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

This guide teaches section 4 (topics 4.1 to 4.5) of Unit P2: Pure Mathematics 2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification states that P2 is a compulsory unit for both IAS Mathematics and IAL Mathematics (and Pure Mathematics), so there is no optional route around it. The specification also says knowledge of P1 may be tested, so keep your P1 algebra sharp.

Use it with the revision notes and the practice questions. For the wider course see the Edexcel A Level Mathematics hub, the printable checklist, the Unit P2 overview and all free 10-minute diagnostics.

What the section covers

Topic	What you must be able to do
4.1	Work with sequences given by a formula for the nth term, or by a relation xₙ₊₁ = f(xₙ)
4.2	Arithmetic sequences and series: nth term, sum of a finite series (know the proof), sum of the first n natural numbers, Σ notation
4.3	Increasing, decreasing and periodic sequences
4.4	Geometric sequences and series: nth term, finite sum (know the proof), sum to infinity S∞ when |r| < 1, using logs to find n
4.5	Binomial expansion of (a + bx)ⁿ for positive integer n; the notations n!, ⁿCᵣ and the column form may be used

4.1 Sequences

A sequence is an ordered list of terms. You write the terms u₁, u₂, u₃, … and the general term uₙ.

By formula. If uₙ = n² − 6n + 10, substitute n = 1, 2, 3, 4 to get 5, 2, 1, 2.

By recurrence. A relation xₙ₊₁ = f(xₙ) needs a starting value. You build the terms one at a time.

Worked example 1. The sequence uₙ = n² − 6n + 10 is given. Which term equals 65?

n² − 6n + 10 = 65
n² − 6n − 55 = 0
(n − 11)(n + 5) = 0
n = 11   (n = −5 is rejected: n must be a positive integer)

The 11th term is 65.

Worked example 2. x₁ = 4 and xₙ₊₁ = xₙ/2 + 3. Find x₂, x₃ and x₄.

x₂ = 4/2 + 3 = 5
x₃ = 5/2 + 3 = 11/2
x₄ = (11/2)/2 + 3 = 23/4

4.2 Arithmetic sequences and series

In an arithmetic sequence each term is the previous term plus a constant common difference d. With first term a:

uₙ = a + (n − 1)d
Sₙ = (n/2)(2a + (n − 1)d) = (n/2)(a + l)     where l is the last term

Proof of the sum formula

Write the sum forwards and backwards, then add.

Sₙ = a + (a + d) + … + (a + (n − 1)d)
Sₙ = (a + (n − 1)d) + (a + (n − 2)d) + … + a
2Sₙ = [2a + (n − 1)d] + [2a + (n − 1)d] + … (n times)
2Sₙ = n[2a + (n − 1)d]

Divide by 2 to finish. Every pair adds to the same total, which is why the method works.

The first n natural numbers

Put a = 1, d = 1: 1 + 2 + … + n = n(n + 1)/2.

Sigma notation

Σ (from r = 1 to n) of f(r) means f(1) + f(2) + … + f(n). The limits tell you the first and last value of r.

Worked example 3. The 3rd term of an arithmetic sequence is 16 and the 9th term is 40. Find a and d, then S₂₀.

a + 2d = 16
a + 8d = 40
Subtract: 6d = 24, so d = 4 and a = 8
S₂₀ = (20/2)(2×8 + 19×4) = 10 × 92 = 920

Worked example 4. Find the sum of all multiples of 7 between 100 and 500.

Multiples run 105, 112, …, 497, so a = 105, d = 7, l = 497
497 = 105 + (n − 1)×7, so n − 1 = 56 and n = 57
S₅₇ = (57/2)(105 + 497) = (57/2)(602) = 17157

Worked example 5. Evaluate Σ (r = 1 to 30) of (4r − 3).

Σ(4r − 3) = 4 Σr − 3 × 30
Σr = 30 × 31/2 = 465
Result: 4 × 465 − 90 = 1770

Worked example 6. An arithmetic series has a = −12 and d = 3. Find the least n for which Sₙ > 200.

Sₙ = (n/2)(−24 + 3(n − 1)) = n(3n − 27)/2
n(3n − 27)/2 > 200
3n² − 27n − 400 > 0
Roots: n = (27 ± √5529)/6, and the positive root is about 16.89

So n ≥ 17. Check: S₁₆ = 168 (too small) and S₁₇ = 204.

4.3 Increasing, decreasing and periodic sequences

A sequence is increasing if uₙ₊₁ > uₙ for every n, and decreasing if uₙ₊₁ < uₙ for every n. It is periodic if the terms repeat in a fixed cycle: uₙ₊ₖ = uₙ for all n, where k is the order of the sequence (the smallest such k).

Worked example 7. For uₙ = n² − 8n, show how the sequence behaves.

uₙ₊₁ − uₙ = (n + 1)² − 8(n + 1) − n² + 8n = 2n − 7

This is negative for n = 1, 2, 3 and positive from n = 4. The terms are −7, −12, −15, −16, −15, −12, −7, … The sequence decreases up to u₄ and then increases. It is neither increasing nor decreasing overall.

Worked example 8. x₁ = 2 and xₙ₊₁ = 1/(1 − xₙ). Show it is periodic and find the sum of the first 100 terms.

x₂ = 1/(1 − 2) = −1
x₃ = 1/(1 + 1) = 1/2
x₄ = 1/(1 − 1/2) = 2 = x₁

The cycle 2, −1, 1/2 repeats, so the order is 3. One cycle sums to 3/2. In 100 terms there are 33 complete cycles and one extra term, which is x₁ = 2. Sum = 33 × 3/2 + 2 = 51.5.

4.4 Geometric sequences and series

In a geometric sequence each term is the previous term times a constant common ratio r.

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

Proof of the finite sum

Sₙ = a + ar + ar² + … + arⁿ⁻¹
rSₙ = ar + ar² + … + arⁿ⁻¹ + arⁿ
Sₙ − rSₙ = a − arⁿ
Sₙ(1 − r) = a(1 − rⁿ)

Divide by (1 − r). If |r| < 1 then rⁿ → 0 as n grows, which leaves S∞ = a/(1 − r). If |r| ≥ 1 the terms do not shrink to zero and there is no sum to infinity.

Worked example 9. A geometric sequence has u₂ = 6 and u₅ = 162. Find a, r, u₈ and S₈.

ar = 6 and ar⁴ = 162
Divide: r³ = 27, so r = 3 and a = 2
u₈ = 2 × 3⁷ = 4374
S₈ = 2(3⁸ − 1)/(3 − 1) = 6560

Worked example 10. Find S∞ for 18 + 12 + 8 + … Then find the least n for which Sₙ > 53.

r = 12/18 = 2/3, which satisfies |r| < 1
S∞ = 18/(1 − 2/3) = 54
Sₙ = 54(1 − (2/3)ⁿ) > 53
(2/3)ⁿ < 1/54
n ln(2/3) < −ln 54
n > ln 54 / ln(3/2) = 9.84…     (the inequality flips because ln(2/3) < 0)

So n = 10.

Worked example 11. A geometric sequence starts 5, 6, 7.2, … Find the first term greater than 1000.

r = 1.2, uₙ = 5(1.2)ⁿ⁻¹ > 1000
(1.2)ⁿ⁻¹ > 200
(n − 1) ln 1.2 > ln 200, so n − 1 > 29.06…
n = 31

The 30th term is about 989 and the 31st is about 1187.

Worked example 12. A series has first term 12 and sum to infinity 48. Find r.

12/(1 − r) = 48, so 1 − r = 1/4 and r = 3/4

4.5 The binomial expansion

n! = n × (n − 1) × … × 1, with 0! = 1. The binomial coefficient is ⁿCᵣ = n!/(r!(n − r)!), also written with n above r in brackets. For a positive integer n:

(a + bx)ⁿ = aⁿ + ⁿC₁ aⁿ⁻¹(bx) + ⁿC₂ aⁿ⁻²(bx)² + … + (bx)ⁿ

There are n + 1 terms. The general term is ⁿCᵣ aⁿ⁻ʳ(bx)ʳ. Keep the bracket round bx, so that b is raised to the power r as well.

Worked example 13. Find the first four terms of (2 − 3x)⁶ in ascending powers of x.

6C0 = 1, 6C1 = 6, 6C2 = 15, 6C3 = 20
Term 1: 2⁶ = 64
Term 2: 6 × 2⁵ × (−3x) = −576x
Term 3: 15 × 2⁴ × (−3x)² = 2160x²
Term 4: 20 × 2³ × (−3x)³ = −4320x³

Worked example 14. Find the coefficient of x² in (1 + 2x)(3 − x)⁵.

(3 − x)⁵ = 243 − 405x + 270x² − …
x² terms: 1 × 270x² + 2x × (−405x) = 270x² − 810x²
Coefficient: −540

Worked example 15. The coefficient of x³ in (2 + kx)⁶ is −1280. Find k.

6C3 × 2³ × k³ = 20 × 8 × k³ = 160k³
160k³ = −1280, so k³ = −8 and k = −2

Common errors

• Using Sₙ with the wrong n after finding a “number of terms” that is one out. Count with n = (l − a)/d + 1.
• Applying S∞ with |r| ≥ 1.
• Forgetting to reverse the inequality when dividing by a negative logarithm.
• Writing (bx)ʳ as bxʳ in a binomial term.
• Reading “increasing” from the first two terms only. Test uₙ₊₁ − uₙ for general n.

Check your skills on the practice set and recall them with the revision notes.

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).