📚 Geometric Series | 等比级数
Geometric series are a central topic in A-Level Mathematics, building on geometric sequences. A geometric series is the sum of the terms of a geometric sequence, where each term after the first is found by multiplying the previous term by a fixed non-zero constant called the common ratio, r. Understanding how to sum a geometric series—both finite and infinite—is essential for solving problems in pure mathematics and in applied contexts such as finance and recurring decimals.
等比级数是A-Level数学的核心内容，建立在等比数列的基础之上。等比级数即等比数列各项之和，其中从第二项起每一项都等于前一项乘以一个固定的非零常数，称为公比 r。掌握如何对有限和无限等比级数求和，对于解决纯数学问题以及金融、循环小数等实际应用至关重要。
1. What is a Geometric Series? | 什么是等比级数？
A geometric sequence has the form a, ar, ar², ar³, … mmon ratio. A geometric series is simply the sum of these terms. For example, 2 + 6 + 18 + 54 + … . If we add only the first n terms, we obtain a finite geometric series. If the series continues indefinitely, it is an infinite geometric series.
等比数列的形式为 a, ar, ar², ar³, … 其中 a 为首项，r 为公比。等比级数就是这些项的总和。例如，2 + 6 + 18 + 54 + … 是一个等比级数，首项 a = 2，公比 r = 3。如果仅将前 n 项相加，我们得到有限等比级数。如果级数无限延续下去，则为无穷等比级数。
2. General Term and Common Ratio | 通项与公比
The nᵗʰ /u₁ = u₃/u₂, and so on. The ratio must be constant for the sequence to be geometric. In series problems, you may need to find a or r given two terms, often by setting up simultaneous equations.
等比数列的第 n 项（通项）公式为 uₙ = arⁿ⁻¹，其中 uₙ 代表第 n 项。要求出公比 r，可将任意一项除以前一项：r = u₂/u₁ = u₃/u₂，以此类推。数列必须保持比值恒定才是等比数列。在级数问题中，你可能需要根据给定的两项求出 a 或 r，通常需要列出联立方程求解。
3. Sum of a Finite Geometric Series | 有限等比级数的和
The sum of the first n terms of a geometric series is denoted Sₙ. The standard formula is:
Sₙ = a(1 − rⁿ)/(1 − r) for r ≠ 1
rms are equal to a. You must be comfortable applying this formula directly and after rearranging to find a, r or n. Often you will need to use logarithms when solving for n.
有限等比级数的前 n 项和记作 Sₙ。标准求和公式为：Sₙ = a(1 − rⁿ)/(1 − r)，其中 r ≠ 1。当 r = 1 时，各项均为 a，和为 n × a。你必须能够熟练直接套用该公式，并在重新整理后灵活求 a、r 或 n。在求解 n 时通常需要用到对数。
4. Derivation of the Sum Formula | 求和公式的推导
first: Sₙ − rSₙ = a − arⁿ. Factor out Sₙ: Sₙ(1 − r) = a(1 − rⁿ). Finally, divide by (1 − r) to obtain Sₙ = a(1 − rⁿ)/(1 − r). This derivation is frequently examined, and you should be able to reproduce it clearly, showing each step and stating the condition r ≠ 1.
推导公式时，写出 Sₙ = a + ar + ar² + ... + arⁿ⁻¹。两边乘以 r 得到 rSₙ = ar + ar² + ... + arⁿ⁻¹ + arⁿ。将第一式减去第二式得：Sₙ − rSₙ = a − arⁿ。提取公因式 Sₙ 得 Sₙ(1 − r) = a(1 − rⁿ)。最后两边除以 (1 − r) 得到 Sₙ = a(1 − rⁿ)/(1 − r)。该推导在考试中经常出现，你应该能够清晰地写出每一步，并说明条件 r ≠ 1。
5. Alternative Form and Choosing the Right Formula | 求和公式的变形与选用
r − 1), which is simply obtained by multiplying numerator and denominator of the standard formula by −1. This version is particularly useful when r > 1, as it keeps the numerator positive and reduces sign errors. Both formulas are equivalent; choose whichever makes the calculation cleaner in the given context. Always check the value of r before deciding which form to use.
另一种形式为 Sₙ = a(rⁿ − 1)/(r − 1)，只需将标准公式的分子分母同乘以 −1 即可得到。当 r > 1 时，该形式尤为实用，可使分子保持正号，减少符号错误。两个公式等价，应选择使运算更清晰的形式。在决定使用哪种形式前，务必先判断 r 的值。
6. Infinite Geometric Series and Convergence | 无穷等比级数与收敛条件
rges to a finite sum only when the absolute value of the common ratio is less than 1, i.e. |r| < 1. If |r| ≥ 1, the series diverges—its terms do not approach zero, and the sum either grows without bound or oscillates. For |r| is convergence condition is crucial: applying the sum to infinity formula without checking |r| < 1 is a common and serious mistake.
无穷等比级数只有在公比的绝对值小于 1，即 |r| < 1 时才能收敛于一个有限和。如果 |r| ≥ 1，级数发散——其项不趋近于零，总和要么无限增大，要么振荡。当 |r| < 1 时，随着 n → ∞，rⁿ → 0，使得无穷求和成为可能。这个收敛条件至关重要：在未检查 |r| < 1 的情况下直接套用无穷和公式是一个常见且严重的错误。
7. Sum to Infinity | 无穷和公式
The sum to infinity, denoted S∞, is given by:
S∞ = a/(1 − r) for |r| < 1
antly simple and is widely used in pure contexts such as converting recurring decimals and in applied contexts like long-term financial modelling. Always check the condition before using it.
无穷和 S∞ 的公式为：S∞ = a/(1 − r)，其中 |r| < 1。它由 Sₙ 的极限推出：Sₙ = a(1 − rⁿ)/(1 − r) → a(1 − 0)/(1 − r) = a/(1 − r)。该公式形式简洁，在循环小数化分数等纯数学场景和长期金融建模等应用场景中广泛使用。使用前务必先验证条件。
8. Applications: Converting Recurring Decimals | 应用：循环小数化分数
A recurring decimal can be expressed as an infinite geometric series. For instance, 0.2̇7̇ = 0.27 + 0.0027 + 0.000027 + … Here the first term a = 0.27 and the common ratio r = 0.01. Since |0.01| < 1, S∞ = 0.27/(1 − 0.01) = 0.27/0.99 = 27/99 = 3/11. Similarly, 0.1̇2̇3̇ s a classic exam question; always simplify the fraction fully and verify by long division.
循环小数可以表示为无穷等比级数。例如，0.2̇7̇ = 0.27 + 0.0027 + 0.000027 + … 其中首项 a = 0.27，公比 r = 0.01。因为 |0.01| < 1，S∞ = 0.27/(1 − 0.01) = 0.27/0.99 = 27/99 = 3/11。类似地，0.1̇2̇3̇ 可写作 123/999。此方法为经典考题；务必将分数化至最简，并用长除法验证。
9. Applications: Financial Mathematics | 应用：金融数学
Geometric series naturally model compound interest, depreciation, and regular savings plans. For example, if you invest £P at the beginning of each year into an account paying compound interest at a rate of i per annum, the total value after n years is: P(1+i) + P(1+i)² + ... + P(1+i)ⁿ. This is a geometric
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