📚 Geometric Series | 几何级数
A geometric series is one of the foundational concepts in IB Mathematics, linking sequences, exponential growth, and real‑world applications. Understanding how to sum its terms – whether finite or infinite – is essential for both analysis and problem‑solving.
几何级数是IB数学中的一个基础概念，它将数列、指数增长与现实应用紧密相连。掌握如何对有限项或无限项进行求和，对于数学分析和解题都至关重要。
1. Understanding Sequences vs Series | 理解数列与级数
A sequence is an ordered list of numbers; a series is the sum of the terms of a sequence. In the context of geometric patterns, we speak of a geometric sequence when each term is obtained by multiplying the previous term by a constant, and a geometric series when we add those terms together.
数列是一组有序排列的数字；级数是数列各项的和。在等比模式下，每当每一项都由前一项乘以一个常数得到时，我们称之为等比数列，而当我们将这些项相加时，则称之为几何级数。
The distinction is crucial because the convergence behaviour of the sequence (does it approach zero?) directly determines whether an infinite series can have a finite sum.
这种区分至关重要，因为数列的收敛行为（它是否趋近于零？）直接决定了无穷级数是否能够有一个有限的和。
2. Definition of a Geometric Sequence | 几何数列的定义
A geometric sequence is one in which the ratio between any term and its preceding term is constant. If we denote the first term by a and the common ratio by r, the sequence looks like: a, ar, ar², ar³, …
几何数列是指任意一项与前一项的比值恒定的数列。如果我们用 a 表示首项，r 表示公比，则该数列形如：a, ar, ar², ar³, …
For example, the sequence 3, 6, 12, 24, … has a = 3 and r = 2. The sequence 81, 27, 9, 3, … has a = 81 and r = ⅓.
例如，数列 3, 6, 12, 24, … 有 a = 3, r = 2。数列 81, 27, 9, 3, … 有 a = 81, r = ⅓。
3. The Common Ratio and Its Role | 公比及其作用
The common ratio r can be any real number, and its value determines the behaviour of both the sequence and the series. If |r| > 1, the terms grow in magnitude; if 0 < |r| < 1, the terms decay towards zero; if r is negative, the terms alternate in sign.
公比 r 可以是任何实数，它的值决定了数列和级数的行为。若 |r| > 1，项的绝对值会不断增大；若 0 < |r| < 1，项会逐渐衰减趋于零；若 r 为负，项的正负号交替变化。
When working with infinite geometric series, the condition |r| < 1 is the deciding factor for convergence; otherwise, the sum diverges and we cannot assign a finite value to it.
在处理无穷几何级数时，条件 |r| < 1 是判定收敛的决定性因素；否则级数发散，我们无法为其赋予一个有限的和。
4. The nth Term Formula | 第 n 项公式
For a geometric sequence, the nth term is given by un = a · rn−1. This formula allows us to find any term without listing all the preceding ones.
对于几何数列，第 n 项由公式 un = a · rn−1 给出。这个公式使我们无需列出前面的所有项就能直接求出任何项。
83 = 98415. This exponential nature makes geometric sequences grow or shrink very rapidly.
例如，若 a = 5 且 r = 3，则第 10 项为 5 × 3⁹ = 5 × 19683 = 98415。这种指数特性使得几何数列增长或衰减得非常迅速。
The index shift is a common source of error: the first term uses exponent 0, the second term exponent 1, so the nth term indeed has exponent n−1.
指标的偏移是常见的错误来源：第一项使用指数 0，第二项指数为 1，因此第 n 项确实具有指数 n−1。
5. Geometric Series Defined | 几何级数的定义
A geometric series is the sum of the terms of a geometric sequence. It can be expressed in summation notation as Σk=0n−1 a rk or Σk=1n a rk−1.
几何级数是几何数列各项的和。它可以用求和符号表示为 Σk=0n−1 a rk 或 Σk=1n a rk−1。
We distinguish between a partial sum Sn (the sum of the first n terms) and the infinite sum S∞ (the limit of Sn as n → ∞, provided it exists).
我们区分部分和 Sn（前 n 项的和）与无穷和 S∞（当 n → ∞ 时 Sn 的极限，如果存在的话）。
6. Sum of a Finite Geometric Series | 有限几何级数的求和
The sum of the first n terms of a geometric series is given by the formula:
几何级数前 n 项的和由以下公式给出：
Sn = a (1 − rn) / (1 − r) for r ≠ 1
If r = 1, all terms are equal to a, so Sn = n a. This formula is derived by subtracting r Sn from Sn and solving for Sn.
如果 r = 1，所有项都等于 a，因此 Sn = n a。这个公式是通过从 Sn 中减去 r Sn 并求解 Sn 推导出来的。
An equivalent form is Sn = a (rn − 1) / (r − 1), which is often more convenient when r > 1.
等价的公式为 Sn = a (rn − 1) / (r − 1)，当 r > 1 时使用这个形式往往更方便。
7. Derivation of the Sum Formula | 求和公式的推导
Write the sum Sn = a + a r + a r² + … + a rn−1. Multiply both sides by r: r Sn = a r + a r² + a r³ + … + a rn.
写出和为 Sn = a + a r + a r² + … + a rn−1。将两边乘以 r：r Sn = a r + a r² + a r³ + … + a rn。
Subtract: Sn − r Sn = a − a rn. Factor the left side: Sn(1 − r) = a (1 − rn). Dividing by (1 − r) yields the formula, provided r ≠ 1.
相减：Sn − r Sn = a − a rn。左边因式分解：Sn(1 − r) = a (1 − rn)。除以 (1 − r) 即得到公式，前提是 r ≠ 1。
This elegant cancellation is the reason the geometric series sum has a closed form, unlike many other series.
由于这种巧妙的相消，几何级数的和才能拥有闭形表达式，这与许多其他级数不同。
8. Infinite Geometric Series | 无穷几何级数
An infinite geometric series sums all terms of the form a rk−1 for k = 1, 2, 3, … . The series converges if the terms become arbitrarily small, which happens exactly when |r| < 1.
无穷几何级数对 k = 1, 2, 3, … 的所有形如 a rk−1 的项求和。如果项变得任意小，级数就收敛，这恰好发生在 |r| < 1 时。
Under this condition, rn → 0 as n → ∞, so the partial sum formula simplifies to S∞ = a / (1 − r).
在此条件下，当 n → ∞ 时 rn → 0，因此部分和公式简化为 S∞ = a / (1 − r)。
If |r| 1 and r ≠ 1, the series oscillates or grows without bound and does not have a finite sum.
若 |r| ≥ 1 且 r ≠ 1，级数要么振荡，要么无限增长，不存在有限和。
9. Convergence Condition in Detail | 收敛条件详解
The condition |r| sum approaches a limit; with r = 2, the terms double, and the series diverges to infinity.
条件 |r| < 1 是无穷几何级数收敛的充要条件。例如，当 r = 0.5 时，各项每次减半，和趋近一个极限；当 r = 2 时，各项加倍，级数发散至无穷。
When r = −0.5, the terms alternate: a, −0.5a, 0.25a, −0.125a, … . The absolute values shrink, and the series converges to a/(1 − (−0.5)) = a/1.5.
当 r = −0.5 时，各项正负交替：a, −0.5a, 0.25a, −0.125a, …。绝对值不断缩小，级数收敛到 a/(1 − (−0.5)) = a/1.5。
IB exam questions often ask you to state the range of r for convergence or to find the sum to infinity only after verifying the condition.
IB 考试题目经常要求你陈述使级数收敛的 r 的范围，或者只有在验证条件之后才求无穷和。
10. Sum to Infinity – Worked Examples | 无穷项求和 – 示例解析
Example 1: Find the sum to infinity of the series 12 + 6 + 3 + 1.5 + … . Here a = 12, r = ½. Since |r| < 1, S∞ = 12/(1 − ½) = 24.
例 1：求级数 12 + 6 + 3 + 1.5 + … 的无穷和。这里 a = 12，r = ½。因为 |r| < 1，所以 S∞ = 12/(1 − ½) = 24。
geometric series has first term 5 and sum to infinity 20. Find r. We set up 5/(1 − r) = 20 ⇒ 1 − r = 0.25 ⇒ r = 0.75.
例 2：一个几何级数的首项为 5，无穷和为 20。求 r。我们列出 5/(1 − r) = 20 ⇒ 1 − r = 0.25 ⇒ r = 0.75。
Example 3: An infinite geometric series has r = −2/3 and S∞ = 9. Find a. Since 1 − r = 1 − (−2/3) = 5/3, we obtain a = 9 × 5/3 = 15.
例 3：一个无穷几何级数的 r = −2/3，S∞ = 9。求 a。因为 1 − r = 1 − (−2/3) = 5/3，所以 a = 9 × 5/3 = 15。
11. Applications – Finance, Physics, and More | 应用 – 金融、物理等
Geometric series model compound interest: an amount P invested at an annual rate i becomes P(1 + i)n after n years. The accumulation of regular savings forms a geometric series whose sum can be found with the finite sum formula.
几何级数可以模拟复利：以年利率 i 投入的本金 P 在 n 年后变为 P(1 + i)n。定期储蓄的累积构成一个几何级数，其总和可用有限求和公式计算。
In physics, geometric series describe exponential decay – for example, the total distance travelled by a bouncing ball, where each bounce reaches a height that is a fixed fraction of the previous one.
在物理学中，几何级数描述指数衰减——例如，一个弹跳球每次弹起的高度是前一次的固定比例，其所经过的总路程就是一个几何级数。
nderpin power series expansions and are used to derive formulas for sums like 1 + x + x² + … = 1/(1 − x), valid for |x| < 1.
在微积分中，几何级数是幂级数展开的基础，用于推导诸如 1 + x + x² + … = 1/(1 − x)（在 |x| < 1 时成立）的公式。
12. Common Pitfalls and Exam Strategies | 常见错误与应试策略
Do not confuse the sequence term formula un = a rn−1 with the series sum formula Sn. Many marks are lost by misplacing brackets or misidentifying the first term.
不要混淆数列通项公式 un = a rn−1 与级数求和公式 Sn。许多失分都是由于括号位置错误或首项识别不当造成的。
Always check whether a series is arithmetic or geometric before applying a formula. In geometric problems, explicitly state the values of a and r as your first step.
在应用公式之前，务必先检查级数是等差级数还是等比级数。在几何问题中，第一步要明确写出 a 和 r 的值。
For sums to infinity, verify |r| < 1; otherwise state that the sum does not converge. When r is negative, use the form with (1 − r) carefully to avoid sign errors.
对于无穷项求和，要验证 |r| < 1；否则应说明其和不收敛。当 r 为负时，要小心使用含有 (1 − r) 的公式，以免出现符号错误。
Finally, if a geometric series is given in summation notation, interpret it correctly: Σ 2·3k from k=0 to 4 has a = 2, r = 3, and n = 5 terms.
最后，如果几何级数以求和符号给出，要正确解读：从 k=0 到 4 的 Σ 2·3k 有 a = 2, r = 3, 且一共有 n = 5 项。
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