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 — and the A-level extension: the expansion of (1 + x)ⁿ for negative and fractional n, valid only when |x| < 1.

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 — that is where the marks get thrown away. Press Start when you are ready.

Sequences · arithmetic

Arithmetic sequences and series

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

u_n = a + (n − 1)d · S_n = n/2 [2a + (n − 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 + (L−d) + … + a.

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

Worked example — a = 7, d = 4

u₂₀ = 7 + 19 × 4 = 7 + 76 = 83.

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

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

Calculate

Your turn — nth term of an AP

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

Set-up

u₂₀ = 7 + (20 − 1) × 4 = 7 + 76 = 83.

Hint: u_n = a + (n − 1)d, so u₂₀ = 7 + 19 × 4.

Calculate

Your turn — 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 × 130 = 1950.

Check with S = n/2(a + L): u₃₀ = 7 + 29(4) = 123, and 15 × (7 + 123) = 15 × 130 = 1950 ✓

Hint: S_n = n/2 [2a + (n − 1)d] = 15[14 + 116].

Sequences · geometric

Geometric sequences and series

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

u_n = ar^(n−1) · S_n = a(1 − rⁿ)/(1 − r)Sum to infinity: S∞ = a/(1 − r), and this exists only if |r| < 1

Worked example — a = 24, r = 1/3

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

S∞ = 24 / (1 − 1/3) = 24 / (2/3) = 24 × 3/2 = 36.

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

Exam trap: |r| < 1 means r can be negative (e.g. r = −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 — sum to infinity

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

Set-up

1 − r = 1 − 1/3 = 2/3.

S∞ = 24 ÷ (2/3) = 24 × 3/2 = 36.

Hint: S∞ = a/(1 − r) = 24 ÷ (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 · sigma

Sigma notation

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

Σ (from r = 1 to n) of (3r + 2)= 5 + 8 + 11 + … + (3n + 2) — an AP with a = 5, d = 3

Worked example

Σ (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 × 37 = 185.

Exam trap: count the terms carefully. Σ from r = 0 to n has n + 1 terms, not n. And Σ from r = 5 to 20 has 20 − 5 + 1 = 16 terms.

Sequences · 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 — u_(n+1) = 3u_n − 5, u₁ = 4

u₂ = 3(4) − 5 = 7 · u₃ = 3(7) − 5 = 16 · u₄ = 3(16) − 5 = 43 — increasing.

Periodic example — u_(n+1) = 1/u_n, u₁ = 2

u₂ = 1/2, u₃ = 2, u₄ = 1/2 … period (order) 2.

Calculate

Your turn — recurrence

4A sequence is defined by u₁ = 4 and u_(n+1) = 3u_n − 5. Find u₄.

Set-up

u₂ = 3(4) − 5 = 7

u₃ = 3(7) − 5 = 16

u₄ = 3(16) − 5 = 43

u₄ =

Hint: Work up one term at a time: u₂, then u₃, then u₄.

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 — or neither?

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

➕ Arithmetic

✖️ Geometric

❓ Neither

Sequences · binomial

The binomial expansion (positive integer n)

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

(a + b)ⁿ = Σ nCr · a^(n−r) · b^rnCr = n! / [r!(n − r)!] — the numbers in Pascal’s triangle

Worked example — the coefficient of x³ in (2 + 3x)⁵

Term = ⁵C₃ × 2^(5−3) × (3x)³ = 10 × 2² × 27x³ = 10 × 4 × 27 x³ = 1080x³.

So the coefficient of x³ is 1080.

Exam trap: the whole bracket is cubed — (3x)³ = 27x³, not 3x³. Forgetting to cube the 3 is the classic error. Also, the coefficient of x³ is 1080; the term is 1080x³. Read which is wanted.

Calculate

Your turn — binomial coefficient

5Find the coefficient of x³ in the expansion of (2 + 3x)⁵.

Set-up

⁵C₃ = 10, 2^(5−3) = 4, (3x)³ = 27x³.

10 × 4 × 27 = 1080, so the term is 1080x³ and the coefficient is 1080.

Hint: Term = ⁵C₃ × 2² × (3x)³, and ⁵C₃ = 10.

Sequences · 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(n−1)/2! x² + n(n−1)(n−2)/3! x³ + …is an infinite series, valid only for |x| < 1

Worked example — expand (1 + 4x)^½ up to the x² term

Here n = ½ and the “x” of the formula is 4x.

Term in x: ½ (4x) = 2x.

Term in x²: [½ × (−½)]/2! × (4x)² = (−¼/2) × 16x² = (−1/8)(16x²) = −2x².

So (1 + 4x)^½ ≈ 1 + 2x − 2x², valid for |4x| < 1, i.e. |x| < ¼.

Check at x = 0.05: 1 + 0.1 − 0.005 = 1.095, and √1.2 = 1.0954 ✓

Exam trap: the formula needs the bracket to start with a 1. For (4 + x)^½ you must factor first: (4 + x)^½ = 2(1 + x/4)^½, valid for |x| < 4. And you must always state the range of validity — it is a mark on its own.

Calculate

Your turn — fractional index

6Expand (1 + 4x)^½ in ascending powers of x. What is the coefficient of x²?

Set-up

n(n−1)/2! = (½)(−½)/2 = −1/8.

(4x)² = 16x².

Coefficient = −1/8 × 16 = −2. Expansion: 1 + 2x − 2x² + …, valid for |x| < ¼.

Hint: Use n(n − 1)/2! × (4x)² with n = ½ — and remember to square the 4.

Quick check

Range of validity

?For which values of x is the binomial expansion of (1 − 3x)^(−2) valid?

Match it

Series formula flashcards

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

Quantity

Formula

Sequences · applications

Series in context

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

“Increases by £200 each year” → arithmetic, d = 200.
“Increases by 4% each year” → geometric, r = 1.04.
“Total earned over 20 years” → a sum, S₂₀ — not the 20th term.

Worked example

A salary starts at £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 − 1)/(1.04 − 1) = 24000 × (1.480244 − 1)/0.04 = 24000 × 12.00611 = £288 147 (nearest £).

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

Quick check

Term or sum?

?A savings scheme pays in £500 in year 1 and £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 − 1)d · S_n = n/2 [2a + (n − 1)d]

Geometric: u_n = ar^(n−1) · S_n = a(1 − rⁿ)/(1 − r) · S∞ = a/(1 − r) only if |r| < 1

Sigma: count the terms carefully — 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.

🏆

Mini-lesson complete!

⭐⭐⭐

You have worked through Sequences and series for Edexcel A-level Mathematics (9MA0). 🎉

Your stars: 0 / 0

Next: test yourself in the Verify stage.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to all subjects