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Edexcel A-level Mathematics (9MA0) Β· Sequences and series
Mini-Lesson

Sequences and series

Topic 4 covers arithmetic and geometric sequences and series, sigma notation, recurrence relations (increasing, decreasing, periodic), the binomial expansion of (a + bx)ⁿ for positive integer n β€” and the A-level extension: the expansion of (1 + x)ⁿ for negative and fractional n, valid only when |x| < 1.

AP & GPseriessigma &recurrencebinomialexpansionTopic 4: the sum to infinity and the |x| < 1 binomial are new at A-level

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes β€” that is where the marks get thrown away. Press Start when you are ready.

Sequences Β· arithmetic

Arithmetic sequences and series

An arithmetic progression adds a constant common difference d each time.

u_n = a + (n βˆ’ 1)d Β· S_n = n/2 [2a + (n βˆ’ 1)d]equivalently S_n = n/2 (a + L), where L is the last term
Why the sum formula works (a proof you can be asked for)

Write S = a + (a+d) + … + L, then write it backwards: S = L + (Lβˆ’d) + … + a.

Adding the two lines pairs every term: 2S = n(a + L), so S = n/2 (a + L). ∎

Worked example β€” a = 7, d = 4

uβ‚‚β‚€ = 7 + 19 Γ— 4 = 7 + 76 = 83.

S₃₀ = 30/2 [2(7) + 29(4)] = 15[14 + 116] = 15 Γ— 130 = 1950.

Exam trap: it is (n βˆ’ 1)d, not nd. The 20th term has only 19 gaps in front of it.

Calculate

Your turn β€” nth term of an AP

1An arithmetic sequence has first term a = 7 and common difference d = 4. Find the 20th term.
Set-up

uβ‚‚β‚€ = 7 + (20 βˆ’ 1) Γ— 4 = 7 + 76 = 83.

Hint: u_n = a + (n βˆ’ 1)d, so uβ‚‚β‚€ = 7 + 19 Γ— 4.
Calculate

Your turn β€” sum of an AP

2For the same sequence (a = 7, d = 4), find the sum of the first 30 terms.
Set-up

S₃₀ = 30/2 [2(7) + 29(4)] = 15 [14 + 116] = 15 Γ— 130 = 1950.

Check with S = n/2(a + L): u₃₀ = 7 + 29(4) = 123, and 15 Γ— (7 + 123) = 15 Γ— 130 = 1950 βœ“

Hint: S_n = n/2 [2a + (n βˆ’ 1)d] = 15[14 + 116].
Sequences Β· geometric

Geometric sequences and series

A geometric progression multiplies by a constant common ratio r each time.

u_n = ar^(nβˆ’1) Β· S_n = a(1 βˆ’ rⁿ)/(1 βˆ’ r)Sum to infinity: S∞ = a/(1 βˆ’ r), and this exists only if |r| < 1
Worked example β€” a = 24, r = 1/3

|r| = 1/3 < 1, so the series converges.

S∞ = 24 / (1 βˆ’ 1/3) = 24 / (2/3) = 24 Γ— 3/2 = 36.

Sanity check: 24 + 8 + 2.667 + 0.889 + … is creeping up towards 36 βœ“

Exam trap: |r| < 1 means r can be negative (e.g. r = βˆ’0.5 converges, with terms alternating in sign). Writing 0 < r < 1 loses the mark. And r = 1 makes the S_n formula divide by zero.

Calculate

Your turn β€” sum to infinity

3A geometric series has first term 24 and common ratio 1/3. Find its sum to infinity.
Set-up

1 βˆ’ r = 1 βˆ’ 1/3 = 2/3.

S∞ = 24 Γ· (2/3) = 24 Γ— 3/2 = 36.

Hint: S∞ = a/(1 βˆ’ r) = 24 Γ· (2/3).
Quick check

When does it converge?

?A geometric series has a sum to infinity. Which condition on the common ratio r must hold?
Sequences Β· sigma

Sigma notation

Ξ£ is a compact instruction: substitute each integer from the bottom to the top and add.

Ξ£ (from r = 1 to n) of (3r + 2)= 5 + 8 + 11 + … + (3n + 2) β€” an AP with a = 5, d = 3
Worked example

Ξ£ (r = 1 to 10) of (3r + 2): first term 3(1) + 2 = 5, last term 3(10) + 2 = 32, n = 10 terms.

S = 10/2 (5 + 32) = 5 Γ— 37 = 185.

Exam trap: count the terms carefully. Ξ£ from r = 0 to n has n + 1 terms, not n. And Ξ£ from r = 5 to 20 has 20 βˆ’ 5 + 1 = 16 terms.

Sequences Β· recurrence

Recurrence relations

A recurrence relation defines each term from the one before: u_(n+1) = f(u_n), with a stated first term.

  • Increasing: u_(n+1) > u_n for all n.
  • Decreasing: u_(n+1) < u_n for all n.
  • Periodic: the terms repeat in a cycle of fixed order k, so u_(n+k) = u_n.
Worked example β€” u_(n+1) = 3u_n βˆ’ 5, u₁ = 4

uβ‚‚ = 3(4) βˆ’ 5 = 7 Β· u₃ = 3(7) βˆ’ 5 = 16 Β· uβ‚„ = 3(16) βˆ’ 5 = 43 β€” increasing.

Periodic example β€” u_(n+1) = 1/u_n, u₁ = 2

uβ‚‚ = 1/2, u₃ = 2, uβ‚„ = 1/2 … period (order) 2.

Calculate

Your turn β€” recurrence

4A sequence is defined by u₁ = 4 and u_(n+1) = 3u_n βˆ’ 5. Find uβ‚„.
Set-up

uβ‚‚ = 3(4) βˆ’ 5 = 7

u₃ = 3(7) βˆ’ 5 = 16

uβ‚„ = 3(16) βˆ’ 5 = 43

uβ‚„ =
Hint: Work up one term at a time: uβ‚‚, then u₃, then uβ‚„.
Quick check

Describe the sequence

?A sequence is given by u₁ = 2 and u_(n+1) = 1/u_n. How is it best described?
Sort it

Arithmetic, geometric β€” or neither?

Tap a sequence, then its type. Check the differences and the ratios.

βž• Arithmetic

βœ–οΈ Geometric

❓ Neither

Sequences Β· binomial

The binomial expansion (positive integer n)

For a positive integer n the expansion terminates and is exact:

(a + b)ⁿ = Ξ£ nCr Β· a^(nβˆ’r) Β· b^rnCr = n! / [r!(n βˆ’ r)!] β€” the numbers in Pascal’s triangle
Worked example β€” the coefficient of xΒ³ in (2 + 3x)⁡

Term = ⁡C₃ Γ— 2^(5βˆ’3) Γ— (3x)Β³ = 10 Γ— 2Β² Γ— 27xΒ³ = 10 Γ— 4 Γ— 27 xΒ³ = 1080xΒ³.

So the coefficient of xΒ³ is 1080.

Exam trap: the whole bracket is cubed β€” (3x)Β³ = 27xΒ³, not 3xΒ³. Forgetting to cube the 3 is the classic error. Also, the coefficient of xΒ³ is 1080; the term is 1080xΒ³. Read which is wanted.

Calculate

Your turn β€” binomial coefficient

5Find the coefficient of x³ in the expansion of (2 + 3x)⁡.
Set-up

⁡C₃ = 10, 2^(5βˆ’3) = 4, (3x)Β³ = 27xΒ³.

10 Γ— 4 Γ— 27 = 1080, so the term is 1080xΒ³ and the coefficient is 1080.

Hint: Term = ⁡C₃ Γ— 2Β² Γ— (3x)Β³, and ⁡C₃ = 10.
Sequences Β· binomial (A-level)

Binomial expansion when n is negative or fractional

This is the big A-level extension. For any real n the series

(1 + x)ⁿ = 1 + nx + n(nβˆ’1)/2! xΒ² + n(nβˆ’1)(nβˆ’2)/3! xΒ³ + …is an infinite series, valid only for |x| < 1
Worked example β€” expand (1 + 4x)^Β½ up to the xΒ² term

Here n = Β½ and the “x” of the formula is 4x.

Term in x: Β½ (4x) = 2x.

Term in xΒ²: [Β½ Γ— (βˆ’Β½)]/2! Γ— (4x)Β² = (βˆ’ΒΌ/2) Γ— 16xΒ² = (βˆ’1/8)(16xΒ²) = βˆ’2xΒ².

So (1 + 4x)^Β½ β‰ˆ 1 + 2x βˆ’ 2xΒ², valid for |4x| < 1, i.e. |x| < ΒΌ.

Check at x = 0.05: 1 + 0.1 βˆ’ 0.005 = 1.095, and √1.2 = 1.0954 βœ“

Exam trap: the formula needs the bracket to start with a 1. For (4 + x)^Β½ you must factor first: (4 + x)^Β½ = 2(1 + x/4)^Β½, valid for |x| < 4. And you must always state the range of validity β€” it is a mark on its own.

Calculate

Your turn β€” fractional index

6Expand (1 + 4x)^Β½ in ascending powers of x. What is the coefficient of xΒ²?
Set-up

n(nβˆ’1)/2! = (Β½)(βˆ’Β½)/2 = βˆ’1/8.

(4x)Β² = 16xΒ².

Coefficient = βˆ’1/8 Γ— 16 = βˆ’2. Expansion: 1 + 2x βˆ’ 2xΒ² + …, valid for |x| < ΒΌ.

Hint: Use n(n βˆ’ 1)/2! Γ— (4x)Β² with n = Β½ β€” and remember to square the 4.
Quick check

Range of validity

?For which values of x is the binomial expansion of (1 βˆ’ 3x)^(βˆ’2) valid?
Match it

Series formula flashcards

Tap a card on the left, then the correct formula.

Quantity
Formula
Sequences Β· applications

Series in context

Modelling questions are worth big marks and hinge on spotting AP vs GP.

  • “Increases by Β£200 each year” β†’ arithmetic, d = 200.
  • “Increases by 4% each year” β†’ geometric, r = 1.04.
  • “Total earned over 20 years” β†’ a sum, Sβ‚‚β‚€ β€” not the 20th term.
Worked example

A salary starts at Β£24 000 and rises 4% a year. Total earned in the first 10 years?

GP with a = 24000, r = 1.04, n = 10.

S₁₀ = 24000(1.04¹⁰ βˆ’ 1)/(1.04 βˆ’ 1) = 24000 Γ— (1.480244 βˆ’ 1)/0.04 = 24000 Γ— 12.00611 = Β£288 147 (nearest Β£).

Exam trap: when r > 1 use S_n = a(rⁿ βˆ’ 1)/(r βˆ’ 1) to keep everything positive β€” it is the same formula, just tidied.

Quick check

Term or sum?

?A savings scheme pays in Β£500 in year 1 and Β£75 more each subsequent year. Which expression gives the total paid in over 12 years?
Recap

The big ideas to take away

Arithmetic: u_n = a + (n βˆ’ 1)d Β· S_n = n/2 [2a + (n βˆ’ 1)d]

Geometric: u_n = ar^(nβˆ’1) Β· S_n = a(1 βˆ’ rⁿ)/(1 βˆ’ r) Β· S∞ = a/(1 βˆ’ r) only if |r| < 1

Sigma: count the terms carefully β€” r = 0 to n means n + 1 terms

Recurrence: u_(n+1) = f(u_n); describe as increasing, decreasing or periodic of order k

Binomial: positive integer n: finite, use nCr. Negative or fractional n: infinite, and state |x| < 1 validity

Topic 4 is formula-heavy but the marks are in the set-up: identify AP or GP, term or sum, and check the validity. Press Finish to see your score.

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