2=5-1=4, x3=54=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.

Tier 1 · Easy

ORIGINAL

1.

The sequence (xn)(x_n) is defined by xn=73nx_n=7-3n for n1n\ge1. Write down its first three terms and state whether it is increasing or decreasing.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

The sequence is defined by xn=nn+1x_n=\dfrac{n}{n+1} for n1n\geq1. Prove that the sequence is increasing.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A sequence is defined by x1=2x_1=2 and xn+1=(xn+6)/2x_{n+1}=(x_n+6)/2. Let yn=6xny_n=6-x_n. Show that (yn)(y_n) is geometric, find a formula for xnx_n, and hence show that (xn)(x_n) is increasing.

(5)

(Total for Question 1 is 5 marks)

4.3

Understand and use sigma notation for sums of series.

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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)=ar=1nr+br=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.

Tier 1 · Easy

ORIGINAL

1.

Evaluate r=14(2r+1)\sum_{r=1}^{4}(2r+1).

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Evaluate r=18(3r22r)\displaystyle\sum_{r=1}^{8}(3r^2-2r).

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Given that r=1n(4r1)=210\sum_{r=1}^{n}(4r-1)=210, find the positive integer nn. Show all stages of your working.

(5)

(Total for Question 1 is 5 marks)

4.4

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An arithmetic sequence has constant difference dd and nth term un=a+(n1)du_n=a+(n-1)d, where aa is the first term.
  • The sum of its first nn terms is Sn=n[2a+(n1)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+(n1)8151=7+(n-1)8, giving 144=8(n1)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.

Tier 1 · Easy

ORIGINAL

1.

Find the 1818th term of the arithmetic sequence 11,17,23,11,17,23,\ldots.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

The fifth term of an arithmetic sequence is 1818 and the twelfth term is 4646. Find the first term, the common difference and the sum of the first 2020 terms.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

An arithmetic sequence has first term aa and common difference dd. Its 88th term is 3131 and the sum of its first 2020 terms is 730730. Find aa and dd, and find the least value of nn for which the nth term exceeds 100100.

(5)

(Total for Question 1 is 5 marks)

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.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A geometric sequence has constant ratio rr and nth term un=arn1u_n=ar^{n-1}, where aa is the first term.
  • For r1r\ne1, subtracting rSnrS_n from SnS_n proves Sn=a(1rn)/(1r)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/(1r)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 123+0.7512-3+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 arn1ar^{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.

Tier 1 · Easy

ORIGINAL

1.

Find the 88th term of the geometric sequence 3,6,12,3,6,12,\ldots.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A geometric sequence has positive common ratio. Its third term is 1818 and its sixth term is 486486. Find the first term, the common ratio and the sum of the first five terms.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A geometric series has first term 800800 and common ratio 0.90.9. Find the least value of nn for which the sum of the first nn terms exceeds 70007000.

(5)

(Total for Question 1 is 5 marks)

4.6

Use sequences and series in modelling.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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/(10.65)]=56.571412+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.

Tier 1 · Easy

ORIGINAL

1.

Mina saves £25\pounds 25 in the first month and increases the amount saved by £5\pounds 5 each month. Find the total she saves in the first 66 months.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A theatre has 2525 rows of seats. The first row has 1616 seats and each successive row has 22 more seats than the preceding row. Find the total number of seats.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

An account initially contains £5000\pounds 5000. At the end of each year, after interest of 3%3\% has been added, £400\pounds 400 is withdrawn. The model is B0=5000B_0=5000 and Bn+1=1.03Bn400B_{n+1}=1.03B_n-400. Derive a formula for BnB_n and find the first value of nn for which the model predicts a negative balance. State one limitation of the model.

(6)

(Total for Question 1 is 6 marks)

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