Mini-Lesson

Sequences and series

OCR section 1.04 covers arithmetic and geometric sequences and series, sigma notation, recurrence relations, and the binomial expansion — including the A-level extension to negative and fractional indices, which only works when |x| is small.

The binomial validity condition is where most marks are dropped — we will nail it. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.

Sequences · sigma notation

Sigma notation

Σ is an instruction to add. Σ from r = 1 to n of u_r means: substitute r = 1, 2, 3, …, n and total the results.

Σ_r=1..n (ar + b) = a · Σr + b · nuseful: Σ from r = 1 to n of r = n(n + 1)/2

Worked example — Σ from r = 1 to 10 of (3r + 2)

= 3 × (Σr) + 2 × 10 = 3 × (10 × 11 ÷ 2) + 20

= 3 × 55 + 20 = 165 + 20 = 185.

Check — it is an arithmetic series: first term 5, last term 32, 10 terms. Sum = (10/2)(5 + 32) = 5 × 37 = 185. ✓

Calculate

Your turn — sigma notation

1Evaluate Σ from r = 1 to 10 of (3r + 2).

Hint: 3(1 + 2 + … + 10) + 2 × 10 = 3 × 55 + 20.

Sequences · arithmetic

Arithmetic sequences and series

u_n = a + (n − 1)d · S_n = (n/2)[2a + (n − 1)d] = (n/2)(a + L)a = first term, d = common difference, L = last term

Worked example — a = 5, d = 3. Find S₂₀.

S₂₀ = (20/2)[2(5) + 19(3)] = 10[10 + 57] = 10 × 67 = 670.

Check with the other formula: u₂₀ = 5 + 19(3) = 62, so S₂₀ = (20/2)(5 + 62) = 10 × 67 = 670. ✓

Classic slip: it is (n − 1)d, not nd. The 10th term of 5, 8, 11, … is 5 + 9(3) = 32, not 5 + 10(3) = 35.

Sequences · geometric

Geometric sequences and series

u_n = arⁿ⁻¹ · S_n = a(1 − rⁿ)/(1 − r) · S_∞ = a/(1 − r)the sum to infinity exists ONLY if |r| < 1 — the series is then said to converge

Worked example — a = 8, r = 0.5

|r| = 0.5 < 1, so the series converges and S_∞ exists.

S_∞ = 8 / (1 − 0.5) = 8 / 0.5 = 16.

Sanity check: 8 + 4 + 2 + 1 + 0.5 + … is creeping up on 16 and never passes it. ✓

Divergence: if |r| ≥ 1 the terms do not shrink, so there is no sum to infinity. Writing S_∞ = a/(1 − r) for r = 2 gives a nonsense negative answer — a favourite examiner trap.

Calculate

Your turn — sum to infinity

2A geometric series has first term a = 8 and common ratio r = 0.5. Find its sum to infinity.

Hint: |r| < 1, so S_∞ = a/(1 − r) = 8/(1 − 0.5).

Quick check

Does it converge?

?For which common ratio does a geometric series have a sum to infinity?

Sequences · recurrence

Recurrence relations

A recurrence relation defines each term from the previous one, e.g. u_n+1 = 2u_n − 1 with u₁ = 3.

Worked example — generate the terms

u₁ = 3 → u₂ = 2(3) − 1 = 5 → u₃ = 2(5) − 1 = 9 → u₄ = 2(9) − 1 = 17 → u₅ = 2(17) − 1 = 33.

OCR wants the vocabulary too:

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. u_n+1 = 1/u_n with u₁ = 4 gives 4, ¼, 4, ¼, … — period 2.

Calculate

Your turn — recurrence relation

3A sequence is defined by u₁ = 3 and u_n+1 = 2u_n − 1. Find u₅.

u₅ =

Hint: 3 → 5 → 9 → 17 → ? Each time, double and subtract 1.

Sequences · binomial

The binomial expansion (positive integer n)

(a + b)ⁿ = Σ ⁿC_r a^n−r b^r · ⁿC_r = n! / (r!(n − r)!)for n a positive integer the expansion terminates — it has exactly n + 1 terms

Worked example — the coefficient of x³ in (1 + 2x)⁶

The x³ term is ⁶C₃ × 1³ × (2x)³.

⁶C₃ = 6!/(3!3!) = 720/(6 × 6) = 20, and (2x)³ = 8x³.

Coefficient = 20 × 8 = 160.

The classic error: forgetting to cube the 2. Writing ‘coefficient = 20’ loses the mark — the 2 belongs to the bracket and gets raised to the power too.

Calculate

Your turn — binomial coefficient

4Find the coefficient of x³ in the expansion of (1 + 2x)⁶.

Hint: ⁶C₃ × (2)³ = 20 × 8. Do not forget to cube the 2.

Sequences · binomial

Negative and fractional indices — and validity

When n is not a positive integer, the expansion never terminates. It becomes an infinite series, and it is only valid — meaning it actually converges to the function — for small enough x.

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

Worked example — expand (1 + 2x)⁻³ up to the x² term

Here n = −3 and the bracket contains 2x, so substitute X = 2x.

x-term: n × X = −3(2x) = −6x.

x²-term: [n(n − 1)/2!] × X² = [(−3)(−4)/2] × (2x)² = 6 × 4x² = 24x².

∴ (1 + 2x)⁻³ ≈ 1 − 6x + 24x² − …

Validity: we need |X| < 1, i.e. |2x| < 1, so |x| < ½.

Two marks people forget. (1) To expand (4 + x)^1/2 you must first factor out the 4: 2(1 + x/4)^1/2, valid for |x| < 4. (2) Always state the validity condition — it is an explicit OCR requirement.

Calculate

Your turn — negative index

5Expand (1 + 2x)⁻³. Find the coefficient of x².

Hint: [n(n − 1)/2!] × (2x)² with n = −3, so [(−3)(−4)/2] × 4x² = 6 × 4x².

Quick check

When is the expansion valid?

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

Sort it

Arithmetic, geometric or neither?

Tap a sequence, then tap the group it belongs to.

➕ Arithmetic

✖️ Geometric

❓ Neither

Match it

Name that formula

Tap a description on the left, then the formula it names.

Description

Formula

Quick check

Spot the sequence type

?The sequence 81, 27, 9, 3, … is:

Quick check

Periodic sequence

?The sequence defined by u_n+1 = 1/u_n with u₁ = 4 is best described as:

Recap

The big ideas to know

Sigma: Σ is an instruction to add; Σr from 1 to n = n(n + 1)/2

Arithmetic: u_n = a + (n − 1)d; S_n = (n/2)[2a + (n − 1)d]

Geometric: u_n = arⁿ⁻¹; S_n = a(1 − rⁿ)/(1 − r)

Sum to infinity: S_∞ = a/(1 − r), and it exists only when |r| < 1

Recurrence: generate term by term; know increasing, decreasing and periodic

Binomial: ⁿC_r for positive integer n; for negative or fractional n the series is infinite

Validity: (1 + kx)ⁿ is valid for |kx| < 1 — always state it

That is OCR 1.04 — and partial fractions plus the binomial series is a favourite combined question. Press Finish to see your score.

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