IGCSE Additional Maths · Cambridge 0606 · Edexcel 4PM1

Series and the binomial expansion: notes and questions

Revision for the binomial expansion, arithmetic and geometric progressions, sum to infinity, sigma notation and (4PM1) the binomial series for rational powers. Both boards. Edexcel 4PM1 also uses Σ notation and expands (1 + x)ⁿ for rational n with its validity condition.

Cambridge 0606: section 12
Edexcel 4PM1: section 5, 6
14 questions

Practise series and the binomial expansion →

Key points

(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ for r = 0 to n. Keep the sign inside the bracket with the b term.
For a single term, write the general term and choose r to get the power you need.
Arithmetic: uₙ = a + (n − 1)d and Sₙ = n/2 (2a + (n − 1)d) = n/2 (a + l).
Geometric: uₙ = arⁿ⁻¹ and Sₙ = a(1 − rⁿ)/(1 − r). If |r| < 1, S∞ = a/(1 − r).
(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + … For n that is not a positive integer it is valid only for |x| < 1.
To find the least n for which a GP sum passes a value, form an inequality and use logarithms.

Watch out: In (3 − 2x)⁶ the second term is (−2x), so powers of −2 alternate in sign. Brackets round (−2x)³ stop sign errors.

Worked example · 3 marks

Expand \((2 + x)^4\) fully, simplifying each term.

M1 binomial coefficients 1, 4, 6, 4, 1 with powers of 2
A1 two terms correct
A1 16 + 32x + 24x² + 8x³ + x⁴

Coefficients from Pascal's triangle: 1, 4, 6, 4, 1.
2⁴ + 4(2³)x + 6(2²)x² + 4(2)x³ + x⁴.
= 16 + 32x + 24x² + 8x³ + x⁴.

Answer: \(16 + 32x + 24x^2 + 8x^3 + x^4\)

Series and the binomial expansion questions

Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).

Question 1 · 2 marks06064PM1no calculator

In the expansion of \((2 - 3x)^5\), find the coefficient of \(x^3\).

Show the mark scheme and worked solution

M1 ⁵C₃ × 2² × (−3)³
A1 −1080

The x³ term is ⁵C₃ × 2² × (−3x)³.
= 10 × 4 × (−27)x³ = −1080x³.
The coefficient is −1080.

Answer: \(-1080\)

Question 2 · 4 marks06064PM1no calculator

(a) Find the first three terms, in ascending powers of \(x\), of the expansion of \(\left(1 + \frac{x}{2}\right)^8\).

(b) Use your answer to estimate the value of \(1.005^8\).

Show the mark scheme and worked solution

B1 (a) 1 + 4x
B1 (a) 7x²
M1 (b) x = 0.01 substituted
A1 (b) 1.0407

(1 + x/2)⁸ = 1 + 8(x/2) + 28(x/2)² + … = 1 + 4x + 7x² + …
1.005 = 1 + x/2 when x = 0.01.
1 + 0.04 + 0.0007 = 1.0407.

Answer: (a) \(1 + 4x + 7x^2\) (b) \(1.0407\)

Question 3 · 4 marks06064PM1no calculator

The 5th term of an arithmetic progression is 18 and the 12th term is 46.

(a) Find the first term and the common difference.

(b) Find the least number of terms needed for the sum to exceed 500.

Show the mark scheme and worked solution

M1 (a) a + 4d = 18, a + 11d = 46
A1 (a) d = 4, a = 2
M1 (b) Sₙ = 2n² > 500
A1 (b) 16

7d = 46 − 18 = 28, so d = 4 and a = 18 − 16 = 2.
Sₙ = n/2 (2 × 2 + (n − 1) × 4) = n/2 × 4n = 2n².
2n² > 500 means n² > 250. 15² = 225 is too small and 16² = 256 works.
The least number of terms is 16.

Answer: (a) first term \(2\), common difference \(4\) (b) \(16\)

Question 4 · 2 marks06064PM1no calculator

Find the sum of the first 10 terms of the geometric progression \(3 + 6 + 12 + \ldots\)

Show the mark scheme and worked solution

M1 3(2¹⁰ − 1)/(2 − 1)
A1 3069

a = 3 and r = 2.
S₁₀ = 3(2¹⁰ − 1)/(2 − 1) = 3 × 1023 = 3069.

Answer: \(3069\)

9 more series and the binomial expansion questions

Each with a mark scheme and a worked solution. Included with IGCSE plans, free trials and school licences.

Next steps

Cambridge 0606: papers, syllabus map and official past papers
Edexcel 4PM1: papers, syllabus map and official past papers
Going on to A Level? Series and the binomial expansion continues in A Level sequences and series (A Level Math Revision). See also from Additional Maths to A Level.