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第 14 週 · 等差與等比數列 · 第 4 節

等比數列求和 Sₙ：錯位相減與公式

2+4+8+16+32 加起來是多少？等比求和有一個漂亮的推導「錯位相減」，把中間項全部消掉，得到 Sₙ = a(rⁿ−1)/(r−1)。從零講透，配滿圖例。

▶ 第一關： 求和公式與錯位相減推導
Sₙ = a ( rⁿ − 1 ) / ( r − 1 )first term aexponent is ndenominator r − 1 (do not drop)adds a + ar + ar² + … + arⁿ⁻¹ all at once
Sₙ = a ( 1 − rⁿ ) / ( 1 − r )same value — use this form when r < 1 (no minus signs)
S =aarar²…arⁿ⁻¹rS =arar²…arⁿ⁻¹arⁿmiddle terms line up and cancel in S − rSS − rS = a − arⁿ→S(1 − r) = a(1 − rⁿ)Sₙ = a(1 − rⁿ) / (1 − r)
pick the form with NO minus signr > 1Sₙ = a(rⁿ−1)/(r−1)e.g. r = 2, 3r < 1Sₙ = a(1−rⁿ)/(1−r)e.g. r = ½, ⅓
▶ 第二關： 套公式（r > 1）
worked example: 2 + 4 + 8 + 16 + 32a = 2, r = 2, n = 5S₅ = 2 ( 2⁵ − 1 ) / ( 2 − 1 ) = 2 ( 32 − 1 ) / 1 = 2 × 31S₅ = 62
2481632 (last)doubling each step (r = 2): the last bar dwarfs the rest
2 + 4 + 8 + 16 + 32 = 62≈2 × 32 = 64sum of all earlier terms ≈ the last term itself (off by 2)
▶ 第三關： r < 1、r 為負、r = 1 特例
r = ½ : 8 + 4 + 2 + 18421S₄ = 8(1−(½)⁴)/(1−½) = 15each block is half the previous one — just add: 8+4+2+1 = 15
r = −2 : 1 − 2 + 4 − 8 (signs alternate)+1−2+4−8S₄ = 1((−2)⁴−1)/((−2)−1) = 15/(−3) = −5(−2)⁴ = +16 (even power is positive) — keep the brackets!
special case r = 1 : 5 + 5 + 5 + 5formula breaks: a(rⁿ−1)/(r−1) → denominator 1−1 = 0 ✗every term equals a, so just multiply: Sₙ = n·a = 4 × 5 = 20Sₙ = na when r = 1
⚔ BOSS： 先求 r 或 n，再求和
given T₁ = 3, T₃ = 12 → find S₄① r² = T₃/T₁ = 12/3 = 4 → r = 2② a = 3 : S₄ = 3(2⁴−1)/(2−1) = 3 × 15S₄ = 45