← All notesMaths · Series: binomial expansion, AP and GP⬇ Download PDF
A Level Maths 9709 · Paper 1 Pure Mathematics 1 · 1.6 · ~8 min read

Series: binomial expansion, AP and GP

The binomial expansion for positive integer powers, specific terms and terms independent of x, arithmetic and geometric progressions, sums to infinity, and problems that link the two.

Binomial expansionⁿCᵣ & PascalSpecific coefficientsTerm independent of xArithmetic progressionsGeometric progressionsSum to infinityLinked AP & GP
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • (a + b)ⁿ has n + 1 terms. The term containing bʳ is ⁿCᵣ aⁿ⁻ʳ bʳ. Put brackets round every term before raising it to a power.
  • Arithmetic progression (AP): add the same d each time. uₙ = a + (n − 1)d and Sₙ = ½n[2a + (n − 1)d].
  • Geometric progression (GP): multiply by the same r each time. uₙ = arⁿ⁻¹ and Sₙ = a(1 − rⁿ)/(1 − r). If ǀrǀ < 1, the sum to infinity is a/(1 − r).
  • Three numbers a, b, c are in AP if 2b = a + c, and in GP if b² = ac.

By the end you can

  • Expand (a + b)ⁿ and find any specific term or coefficient, including in products.
  • Find the term independent of x.
  • Solve problems with APs and GPs, including sums between two terms and the smallest n for a sum.
  • Use the condition for convergence and the sum to infinity.

1The binomial expansion

Where the coefficients come from

  • (a + b)ⁿ = (a + b)(a + b)…(a + b). To get the term in aⁿ⁻ʳbʳ, you choose b from r of the n brackets. There are ⁿCᵣ ways to do that.
(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + bⁿ FORMULA LIST (MF19)
n a positive integer; ⁿCᵣ = n!/(r!(n − r)!)
  • Pascal's triangle gives the same numbers for small n: row 5 is 1, 5, 10, 10, 5, 1.
  • Symmetry: ⁿCᵣ = ⁿCₙ₋ᵣ, e.g. ⁸C₃ = ⁸C₅ = 56.
Worked example

Find the first three terms, in ascending powers of x, of (2 − x)⁵.

  1. 2⁵ + ⁵C₁(2⁴)(−x) + ⁵C₂(2³)(−x)² M1
  2. = 32 − 80x + 80x² A1 A1
Common mistakeLosing brackets. (3x)² is 9x², not 3x². (−x)³ is −x³. The June 2024 report flagged expanding (ax)³ as ax³.

Specific terms and products

Worked example

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

  1. from (2 − x)⁵: 32 − 80x + 80x² + … B1
  2. x² terms come from 1 × 80x² and 3x × (−80x) M1
  3. coefficient = 80 − 240 = −160 A1
Worked example

Find the term independent of x in the expansion of (x² + 2/x)⁶.

  1. general term ⁶Cᵣ(x²)⁶⁻ʳ(2/x)ʳ = ⁶Cᵣ 2ʳ x12 − 3r M1
  2. independent of x ⇒ 12 − 3r = 0 ⇒ r = 4 M1
  3. ⁶C₄ × 2⁴ = 15 × 16 = 240 A1
Common mistakeMultiplying the wrong coefficients. In a product like (1 + 3x)(2 − x)⁵, pair every term from the first bracket with the term that makes the right power. Students lost marks in 2024 by multiplying only one pair.

2Arithmetic progressions

The formulas

uₙ = a + (n − 1)d FORMULA LIST (MF19)
a = first term, d = common difference
Sₙ = ½n[2a + (n − 1)d] = ½n(a + l) FORMULA LIST (MF19)
l = last term
  • Why ½n(a + l)? Write the sum forwards and backwards and add: every pair adds to a + l, and there are n pairs, so 2Sₙ = n(a + l). This is Gauss's trick.
  • Sum of the terms from the 11th to the 40th = S₄₀ − S₁₀, not S₄₀ − S₁₁.
Worked example

The 3rd term of an AP is 11 and the 10th term is 32. Find the sum of the first 20 terms.

  1. a + 2d = 11, a + 9d = 32 M1
  2. 7d = 21 ⇒ d = 3, a = 5 A1
  3. S₂₀ = 10[10 + 19 × 3] = 670 M1 A1
Worked example

An AP has first term 5 and common difference 3. Find the least number of terms for the sum to exceed 1000.

  1. ½n[10 + 3(n − 1)] > 1000 ⇒ 3n² + 7n − 2000 > 0 M1
  2. roots: n = (−7 ± √24049)/6 ⇒ positive root n = 24.68 M1
  3. n must be a whole number above 24.68: n = 25 A1
  4. check: S₂₄ = 948 and S₂₅ = 1025 ✓
Common mistakeRounding n to the nearest integer. 24.68 means 24 terms are not enough, so n = 25. Always check the two whole numbers on either side.

3Geometric progressions

The formulas

uₙ = arⁿ⁻¹ FORMULA LIST (MF19)
r = common ratio
Sₙ = a(1 − rⁿ)/(1 − r) FORMULA LIST (MF19)
use the form a(rⁿ − 1)/(r − 1) when r > 1
S∞ = a/(1 − r), for ǀrǀ < 1 FORMULA LIST (MF19)
the condition for convergence
  • Why the sum formula works: Sₙ = a + ar + … + arⁿ⁻¹ and rSₙ = ar + … + arⁿ. Subtract: Sₙ − rSₙ = a − arⁿ.
  • Why ǀrǀ < 1? Then rⁿ → 0 as n grows, so Sₙ → a/(1 − r). If ǀrǀ ≥ 1, the terms don't shrink and the sum has no limit.
Worked example

A GP has first term 24 and common ratio −½. Find the sum of the first 5 terms and the sum to infinity.

  1. S₅ = 24(1 − (−½)⁵)/(1 − (−½)) = 24 × (33/32)/(3/2) = 16.5 M1 A1
  2. S∞ = 24/(1 + ½) = 16 M1 A1
Worked example

The second term of a GP is 12 and its sum to infinity is 64. Find the possible values of r.

  1. ar = 12 and a/(1 − r) = 64 B1
  2. a = 12/r ⇒ 12/(r(1 − r)) = 64 ⇒ 16r² − 16r + 3 = 0 M1
  3. (4r − 1)(4r − 3) = 0 ⇒ r = ¼ or r = ¾ A1 (both satisfy ǀrǀ < 1)
Exam tip"Show that the GP is convergent" needs the reason: ǀrǀ < 1, with the value of r shown.

4Linking an AP and a GP

Use 2b = a + c and b² = ac

Worked example

The first, second and fourth terms of an AP (with d ≠ 0) are the first three terms of a GP. Show that a = d, and find the common ratio of the GP.

  1. AP terms: a, a + d, a + 3d B1
  2. in GP: (a + d)² = a(a + 3d) M1
  3. a² + 2ad + d² = a² + 3ad ⇒ d² = ad ⇒ d = a (since d ≠ 0) A1
  4. ratio = (a + d)/a = 2a/a = 2 A1
Common mistakeMixing up the letters. Use different letters for the AP and the GP (for example a, d and A, r), or say clearly which is which. The 2024 report noted confusion between p and d.
Linking questionThe binomial expansion extends to negative and fractional powers in Algebra (Paper 3), and binomial coefficients reappear in Permutations and combinations and the binomial distribution.

Test yourself — cover the answers first

Quick check

  1. Write down ⁷C₂.
  2. Find the coefficient of x³ in (1 + x/2)⁸.
  3. An AP has a = 4, d = −2. Find u₁₀.
  4. For which values of r does a GP have a sum to infinity?
  5. Find S∞ for 9 + 3 + 1 + ….

Exam-style

  1. (a) Find the first three terms in the expansion of (1 + ax)⁶ in ascending powers of x. (b) The coefficient of x² in (2 − x)(1 + ax)⁶ is 36. Find the possible values of a.
  2. An AP has sum of first 10 terms 175 and sum of first 20 terms 650. Find the first term and the common difference.
  3. A ball is dropped from 2 m and each bounce reaches 80% of the previous height. Find the total distance travelled before it comes to rest.
Show answers
  1. 21
  2. ⁸C₃ × (½)³ = 7
  3. −14
  4. −1 < r < 1
  5. 13.5
  6. (a) 1 + 6ax + 15a²x²
    (b) 30a² − 6a = 36 ⇒ 5a² − a − 6 = 0 ⇒ a = 6/5 or a = −1
  7. 5(2a + 9d) = 175, 10(2a + 19d) = 650 ⇒ 2a + 9d = 35, 2a + 19d = 65 ⇒ d = 3, a = 4
  8. 2 + 2 × (1.6/(1 − 0.8)) = 2 + 16 = 18 m
Practise this chapter
A Level practice papers
Next chapterDifferentiation →
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.