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第 14 週 · 等差與等比數列 · 第 6 節

數列應用題：複利、堆疊、幾何 + 本週關卡

存錢（複利）、堆磚、折紙都是數列！先認出「這是等差還是等比」，再套通項或求和。本節收官並挑戰 W14 全週 BOSS 總關卡。

sequences are everywhere in real life🏦savingscompoundgeometric ×🧱brick wallstackingarithmetic +📄folding paperhalvinggeometric ×½
▶ 第一關： 認類型 —— 複利（等比）vs 堆疊（等差）
next term comes by…+ adding× multiplyingARITHMETICadd fixed d each stepe.g. 20,18,16,…brick stacking 🧱GEOMETRICmultiply fixed r each stepe.g. 1000,1100,1210,…compound interest 🏦
P=1000 at 10% — balance grows geometrically (×1.1)1000year 01100year 11210year 21331year 3×1.1×1.1×1.1
brick wall widths: each layer has 2 fewer (d = −2)3rd layer = 162nd layer = 181st layer = 20 (a)20, 18, 16, 14, … → a=20, d=−2bar width represents number of bricks
▶ 第二關： 複利公式 A = P(1+r)ⁿ
A = P ( 1 + r )ⁿprincipalrate r (e.g. 0.1)number of yearsgeometric ratio q = 1+r, so A = Pqⁿ
P=500, r=20% (×1.2), n=2 yearsA = 500 × 1.2²=500 × 1.44=720① identify type → ② set P & r → ③ plug into A = P(1+r)ⁿthe same 3 steps solve every compound-interest problem
solving flow for any sequence word-problem① identify typeadd? mult?→② set a and d/rfrom the problem→③ Tₙ or Sₙ?plug the formulawhich term → Tₙ; the total → Sₙ; endless shrink → S∞
▶ 第三關： 堆疊求和 與 幾何無限等比
10-layer wall, bottom 20, d = −2 → total bricks S₁₀top T₁₀ = 2middle layers shrink by 2bottom = 20 (a)S₁₀ = 10×(20+2)/2 = 110sum first and last, then multiply by layers
square area 1, shade half each step½¼⅛…
1 + ½ + ¼ + ⅛ + … (a=1, r=½, |r|<1)1½never passes 2S∞ = 1/(1−½) = 2
⚔ BOSS： W14 全週總關卡（五大主題）
W14 knowledge map — arithmetic vs geometricARITHMETIC (+d)term: Tₙ = a + (n−1)dsum: Sₙ = n/2[2a+(n−1)d]= n(first+last)/2graph = straight line🧱 stacking, stepsGEOMETRIC (×r)term: Tₙ = a·rⁿ⁻¹sum: Sₙ = a(rⁿ−1)/(r−1)infinite: S∞ = a/(1−r)only if |r| < 1🏦 compound, 📄 foldingfirst ask: add or multiply?
BOSS checklist — one question per topic① arith term a=3, d=5 → T₇ = 3+6×5 = 33② arith sum 2,5,8… → S₂₀ = 10×61 = 610③ geo term a=2, r=3 → T₅ = 2×3⁴ = 162④ geo sum 3,6,12… → S₅ = 3×31 = 93⑤ infinite geo a=8, r=½ → S∞ = 8/½ = 16