AQA A-Level Mathematics: D: Sequences and series (7357)

This guide teaches Section D: Sequences and series of AQA A-Level Mathematics (7357), as set out in the AQA A-level Mathematics specification, version 1.3 (31 January 2018), for exams from June 2018 onwards. It covers every content statement from D1 to D6. The specification lists Section D under Paper 1, and Papers 2 and 3 can assess any Paper 1 content, so these skills can earn marks on all three papers. A calculator is required in every 7357 paper, but most of this section is algebra you must show by hand.

Use it with the revision notes and the practice questions. For the whole course, see the AQA A-Level Mathematics hub and the printable checklist.

What this section covers

Ref	What you must be able to do
D1	Use the binomial expansion of (a + bx)ⁿ for positive integer n; use n!, ⁿCᵣ and (n r) notation; link to binomial probabilities. Extend to any rational n, use it for approximation, and know the expansion is valid for |bx/a| < 1 (proof not required)
D2	Work with sequences given by an nth-term formula or by a relation xₙ₊₁ = f(xₙ); recognise increasing, decreasing and periodic sequences
D3	Understand and use sigma notation for sums of series
D4	Arithmetic sequences and series: nth term and sum to n terms
D5	Geometric sequences and series: nth term, finite sum, sum to infinity of a convergent series, the condition |r| < 1, modulus notation
D6	Use sequences and series in modelling

The specification’s Appendix B lists two results in this section that you must recall without being given them: uₙ = a + (n − 1)d and uₙ = arⁿ⁻¹. Learn the sum formulas just as well.

D1 – The binomial expansion

Factorials and binomial coefficients

n! = n × (n − 1) × … × 2 × 1, with 0! = 1. The binomial coefficient is

ⁿCᵣ = (n r) = n! / (r!(n − r)!)

For a positive integer n,

(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCᵣaⁿ⁻ʳbʳ + … + bⁿ

There are n + 1 terms and the expansion is exact.

Worked example 1. Expand (3 + 2x)⁵ fully.

(3 + 2x)⁵ = 3⁵ + 5(3⁴)(2x) + 10(3³)(2x)² + 10(3²)(2x)³ + 5(3)(2x)⁴ + (2x)⁵
          = 243 + 810x + 1080x² + 720x³ + 240x⁴ + 32x⁵

Bracket the whole term (2x) before you raise it to a power. Writing 2x² instead of (2x)² is the commonest slip.

Link to binomial probabilities

If X ~ B(n, p) and q = 1 − p, then P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ. These are exactly the terms of (q + p)ⁿ, which is why the probabilities add to (q + p)ⁿ = 1ⁿ = 1. For X ~ B(6, 0.3):

P(X = 2) = ⁶C₂ (0.3)² (0.7)⁴ = 15 × 0.09 × 0.2401 = 0.324 (3 s.f.)

Rational n and approximation

When n is negative or a fraction, the expansion of (1 + x)ⁿ never stops:

(1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + …

The general coefficient is n(n − 1)…(n − r + 1)/r!, which is how the specification defines ⁿCᵣ for rational n. The series is valid only when |x| < 1. For (a + bx)ⁿ, take out aⁿ first:

(a + bx)ⁿ = aⁿ(1 + (b/a)x)ⁿ,   valid for |bx/a| < 1

Worked example 2. Expand √(4 − x) in ascending powers of x up to x², state the range of validity, and use it to estimate √3.96.

√(4 − x) = 4^(1/2) (1 − x/4)^(1/2) = 2(1 − x/4)^(1/2)

(1 − x/4)^(1/2) = 1 + (1/2)(−x/4) + (1/2)(−1/2)(−x/4)²/2! + …
                = 1 − x/8 − x²/128 + …

√(4 − x) ≈ 2 − x/4 − x²/64

Valid for |−x/4| < 1, that is |x| < 4. For √3.96, put x = 0.04:

2 − 0.01 − 0.000025 = 1.989975

A calculator gives 1.989975 to 6 d.p. The estimate is good because x is small, so the x³ and later terms are tiny. A value of x near the edge of the valid range gives a poor approximation, and a value outside it gives nonsense.

Worked example 3. Expand (1 + 2x)⁻² up to x³.

1 + (−2)(2x) + (−2)(−3)(2x)²/2! + (−2)(−3)(−4)(2x)³/3!
= 1 − 4x + 12x² − 32x³,   valid for |x| < 1/2

D2 – Sequences: nth term, recurrence, behaviour

A sequence can be defined in two ways.

• By a formula for the nth term, such as uₙ = (n + 3)/n. Substitute n = 1, 2, 3, … to get 4, 5/2, 2, 7/4, …
• By a recurrence relation xₙ₊₁ = f(xₙ) with a starting value. For xₙ₊₁ = 2xₙ − 3 with x₁ = 5, you get 5, 7, 11, 19, 35, …

Describe the behaviour using the specification’s terms.

• Increasing: uₙ₊₁ > uₙ for all n. Show it by proving uₙ₊₁ − uₙ > 0.
• Decreasing: uₙ₊₁ < uₙ for all n. Rewriting uₙ = (n + 3)/n as 1 + 3/n shows it is decreasing, because 3/n gets smaller as n grows.
• Periodic: the terms repeat in a cycle. The order (period) is the length of the cycle.

Worked example 4. u₁ = 2 and uₙ₊₁ = 1/(1 − uₙ). Find u₂ to u₄ and describe the sequence.

u₂ = 1/(1 − 2) = −1
u₃ = 1/(1 − (−1)) = 1/2
u₄ = 1/(1 − 1/2) = 2

u₄ = u₁, so the sequence repeats 2, −1, 1/2, 2, −1, 1/2, … It is periodic with order 3. To find a far-off term, use the cycle: u₁₀₀ = u₁ because 100 = 3 × 33 + 1, so u₁₀₀ = 2.

D3 – Sigma notation

Σ (sigma) means “add up”. The expression

 20
 Σ (3r + 2)
r=1

means substitute r = 1, 2, …, 20 into (3r + 2) and add: 5 + 8 + 11 + … + 62. Count terms carefully: from r = k to r = m there are m − k + 1 terms.

Worked example 5. Evaluate Σ from r = 1 to 20 of (3r + 2).

The terms form an arithmetic series with first term 5, last term 62 and 20 terms:

S = (20/2)(5 + 62) = 10 × 67 = 670

Splitting also works: 3 × Σr + Σ2 = 3 × (20 × 21/2) + 20 × 2 = 630 + 40 = 670. Note that Σ2 over 20 terms is 40, not 2.

D4 – Arithmetic sequences and series

An arithmetic sequence has a common difference d. Using the specification’s notation (a = first term, l = last term, Sₙ = sum to n terms):

uₙ = a + (n − 1)d
Sₙ = (n/2)(2a + (n − 1)d) = (n/2)(a + l)

Proof of the sum formula: write Sₙ forwards and backwards and add. Each pair sums to a + l, there are n pairs, so 2Sₙ = n(a + l).

Worked example 6. The 3rd term of an arithmetic sequence is 17 and the 10th term is 52. Find the 20th term and the sum of the first 20 terms.

a + 2d = 17
a + 9d = 52
Subtract: 7d = 35, so d = 5 and a = 7

u₂₀ = 7 + 19 × 5 = 102
S₂₀ = (20/2)(7 + 102) = 1090

Questions that ask for the least n with Sₙ above a target lead to a quadratic inequality in n. Solve it, then check whole-number values either side.

D5 – Geometric sequences and series

A geometric sequence has a common ratio r: each term is the previous one multiplied by r.

uₙ = arⁿ⁻¹
Sₙ = a(1 − rⁿ)/(1 − r)      (r ≠ 1)
S∞ = a/(1 − r)              only when |r| < 1

|r| is the modulus of r, its size ignoring sign, so |r| < 1 means −1 < r < 1. When |r| < 1, rⁿ → 0 as n → ∞, so Sₙ → a/(1 − r). The series is then convergent. If |r| ≥ 1 the series has no sum to infinity.

Proof of Sₙ: write Sₙ = a + ar + … + arⁿ⁻¹, multiply by r, and subtract: Sₙ − rSₙ = a − arⁿ.

Worked example 7. A geometric series has first term 81 and common ratio 2/3. Find S∞, S₅, and the least n for which S∞ − Sₙ < 1.

S∞ = 81/(1 − 2/3) = 243
S₅ = 81(1 − (2/3)⁵)/(1/3) = 243(1 − 32/243) = 211

S∞ − Sₙ = 243(2/3)ⁿ < 1
(2/3)ⁿ < 1/243
n ln(2/3) < ln(1/243)
n > ln 243 / ln 1.5 = 13.55…     (the inequality flips: ln(2/3) is negative)

So n = 14. Check: 243(2/3)¹³ ≈ 1.25 and 243(2/3)¹⁴ ≈ 0.832.

Worked example 8. For which x does 1 + (2x − 1) + (2x − 1)² + … converge?

Here r = 2x − 1, so you need |2x − 1| < 1, giving −1 < 2x − 1 < 1, so 0 < x < 1. At x = 0.75, r = 0.5 and S∞ = 1/(1 − 0.5) = 2.

D6 – Sequences and series in modelling

A constant amount of change each step gives an arithmetic model. A constant percentage change gives a geometric model. Define what n counts (weeks, years) and whether term 1 is at the start or after one step.

Worked example 9. A car bought for £18 000 loses 15% of its value each year.

Value after n years: Vₙ = 18000 × 0.85ⁿ
After 5 years: 18000 × 0.85⁵ = £7986.70 (nearest penny)
Below £5000 when 0.85ⁿ < 5000/18000
n > ln(5000/18000) / ln 0.85 = 7.88

So the value first drops below £5000 after 8 years (about £4904.83). A limitation: real depreciation is steeper in the first year and depends on mileage and condition, so a fixed 15% is a simplification.

A runner who covers 5 km in week 1 and adds 0.5 km each week follows an arithmetic model. Week 12 is 5 + 11 × 0.5 = 10.5 km, and the 12-week total is (12/2)(5 + 10.5) = 93 km. The model fails eventually, because no runner can increase distance forever. Saying where a model stops being realistic is a skill AQA expects in every section (see overarching themes).

Common errors

• Forgetting to bracket the whole term: (−3x)² is 9x², not −3x².
• Expanding (a + bx)ⁿ for rational n without taking out aⁿ first, then quoting the wrong validity range.
• Treating Sₙ as the nth term, or uₙ as a running total.
• Losing the inequality direction when dividing by the log of a number less than 1.
• Quoting S∞ for a series with |r| ≥ 1.
• Miscounting terms in sigma notation when the lower limit is not 1.
• In models, being one term out because term 1 is “after one year” in one question and “now” in another.

Next steps

Condense this into the Section D revision notes, then test yourself with the Section D practice questions. For timing and paper strategy, read the 7357 exam preparation guide. To find your weakest topics quickly, try the free diagnostics.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for teaching from September 2017 and exams from June 2018 onwards, published by AQA. Section 3.5, D: Sequences and series (D1 to D6).