📚 Geometric Sequences and Series: A High-Efficiency Review Guide | 等比数列与级数高效复习指南
Geometric sequences and series are a core topic in IB Mathematics (both Analysis and Approaches and Applications and Interpretation). This guide consolidates the essential definitions, formulas, convergence conditions, and exam-style problem-solving strategies into one structured review. We emphasise clarity, common pitfalls, and the exact notation you need for exams.
等比数列与级数是 IB 数学（分析与方法、应用与解释）中的核心内容。本指南将核心定义、公式、收敛条件以及考试风格的解题策略整合为一份结构化复习资料。我们特别强调清晰性、常见陷阱以及考试中需要使用的规范记号。
1. Definition and Key Notation | 定义与关键记号
A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed non-zero constant called the common ratio, usually denoted by r. The first term is usually denoted by a or u₁.
等比数列是指从第二项起，每一项与前一项的比值都等于同一个非零常数，这个常数称为公比，通常用 r 表示。首项通常用 a 或 u₁ 表示。
uₙ = a · rⁿ⁻¹
ample, the sequence 3, 6, 12, 24, ... has a = 3 and r = 2. The n-th term is uₙ = 3 · 2ⁿ⁻¹.
例如，数列 3, 6, 12, 24, ... 中，a = 3，r = 2。其通项公式为 uₙ = 3 · 2ⁿ⁻¹。
- r can be positive, negative, or fractional. A negative ratio makes the terms alternate in sign.
- r 可以是正数、负数或分数。负公比会使各项符号交替变化。
- The common ratio is found by dividing any term by the previous term: r = uₙ / uₙ₋₁.
- 公比可通过任一项除以前一项得到：r = uₙ / uₙ₋₁。
- If r = 1, the sequence is constant; if r = -1, it alternates between two values.
- 若 r = 1，数列为常数列；若 r = -1，数列在两个值之间交替。
2. The General Term and Finding Unknowns | 通项公式与求未知量
To solve problems involving geometric sequences, you often need to find a, r, or n using the general term formula. Given two non-adjacent terms, you can set up a system of equations.
在解等比数列问题时，通常需要利用通项公式求 a、r 或 n。若已知两个不相邻的项，可以建立方程组求解。
uₙ = a · rⁿ⁻¹ ⇔ a = uₙ / rⁿ⁻¹ ⇔ r = (uₙ / a)^(1/(n-1))
: If u₃ = 20 and u₆ = 160, then u₆ / u₃ = r³ = 8, so r = 2. Then a = u₃ / r² = 20 / 4 = 5.
例如：若 u₃ = 20，u₆ = 160，则 u₆ / u₃ = r³ = 8，所以 r = 2。于是 a = u₃ / r² = 20 / 4 = 5。
- Always check whether r is positive or negative when taking even roots. If the sequence oscillates, there may be two possible ratios.
- 当开偶次方根时，务必判断 r 的正负。若数列振荡，可能存在两个公比。
- Use logarithms to solve for n when it appears in the exponent: n = 1 + log_r(uₙ / a).
- 当 n 出现在指数位置时，可用对数求解：n = 1 + log_r(uₙ / a)。
3. Geometric Series and the Sum Formula | 等比级数与求和公式
A geometric series is the sum of the terms of a geometric sequence. The sum of the first n terms is denoted Sₙ.
等比级数是等比数列各项之和。前 n 项和记为 Sₙ。
Sₙ = a(1 - rⁿ) / (1 - r) = a(rⁿ - 1) / (r - 1), r ≠ 1
The two forms are equivalent. Choose the one that keeps the denominator positive to reduce sign errors.
两种形式等价。选择使分母为正的形式，可以减少符号错误。
he series 2 + 6 + 18 + 54, a = 2, r = 3, n = 4. S₄ = 2(3⁴ - 1)/(3 - 1) = 2(81 - 1)/2 = 80.
例如：对于级数 2 + 6 + 18 + 54，a = 2，r = 3，n = 4。S₄ = 2(3⁴ - 1)/(3 - 1) = 2(81 - 1)/2 = 80。
- If r = 1, the sum formula cannot be used directly. In that case Sₙ = n · a.
- 若 r = 1，不能直接使用求和公式。此时 Sₙ = n · a。
- The formula for Sₙ requires knowing a, r, and n. If n is unknown but Sₙ is given, use trial or logarithms carefully.
- Sₙ 的公式需要知道 a、r 和 n。若 n 未知但 Sₙ 已知，可小心使用试算或对数方法。
4. Infinite Geometric Series and Convergence | 无穷等比级数与收敛性
An infinite geometric series converges if and only if |r| < 1. In that case, the sum tends to a finite limit as n → ∞.
无穷等比级数收敛的充要条件是 |r| < 1。此时，当 n → ∞ 时，和趋向于有限极限。
S∞ = a / (1 - r), for |r| < 1
If |r| ≥ 1, the infinite series diverges (unless a = 0, which is usually a trivial case).
若 |r| ≥ 1，无穷级数发散（除非 a = 0，这通常是平凡情形）。
Example: For 100 + 50 + 25 + 12.5 + ..., a = 100, r = 0.5, so S∞ = 100/(1 - 0.5) = 200.
例如：对于 100 + 50 + 25 + 12.5 + ...，a = 100，r = 0.5，所以 S∞ = 100/(1 - 0.5) = 200。
- Convergence depends only on r, not on a. Even a huge first term converges if |r| < 1.
- 收敛性仅取决于 r，与 a 无关。即使首项很大，只要 |r| < 1 就会收敛。
- When |r| is close to 1, convergence is slow; the partial sums approach S∞ gradually.
- 当 |r| 接近 1 时，收敛很慢；部分和逐渐逼近 S∞。
5. Sigma Notation and Finite Series | 求和记号与有限级数
a·rⁿ⁻¹. Understanding the index is crucial: the lower limit tells you the first exponent.
在 IB 考试中，等比级数常用求和记号表示：Σ 从 n=1 到 N 的 a·rⁿ⁻¹。理解下标至关重要：下限决定了首项的指数。
Σₖ₌₁ⁿ a·rᵏ⁻¹ = a + ar + ar² + ... + arⁿ⁻¹ = a(1 - rⁿ)/(1 - r)
y. For example, Σ from k=0 to n of a·rᵏ has n+1 terms, and the sum is a(1 - rⁿ⁺¹)/(1 - r).
如果下限不是 1，需要仔细调整表达式。例如，Σ 从 k=0 到 n 的 a·rᵏ 共有 n+1 项，其和为 a(1 - rⁿ⁺¹)/(1 - r)。
- Count the number of terms: upper limit - lower limit + 1.
- 项数 = 上限 - 下限 + 1。
- The exponent in the last term must match the upper limit.
- 最后一项的指数必须与上限一致。
6. Applications: Compound Interest and Depreciation | 应用：复利与折旧
an annual rate r, compounded n times per year. After t years, the amount is A = P(1 + r/n)nt. In the geometric framework, the common ratio is (1 + interest rate per period).
等比数列可以模拟指数增长与衰减。复利是最常见的现实应用：初始金额 P 以年利率 r 增长，每年复利 n 次。t 年后，金额为 A = P(1 + r/n)nt。在等比数列框架中，公比为 (1 + 每期利率)。
with a negative growth rate: value after t years = initial value × (1 - depreciation rate)t.
折旧原理相同，只是增长率为负：t 年后的价值 = 初始价值 × (1 - 折旧率)t。
| Scenario | Common ratio r | Example |
| Compound interest (growth) | 1 + i, where i is the period interest rate | 5% per year → r = 1.05 |
| Depreciation (decay) | 1 - d, where d is the depreciation rate | 12% per year → r = 0.88 |
| Population growth | 1 + g, where g is the growth rate | 2% per year → r = 1.02 |
Always convert the percentage to a decimal before using it in the formula.
使用公式前，务必先将百分数转换为小数。
7. Geometric Means and Interpolation | 等比中项与插值
Given two positive numbers a and b, the geometric mean is √ab. Inserting one geometric mean between a and b creates a three-term geometric sequence: a, √ab, b.
对于两个正数 a 和 b，等比中项为 √ab。在 a 和 b 之间插入一个等比中项，可构成三项等比数列：a, √ab, b。
ic means between a and b, find r = (b/a)^(1/(k+1)), then the terms are a, ar, ar², ..., b.
更一般地，在 a 和 b 之间插入 k 个等比中项时，先求 r = (b/a)^(1/(k+1))，则各项为 a, ar, ar², ..., b。
- If a and b have opposite signs, no real geometric mean exists.
- 若 a 和 b 异号，则不存在实数等比中项。
- When inserting an even number of means, consider both positive and negative r when applicable.
- 当插入偶数个中项时，若适用，应考虑正负两个 r。
8. Recurrence Relations and Explicit Formulas | 递推关系与显式公式
≥ 2. This form is useful for proving properties or constructing sequences computationally.
等比数列可以递归定义：u₁ = a，且当 n ≥ 2 时，uₙ = r · uₙ₋₁。这种形式适用于证明性质或用程序构造数列。
ows by repeated substitution. In IB papers, you may be asked to derive one from the other.
由递推关系，通过反复代入可得显式公式 uₙ = a · rⁿ⁻¹。在 IB 试卷中，可能要求你由递推式推导显式公式，或反过来。
uₙ = r · uₙ₋₁ ⇔ uₙ = u₁ · rⁿ⁻¹
- Recurrence relations emphasize the multiplicative structure.
- 递推关系强调乘法结构。
- Explicit formulas are more efficient for finding any term directly.
- 显式公式更适合直接求任意一项。
9. Common Exam Traps and How to Avoid Them | 常见考试陷阱与规避方法
Many students lose marks on geometric sequences not because of difficulty, but because of small errors in notation or condition-checking. Here are the most common traps.
许多学生在等比数列上失分，不是因为题目难，而是因为记号或条件检验的小错误。以下是最常见的陷阱。
| Trap | 陷阱 | Solution | 对策 |
| Using r = 1 in the sum formula | 在求和公式中使用 r = 1 | Always check r ≠ 1; if r = 1, Sₙ = n·a | 始终检查 r ≠ 1；若 r = 1，Sₙ = n·a |
| Applying the infinite sum formula when |r| ≥ 1 | 当 |r| ≥ 1 时套用无穷求和公式 | Verify |r| < 1 before writing S∞ | 写出 S∞ 前先验证 |r| < 1 |
| Counting the wrong number of terms in sigma notation | 在求和记号中数错项数 | Terms = upper - lower + 1 | 项数 = 上限 - 下限 + 1 |
| Forgetting to convert percentage to decimal | 忘记将百分数转换为小数 | Write 5% as 0.05 before substituting | 代入前将 5% 写作 0.05 |
10. Worked Example: Full Exam-Style Problem | 典型例题完整解析
Problem: The third term of a geometric sequence is 12, and the sixth term is 96. (a) Find the first term and the common ratio. (b) Find the sum of the first 10 terms. (c) Determine whether the infinite series converges, and if so, find its sum.
题目：一个等比数列的第三项为 12，第六项为 96。(a) 求首项和公比。(b) 求前 10 项之和。(c) 判断无穷级数是否收敛；若收敛，求其和。
Solution (a): u₆ / u₃ = r³ = 96/12 = 8, so r = 2. Then u₃ = a · 2² = 12, so a = 3.
解 (a)： u₆ / u₃ = r³ = 96/12 = 8，所以 r = 2。由 u₃ = a · 2² = 12，得 a = 3。
Solution (b): S₁₀ = 3(2¹⁰ - 1)/(2 - 1) = 3(1024 - 1) = 3069.
解 (b)： S₁₀ = 3(2¹⁰ - 1)/(2 - 1) = 3(1024 - 1) = 3069。
Solution (c): Since r = 2, |r| ≥ 1, the infinite series diverges. No finite sum exists.
解 (c)： 因为 r = 2，|r| ≥ 1，无穷级数发散。不存在有限和。
11. Practice Problems with Quick Hints | 练习题与快速提示
Below are three short problems to test your understanding. Try them before checking the hints.
以下三道小题用于测试理解。先尝试作答，再查看提示。
- Problem 1: Find the 8th term of 5, 15, 45, ... | 提示：r = 3; u₈ = 5 · 3⁷ = 10935.
- 题目 1：求 5, 15, 45, ... 的第 8 项 | Hint: r = 3; u₈ = 5 · 3⁷ = 10935。
- Problem 2: Evaluate Σₖ₌₁⁶ 2·3ᵏ⁻¹ | 提示：n = 6, a = 2, r = 3; S₆ = 2(3⁶ - 1)/(3 - 1) = 728.
- 题目 2：计算 Σₖ₌₁⁶ 2·3ᵏ⁻¹ | Hint: n = 6, a = 2, r = 3; S₆ = 2(3⁶ - 1)/(3 - 1) = 728。
- Problem 3: Find S∞ for 27 + 9 + 3 + 1 + ... | 提示：a = 27, r = 1/3; S∞ = 27/(1 - 1/3) = 40.5.
- 题目 3：求 27 + 9 + 3 + 1 + ... 的 S∞ | Hint: a = 27, r = 1/3; S∞ = 27/(1 - 1/3) = 40.5。
12. Final Revision Checklist | 最终复习清单
Use this checklist before your exam to ensure you have mastered the essentials.
考试前使用此清单，确保你已掌握关键要点。
| Skill | 技能 | Status | 掌握情况 |
| Find the common ratio from consecutive terms | 由相邻项求公比 | ☐ |
| Write the general term uₙ = a·rⁿ⁻¹ | 写出通项公式 uₙ = a·rⁿ⁻¹ | ☐ |
| Use the finite sum formula with correct r conditions | 正确使用有限求和公式及 r 条件 | ☐ |
| Apply the infinite sum formula only when |r| < 1 | 仅在 |r| < 1 时使用无穷求和公式 | ☐ |
| Convert between sigma notation and expanded series | 在求和记号与展开式之间转换 | ☐ |
| Solve compound interest and depreciation problems | 解决复利与折旧问题 | ☐ |
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