Lesson 13: Sequences

A sequence is a function whose domain is the counting numbers (Lesson 8 in disguise), and Chinese mathematics mastered its two great families early: Yang Hui stacked spheres in pyramids and summed the layers, producing formulas for Σk and Σk² that are sequence sums. The Gaokao sets one sequence problem every year, typically a proof that a constructed sequence is arithmetic or geometric, followed by a sum. This lesson covers arithmetic and geometric sequences completely — general terms, sums, identification — plus the transformation trick that cracks recursive definitions, and the second-difference method for quadratic sequences.

The two families

Arithmetic: constant difference d. General term aₙ = a₁ + (n − 1)d. Sum of the first n terms: Sₙ = n(a₁ + aₙ)/2 = na₁ + n(n − 1)d/2 — pair first with last (Gauss's schoolroom trick, known in China far earlier). The middle term property: aₘ is the average of terms equidistant from it.

Geometric: constant ratio q. General term aₙ = a₁qⁿ⁻¹ — note the exponent n − 1, because the first term carries q⁰. Sum: Sₙ = a₁(1 − qⁿ)/(1 − q) for q ≠ 1 (and Sₙ = na₁ for q = 1), proved by writing Sₙ, multiplying by q, and subtracting — the shift-and-subtract manoeuvre that also sums infinite series when |q| < 1: S∞ = a₁/(1 − q).

Identification: compute differences (constant → arithmetic) or ratios (constant → geometric); if first differences grow linearly (constant second differences), the sequence is quadratic in n.

Worked examples

Worked example 1 — arithmetic. a₁ = 5, d = 3. Then a₂₀ = 5 + 19 × 3 = 62, and S₂₀ = 20 × (5 + 62)/2 = 670. Check plausibility: twenty terms averaging about (5 + 62)/2 = 33.5 gives 670 ✓.

Worked example 2 — geometric. 3, 6, 12, … has q = 2. a₈ = 3 × 2⁷ = 384. S₈ = 3(2⁸ − 1)/(2 − 1) = 3 × 255 = 765. Partial check: 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765 ✓.

Worked example 3 — second differences. Sequence: 2, 5, 10, 17, 26, … First differences: 3, 5, 7, 9. Second differences: 2, 2, 2 — constant, so aₙ is quadratic: try aₙ = n² + 1: n = 1 gives 2 ✓, n = 2 gives 5 ✓, n = 4 gives 17 ✓. So a₁₀ = 101. Pattern recognition protocol: differences, then ratios, then second differences — in that order.

Worked example 4 — money, both families. (a) Arithmetic saving: deposits of ¥100, ¥200, ¥300, …, ¥1200 in months 1–12 total 100 × (1 + 2 + … + 12) = 100 × 78 = ¥7,800. (b) Geometric growth: ¥10,000 at 5% annual compound interest becomes 10000 × 1.05³ = ¥11,576.25 after 3 years. Interest on interest is a geometric sequence — the reason compound growth outruns every arithmetic saving plan eventually.

Worked example 5 — inserting means. Insert three arithmetic means between 3 and 19 (numbers x₁, x₂, x₃ so that 3, x₁, x₂, x₃, 19 is arithmetic). The common difference: d = (19 − 3)/(3 + 1) = 4 — dividing by the number of gaps (4), not the number of inserted terms (3). Sequence: 3, 7, 11, 15, 19. The means are 7, 11, 15.

Challenge problem — Gaokao standard

Problem. A sequence satisfies a₁ = 1 and aₙ₊₁ = 2aₙ + 1. Find a closed formula for aₙ, and compute a₁₀.

Solution. The "+1" stops the sequence from being purely geometric. Remove it by shifting: seek a constant c with aₙ₊₁ + c = 2(aₙ + c). Expanding: aₙ₊₁ = 2aₙ + c, so c = 1 works. Define bₙ = aₙ + 1. Then bₙ₊₁ = 2bₙ with b₁ = 2, a geometric sequence: bₙ = 2 × 2ⁿ⁻¹ = 2ⁿ. Hence aₙ = 2ⁿ − 1. Verify against the recursion: a₁ = 1 ✓, a₂ = 2 × 1 + 1 = 3 = 2² − 1 ✓, a₃ = 7 ✓. So a₁₀ = 2¹⁰ − 1 = 1023. The shift-to-geometric move handles every recursion aₙ₊₁ = paₙ + r (p ≠ 1): the shift is c = r/(p − 1), the fixed point of the map. This is the Gaokao's favourite sequence construction — recognise it instantly.

Common mistakes

• Exponent off by one. aₙ = a₁qⁿ⁻¹, not a₁qⁿ. Test with n = 1 whenever unsure: the formula must return a₁.
• Sum formula at q = 1. a₁(1 − qⁿ)/(1 − q) divides by zero; the case q = 1 gives Sₙ = na₁ separately. Parameter problems exploit this exact gap.
• Declaring a type from two terms. 2, 4, … could be arithmetic (d = 2) or geometric (q = 2); the third term decides. Never classify on two data points.
• aₙ = Sₙ − Sₙ₋₁ misapplied at n = 1. The formula holds for n ≥ 2; a₁ must be computed as S₁ separately. Sequence problems with Sₙ given trap precisely here.
• Means divided wrongly. Inserting k means creates k + 1 gaps (Worked example 5).

Exam Experience

Multiple choice. In an arithmetic sequence, a₃ = 7 and a₇ = 19. Then a₁₀ equals: A. 25 B. 28 C. 31 D. 34 Answer: B. 4d = a₇ − a₃ = 12 gives d = 3; a₁ = 7 − 2 × 3 = 1; a₁₀ = 1 + 9 × 3 = 28.

Fill in the blank. In a geometric sequence, a₂ = 6 and a₅ = 48. Then a₇ = ____. Answer: 192. a₅/a₂ = q³ = 8, so q = 2 and a₁ = 3; a₇ = 3 × 2⁶ = 192.

Solution question (10 marks). A ball is dropped from a height of 100 m. Each bounce reaches half the previous height. (1) Find the total distance travelled by the ball at the moment it strikes the ground for the fourth time. (2) Find the total distance if it bounces forever. Marking scheme. (1) The falls form a geometric sequence 100, 50, 25, 12.5 (four falls, the fourth ending at the fourth impact) — 2 marks. The rises before that moment are 50, 25, 12.5 (three rises) — 2 marks. Total: (100 + 50 + 25 + 12.5) + (50 + 25 + 12.5) = 187.5 + 87.5 = 275 m — 1 mark. [5 marks] (2) Infinite falls: 100/(1 − 1/2) = 200 m — 2 marks. Infinite rises: 50/(1 − 1/2) = 100 m — 2 marks. Total 300 m — 1 mark. [5 marks] Total: 10 marks. The finite/infinite contrast is the lesson: an endless process with a finite total — Zeno's paradox (Lesson 18) resolved by a geometric series.

Going deeper — the sum of squares, by telescoping

Yang Hui summed stacked spheres; the key formula is Σk² = n(n + 1)(2n + 1)/6, and its cleanest proof is a telescope. Start from (k + 1)³ − k³ = 3k² + 3k + 1 and sum from k = 1 to n. The left side telescopes: every −k³ cancels the previous (k)³ term... precisely, Σ[(k + 1)³ − k³] = (n + 1)³ − 1³ = (n + 1)³ − 1. The right side gives 3Σk² + 3Σk + n. We know Σk = n(n + 1)/2 (Lesson 13's pairing). Solve: 3Σk² = (n + 1)³ − 1 − n − 3n(n + 1)/2. Factor (n + 1): (n + 1)³ − (n + 1) = (n + 1)[(n + 1)² − 1] = (n + 1)(n)(n + 2), so 3Σk² = n(n + 1)(n + 2) − 3n(n + 1)/2 = n(n + 1)[(n + 2) − 3/2] = n(n + 1)(2n + 1)/2, and dividing by 3: Σk² = n(n + 1)(2n + 1)/6. Numerical audit at n = 10: 10 × 11 × 21/6 = 385, and indeed 1 + 4 + 9 + … + 100 = 385 ✓. The telescoping cube is a machine: replace cubes by fourth powers and it yields Σk³ = [n(n + 1)/2]² (check n = 3: 1 + 8 + 27 = 36 = 6² ✓). Whenever a sum resists, ask what difference produces its terms — the answer is usually a power one higher.

Second look — Yang Hui's stacked spheres, and a theatre

Stacked spheres （堆垛）. Cannonballs stacked in a triangular pyramid: layer k (counting from the top) is a triangular array with 1 + 2 + … + k = k(k + 1)/2 balls. How many balls in a stack of 10 layers? The total is Σ k(k + 1)/2 = (1/2)(Σk² + Σk) — and this lesson's extension has already built both sums: Σk² for k = 1..10 is 385, Σk is 55. Total = (385 + 55)/2 = 220. (These are the tetrahedral numbers: 1, 4, 10, 20, 35, 56, 84, 120, 165, 220.) Yang Hui computed with exactly these stacks; the modern packaging C(n + 2, 3) = 220 for n = 10 connects them to Lesson 14's combinations — three lessons, one number.

A theatre. The first row has 20 seats and each subsequent row has 2 more; there are 15 rows. Total seats: an arithmetic series with a₁ = 20, d = 2, n = 15: S₁₅ = (15/2)(2 × 20 + 14 × 2) = (15/2)(68) = 510. The last row holds 20 + 14 × 2 = 48 seats; cross-check by the pairing form: (15/2)(20 + 48) = (15/2)(68) = 510 ✓ — two presentations of one formula, each auditing the other.

One last drill — two terms given, everything recovered. In an arithmetic sequence, a₅ = 13 and a₉ = 25. Find S₁₀. The gap: a₉ − a₅ = 4d = 12, so d = 3. Then a₁ = a₅ − 4d = 13 − 12 = 1, and a₁₀ = 1 + 9 × 3 = 28. Sum: S₁₀ = (10/2)(a₁ + a₁₀) = 5 × 29 = 145. Two data points determine an arithmetic sequence completely — the difference extracts the slope, one substitution extracts the intercept. This is the same two-point determinacy as a line through two points (Lesson 8), because an arithmetic sequence is a line sampled at integer points.