Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 1: Pure Mathematics 1 · official syllabus
Study notes
Use these worked notes to check exact formulas, follow the reasoning and learn the exam wording.
Binomial expansion for positive integer powers 1.6
For a positive integer n, (a + b)ⁿ = Σ from r = 0 to n of C(n,r)aⁿ⁻ʳbʳ. Here C(n,r) = n!/[r!(n-r)!], 0! = 1, and the expansion has n + 1 terms.
The term indexed by r is the (r + 1)th term. For a term in a specified power of x, write the general term first, then solve for r; do not guess from its position.
Example: (2 + x)⁴ = 16 + 32x + 24x² + 8x³ + x⁴. The coefficient of x² is C(4,2)2² = 24.
For (1 - 2x)⁵, the x³ term is C(5,3)(-2x)³ = -80x³. The negative sign belongs to the entire second term and alternates with r.
For the coefficient of x² in (1 + 2x)³(1 - x)², collect all products giving x²: 12 - 12 + 1 = 1. Terms of degree greater than two cannot contribute.
Arithmetic progressions: terms and sums 1.6
An arithmetic progression (AP) has constant difference d. With first term a, its nth term is uₙ = a + (n - 1)d and its first n terms sum to Sₙ = n[2a + (n - 1)d]/2 = n(a + uₙ)/2.
For 7, 11, 15, ... we have a = 7 and d = 4. Then u₂₀ = 7 + 19×4 = 83, and S₂₀ = 20(7 + 83)/2 = 900.
If u₅ = 17 and u₁₂ = 45, subtract a + 4d = 17 from a + 11d = 45: 7d = 28, so d = 4 and a = 1. Always use n - 1 steps from the first term.
Three consecutive terms a, b, c are in AP exactly when 2b = a + c. AP sums are finite; an infinite nonzero arithmetic progression does not have a finite sum.
Geometric progressions: terms and finite sums 1.6
A geometric progression (GP) has constant ratio r, so uₙ = arⁿ⁻¹. If r ≠ 1, Sₙ = a(1 - rⁿ)/(1 - r), equivalently a(rⁿ - 1)/(r - 1). If r = 1, Sₙ = na.
For 3, 6, 12, ... we have a = 3 and r = 2. The eighth term is 3×2⁷ = 384, and S₈ = 3(2⁸ - 1) = 765.
A negative ratio alternates signs. For a = 8 and r = -1/2, the first four terms are 8, -4, 2, -1 and S₄ = 5. Substituting into 8[1 - (-1/2)⁴]/[1 - (-1/2)] also gives 5.
If u₂ = 6 and u₅ = 162, then ar = 6 and ar⁴ = 162. Division gives r³ = 27, so r = 3 and a = 2. Check both original terms.
Consecutive terms a, b, c in a GP satisfy b² = ac. When solving a ratio problem, also check any zero terms rather than dividing by a possible zero.
Convergent geometric series and sums to infinity 1.6
A GP with nonzero first term has a convergent sum to infinity when |r| < 1. Its sum is S∞ = a/(1 - r), because rⁿ tends to zero. r = 1 and r = -1 do not satisfy the condition.
For a = 12 and r = 1/3, S∞ = 12/(2/3) = 18. After four terms, the remaining tail is ar⁴/(1-r) = 2/9. The tail starts at the fifth term, not the fourth.
For a = 6 and r = -1/2, S∞ = 6/(3/2) = 4. Alternating signs do not prevent convergence when |r| < 1.
If a GP has first term 5 and sum to infinity 20, solve 5/(1-r) = 20 to obtain r = 3/4. Verify |r| < 1 before accepting an infinite-sum calculation.
Choosing a model and checking a solution 1.6
An AP adds the same amount each step; a GP multiplies by the same factor. A 10% increase each year is ratio 1.1, not difference 0.1. State what the first term represents before choosing n.
An AP and a GP may share some terms without being the same sequence. Keep their first terms, differences and ratios separate when setting up simultaneous conditions.
If three positive numbers x, 6, y are in GP and x + y = 15, then xy = 36. Thus x and y are the roots of t² - 15t + 36 = 0, giving 3 and 12. Both orders are possible, with ratios 2 and 1/2.
Check that a number of terms is a positive integer, all supplied conditions hold, and an infinite sum uses |r| < 1. A plausible formula without these checks is not a complete solution.
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