Pearson Edexcel International Advanced Level Mathematics: Sequences and series: arithmetic, geometric and the binomial expansion (YMA01) – Revision Notes

These notes condense section 4 (topics 4.1 to 4.5) of Unit P2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification states that P2 is compulsory for IAS Mathematics and IAL Mathematics (and Pure Mathematics). For full explanations and proofs read the study guide, then test yourself on the practice questions.

Related pages: Unit P2 revision notes, the course hub, the printable checklist and all free 10-minute diagnostics.

Key vocabulary

• Term: one entry in a sequence. The nth term is uₙ.
• Common difference d (arithmetic) and common ratio r (geometric).
• Series: the sum of the terms of a sequence. Sₙ is the sum of the first n terms.
• Convergent geometric series: one with |r| < 1, which has a finite sum to infinity S∞.
• Order of a periodic sequence: the number of terms in one repeating cycle.
• Binomial coefficient ⁿCᵣ: the number of ways to choose r items from n, and the coefficient in the expansion.

4.1 Sequences

• Formula: uₙ is given in terms of n. Substitute n = 1, 2, 3, …
• Recurrence: xₙ₊₁ = f(xₙ) plus a starting value. Compute one term at a time and keep exact values (fractions, surds) until the end.
• To find which term equals a value, solve for n and keep only positive integers.

Reminder: uₙ = 2n² − 5 gives u₄ = 27. With x₁ = 3 and xₙ₊₁ = 2xₙ − 1 the terms are 3, 5, 9, 17, …

4.2 Arithmetic

Result	Formula
nth term	uₙ = a + (n − 1)d
Sum, using d	Sₙ = (n/2)(2a + (n − 1)d)
Sum, using last term l	Sₙ = (n/2)(a + l)
First n natural numbers	1 + 2 + … + n = n(n + 1)/2
Number of terms	n = (l − a)/d + 1

Proof of Sₙ (be ready to reproduce it): write the series forwards and backwards, add term by term, and note that each of the n pairs totals 2a + (n − 1)d. So 2Sₙ = n[2a + (n − 1)d].

Sigma notation. Σ (r = p to q) of f(r) adds f(p) up to f(q). Split sums over addition, take out constants, and use Σr = n(n + 1)/2. A sum starting at r = p is the sum to q minus the sum to p − 1.

Reminders: a = 5, d = 4 gives S₁₀ = 5 × 46 = 230. The sum 1 + 2 + … + 50 is 1275. Σ (r = 1 to 12) of (2r + 1) = 2 × 78 + 12 = 168, and Σ (r = 5 to 12) of (2r + 1) = 144.

4.3 Monotonic and periodic sequences

• Increasing means every term is larger than the one before (uₙ₊₁ > uₙ, any n). Decreasing means every term is smaller.
• Periodic with order k: uₙ₊ₖ = uₙ for all n.
• Method: simplify uₙ₊₁ − uₙ and decide its sign for all n, or list terms until the pattern repeats.
• For a periodic sum, divide the number of terms by k. Multiply the quotient by the sum of one cycle, then add the leftover terms from the start of the cycle.

Reminder: uₙ = cos(90n°) gives 0, −1, 0, 1, 0, −1, … and has order 4.

4.4 Geometric

Result	Formula
nth term	uₙ = arⁿ⁻¹
Finite sum	Sₙ = a(1 − rⁿ)/(1 − r), r ≠ 1
Sum to infinity	S∞ = a/(1 − r), valid for |r| < 1
Common ratio	r = u₂/u₁ = u₃/u₂

Proof of Sₙ: subtract rSₙ from Sₙ. Everything cancels except a − arⁿ, so Sₙ(1 − r) = a(1 − rⁿ).

Logs to find n.

1. Reduce the condition to rⁿ (or rⁿ⁻¹) compared with a number
2. Take logs of both sides
3. Bring the power down: n log r
4. Divide by log r. If log r < 0, reverse the inequality
5. Round up (or down) to a whole number and check the neighbouring terms

Reminders: 3, 6, 12, … has u₈ = 384 and S₆ = 189. With a = 9 and r = 1/3, S∞ = 13.5. With a = 20 and r = −1/4, S∞ = 16. For 3(1.5)ⁿ⁻¹ > 500 you need n − 1 > 12.6, so n = 14.

4.5 Binomial expansion

(a + bx)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ (bx)ʳ     r = 0 to n,  n + 1 terms
ⁿCᵣ = n!/(r!(n − r)!)           ⁿC₀ = ⁿCₙ = 1,  ⁿCᵣ = ⁿCₙ₋ᵣ

Pascal’s triangle rows give the coefficients: row 4 is 1, 4, 6, 4, 1.

• Put brackets around the whole of bx and any negative sign before raising to a power.
• To find one coefficient, write the general term and choose r so the power of x is right.
• With a product such as (1 + 3x)(5 − 2x)⁴, find which pairs of terms give the power you want, then add.

Reminders: (2 + x)⁴ = 16 + 32x + 24x² + 8x³ + x⁴. The coefficient of x² in (1 + 3x)⁵ is 10 × 9 = 90. 7C3 = 35 and 6!/4! = 30.

Method in steps

Two conditions on a sequence (simultaneous equations).

1. Turn each fact about the sequence into an equation in the first term and d or r
2. Arithmetic: subtract the equations. Geometric: divide them
3. Solve for d or r, then back-substitute for a
4. Check by generating two or three terms

Worked reminder A (arithmetic). u₂ = 7 and S₆ = 69.

a + d = 7
S₆ = 3(2a + 5d) = 69, so 2a + 5d = 23
Substitute a = 7 − d: 14 − 2d + 5d = 23, so d = 3 and a = 4
Check: 4, 7, 10, 13, 16, 19 sums to 69

Worked reminder B (geometric, two possible ratios). S∞ = 9 and the second term is 2.

ar = 2 and a/(1 − r) = 9
Substitute a = 2/r: 2/(r(1 − r)) = 9
9r² − 9r + 2 = 0, so (3r − 1)(3r − 2) = 0
r = 1/3 with a = 6, or r = 2/3 with a = 3
Both satisfy |r| < 1, so both are valid

Worked reminder C (binomial, unknown constant). The coefficient of x² in (1 + kx)⁵ is 90.

5C2 × k² = 10k² = 90, so k² = 9 and k = 3 or k = −3

Always give both roots unless the question restricts k, for example “k > 0”.

Must-know distinctions

	Arithmetic	Geometric
Rule	add d	multiply by r
nth term	a + (n − 1)d	arⁿ⁻¹
n appears as	multiplier	power (use logs)
Infinite sum	does not exist unless all terms are 0	exists if |r| < 1

Quick self-test

1. Find the 12th term of an arithmetic sequence with a = 3 and d = 7.
2. Find S₁₅ for a = 4, d = 3.
3. Evaluate Σ (r = 1 to 20) of (5r − 2).
4. A geometric sequence has u₃ = 12 and u₆ = 96. Find a and r.
5. Find S∞ for 40 − 10 + 2.5 − …
6. Find the least n for which 2 × 1.1ⁿ⁻¹ > 50.
7. Find the coefficient of x³ in (1 − x)⁸.
8. Expand (3 + x)³.
9. x₁ = 2 and xₙ₊₁ = 7 − xₙ. Find x₄ and state the order of the sequence.
10. The sum of the first n natural numbers is 820. Find n.
11. Show that the sum of the first n odd numbers is n².
12. A geometric series has a = 5 and r = 2. Find the least n for which Sₙ > 10000.

Answers

1. u₁₂ = 3 + 11×7 = 80
2. S₁₅ = (15/2)(8 + 14×3) = (15/2)(50) = 375
3. Σ(5r − 2) = 5×210 − 40 = 1010
4. r³ = 8, so r = 2, and a = 12/4 = 3
5. r = −1/4, S∞ = 40/(1 + 1/4) = 32
6. 1.1ⁿ⁻¹ > 25, n − 1 > ln 25/ln 1.1 = 33.77, so n = 35
7. 8C3 × (−1)³ = −56
8. 27 + 27x + 9x² + x³
9. x₂ = 5, x₃ = 2, x₄ = 5; the terms alternate 2, 5, so the order is 2
10. n(n + 1)/2 = 820, so n² + n − 1640 = 0, giving n = 40
11. a = 1, d = 2: Sₙ = (n/2)(2 + 2(n − 1)) = (n/2)(2n) = n²
12. 5(2ⁿ − 1) > 10000, so 2ⁿ > 2001, n > ln 2001/ln 2 = 10.97, so n = 11

Where marks are usually lost

• Using n when you needed n − 1 in aₙ = a + (n − 1)d, or in the power of r.
• Quoting S∞ for a series with |r| ≥ 1, or forgetting to state that |r| < 1 holds.
• Not reversing an inequality after dividing by a negative logarithm.
• Giving a fractional n as a final answer instead of rounding to a whole number and checking the neighbouring terms.
• Dropping the bracket or the sign in (bx)ʳ, for example writing −3x² instead of 9x² for (−3x)².
• Finding coefficients in a product by multiplying only the first terms and missing the cross terms.
• Declaring a sequence increasing after checking only a few terms.
• Skipping lines in a proof. A proof that jumps to the final formula shows less than the argument asks for.

Next, try the practice questions or revisit the study guide.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Unit P2, section 4 (Pearson Education Limited).