📚 Geometric Series | 几何级数
A geometric series is the sum of the terms of a geometric sequence, where each term is obtained by multiplying the previous term by a constant ratio. This concept is fundamental in A-Level mathematics, appearing in finance, physics, and infinite series analysis.
几何级数是几何数列各项之和，几何数列中每一项都是前一项乘以一个固定公比得到的。这个概念是A-Level数学的基础，出现在金融、物理和无穷级数分析中。
1. Definition and Notation | 定义与记号
A geometric sequence has the form: a, ar, ar², ar³, … where a is the first term and r is the common ratio.
几何数列的形式为：a, ar, ar², ar³, …，其中 a 是首项，r 是公比。
The corresponding geometric series is written as:
Sₙ = a + ar + ar² + … + arⁿ⁻¹
对应的几何级数记为：
Sₙ = a + ar + ar² + … + arⁿ⁻¹
The number r can be any real number: positive, negative, fractional, or even irrational. The behaviour of the series depends critically on the value of r.
公比 r 可以是任何实数：正数、负数、分数甚至无理数。级数的行为关键取决于 r 的取值。
2. Common Ratio and the General Term | 公比与通项
The common ratio is found by dividing any term by the previous term:
r = u₂ / u₁ = u₃ / u₂ = … = uₙ / uₙ₋₁
公比由任意一项除以它前一项得到：
r = u₂ / u₁ = u₃ / u₂ = … = uₙ / uₙ₋₁
The general term (n-th term) of a geometric sequence is:
uₙ = arⁿ⁻¹
几何数列的通项（第 n 项）为：
uₙ = arⁿ⁻¹
For example, if a = 3 and r = 2, then the first five terms are 3, 6, 12, 24, 48, and the 10th term is 3 × 2⁹ = 1536.
例如，若 a = 3，r = 2，则前五项为 3, 6, 12, 24, 48，第10项为 3 × 2⁹ = 1536。
3. Sum of the First n Terms | 前n项和
The sum of the first n terms of a geometric series is given by:
Sₙ = a(1 − rⁿ) / (1 − r) for r ≠ 1
几何级数的前 n 项和为：
Sₙ = a(1 − rⁿ) / (1 − r) （r ≠ 1）
An equivalent form, often useful when |r| < 1, is:
Sₙ = a(rⁿ − 1) / (r − 1)
一个等价形式（当 |r| < 1 时常用）为：
Sₙ = a(rⁿ − 1) / (r − 1)
Both formulas give the same result; the choice merely depends on which denominator is more convenient. If r = 1, all terms are equal to a, and Sₙ = na.
两个公式结果相同；选择哪一个取决于哪个分母更方便。若 r = 1，所有项都等于 a，则 Sₙ = na。
4. Deriving the Sum Formula | 求和公式推导
To derive the formula, write the sum and then multiply by r:
要推导该公式，先写出和，然后乘以 r：
Sₙ = a + ar + ar² + … + arⁿ⁻¹
rSₙ = ar + ar² + ar³ + … + arⁿ
Subtracting the second from the first cancels most terms:
将第二个式子从第一个式子中减去，会抵消大部分项：
Sₙ − rSₙ = a − arⁿ
Factorising gives:
提取公因式得到：
Sₙ(1 − r) = a(1 − rⁿ)
Dividing by (1 − r) yields the standard formula. This derivation is a common exam question asking for proof.
除以 (1 − r) 即得到标准公式。这个推导过程是考试中常见的证明题。
5. Sum to Infinity | 无穷级数和
If |r| < 1, then rⁿ approaches 0 as n tends to infinity. The sum to infinity is therefore:
若 |r| < 1，则当 n 趋于无穷时 rⁿ 趋近于 0。因此无穷级数和为：
S∞ = a / (1 − r) for |r| < 1
例如，1 + ½ + ¼ + ⅛ + … = 1 / (1 − ½) = 2。
For example, 1 + ½ + ¼ + ⅛ + … = 1 / (1 − ½) = 2.
This result is frequently used to express recurring decimals as exact fractions, and to model physical processes such as bouncing balls or radioactive decay.
这一结果常用于将循环小数化为精确分数，也用于模拟弹跳球或放射性衰变等物理过程。
6. Condition for Convergence | 收敛条件
A geometric series converges (has a finite sum to infinity) if and only if |r| < 1. If |r| ≥ 1, the series diverges, meaning its partial sums do not approach a finite limit.
几何级数收敛（存在有限的无穷和）当且仅当 |r| < 1。若 |r| ≥ 1，则级数发散，即部分和不趋于有限极限。
For example, the series 2 + 6 + 18 + 54 + … with r = 3 diverges because each term grows without bound. Conversely, 100 − 50 + 25 − 12.5 + … has r = −0.5 and converges to 100 / (1 − (−0.5)) = 66.666… .
例如，级数 2 + 6 + 18 + 54 + … 中 r = 3，因为每项无限增长而发散。相反，100 − 50 + 25 − 12.5 + … 中 r = −0.5，收敛于 100 / (1 − (−0.5)) = 66.666…。
When r is negative, the terms alternate in sign. Convergence still requires |r| < 1, but the partial sums oscillate around the final value.
当 r 为负数时，项正负交替。收敛仍要求 |r| < 1，但部分和会在最终值附近振荡。
7. Recurring Decimals | 循环小数
A recurring decimal can be written as a geometric series. For example:
循环小数可以写成几何级数。例如：
0.777… = 0.7 + 0.07 + 0.007 + 0.0007 + …
Here a = 0.7 and r = 0.1, so the sum to infinity is:
这里 a = 0.7，r = 0.1，所以无穷和为：
S∞ = 0.7 / (1 − 0.1) = 0.7 / 0.9 = 7/9
Similarly, 0.232323… = 23/99, and 0.123123… chnique works for any purely recurring decimal, and also for mixed decimals such as 0.1666… by separating the non-repeating part.
类似地，0.232323… = 23/99，0.123123… = 123/999 = 41/333。该技巧适用于任何纯循环小数，也适用于 0.1666… 这样的混循环小数（只需分离非循环部分）。
8. Problem-Solving with Geometric Series | 几何级数解题
Exam problems often give two conditions and ask you to find a and r. For instance, the 2nd term is 6 and the 5th term is 48. Then:
考试题常给出两个条件，要求求 a 和 r。例如，第2项为6，第5项为48。则：
ar = 6 and ar⁴ = 48
Dividing the second equation by the first gives r³ = 8, so r = 2, and then a = 3. The series is 3, 6, 12, 24, 48, … .
第二个方程除以第一个方程得 r³ = 8，所以 r = 2，进而 a = 3。该级数为 3, 6, 12, 24, 48, …。
Double-check signs when r could be negative. If the terms alternate, r < 0; an odd number of negative outcomes may appear in equations.
当 r 可能为负数时，务必检查符号。若项正负交替，则 r < 0；方程中可能出现奇数个负号的情况。
9. Applications: Compound Interest and Depreciation | 应用：复利与折旧
Compound interest is a classic geometric-series application. If £1000 is invested at an annual rate of 5% (compounded annually), the amount after n years is a geometric sequence with a = 1000 and r = 1.05.
复利是几何级数的经典应用。若将1000英镑以年利率5%（每年复利）投资，则 n 年后的金额构成几何数列，其中 a = 1000，r = 1.05。
The total value after 10 years is 1000 × 1.05¹⁰ ≈ £ sum of deposits made each year can form a geometric series for future-value calculations.
10年后的总价值为 1000 × 1.05¹⁰ ≈ £1628.89。相反，每年存入的款项之和可构成几何级数，用于终值计算。
Depreciation works similarly: a car losing 15% of its value each year has a multiplier r = 0.85. After 5 years, its value is a × 0.85⁵ ≈ 0.4437 × a.
折旧同理：汽车每年贬值15%相当于乘数 r = 0.85。5年后其价值为 a × 0.85⁵ ≈ 0.4437 × a。
10. Sigma Notation | 求和记号
Geometric series can be written compactly using sigma notation:
几何级数可以用求和记号简洁表示：
Σₖ₌₀ⁿ⁻¹ arᵏ = a + ar + ar² + … + arⁿ⁻¹
Note that when the index starts at k = 0, the first term is a; many students incorrectly start at k = 1, which makes the first term ar. Always check the lower limit.
注意当下标从 k = 0 开始时，首项是 a；很多学生错误地从 k = 1 开始，导致首项变为 ar。务必检查下限。
If the series starts at k orrect form. The upper limit can be infinity for convergent series: Σₖ₌₀∞ arᵏ = a/(1 − r).
若级数从 k = 1 开始，则正确的形式是 Σₖ₌₁ⁿ arᵏ⁻¹。上限可为无穷大（对于收敛级数）：Σₖ₌₀∞ arᵏ = a/(1 − r)。
11. Common Mistakes and Tips | 常见错误与提示
- Using Sₙ = a(1 − rⁿ)/(1 − r) when r = 1: the formula is undefined, so simply add n copies of a.
- 当 r = 1 时使用 Sₙ = a(1 − rⁿ)/(1 − r)：公式无定义，此时应直接把 n 个 a 相加。
- Confusing the n-th term with the sum: uₙ = arⁿ⁻¹, but Sₙ = a(1 − rⁿ)/(1 − r).
- 混淆通项与和：uₙ = arⁿ⁻¹，但 Sₙ = a(1 − rⁿ)/(1 − r)。
- Forgetting to check |r| < 1 before using S∞. If |r| ≥ 1, the infinite sum does not exist.
- 在使用 S∞ 之前忘记检查 |r| < 1。若 |r| ≥ 1，无穷和不存在。
- Mixing up the sign of r in S∞. The formula is a/(1 − r), not a/(r − 1).
- 在 S∞ 中搞错 r 的符号。公式是 a/(1 − r)，不是 a/(r − 1)。
- When solving simultaneous equations such as ar³ = 8 and ar⁶ = 64, divide to eliminate a; but if r could be negative, consider both real roots.
- 解方程组如 ar³ = 8 和 ar⁶ = 64 时，用除法消去 a；但若 r 可能为负，需考虑两个实根。
Always write out the first few terms to check your sequence before applying formulas, especially when the index starts at a value other than 0 or 1.
在应用公式前，务必写出前几项来检查数列，尤其是当下标从0或1以外的值开始时。
12. Past Exam Style Questions | 真题风格练习
Example 1: The third term of a geometric series is 18 and the sixth term is 486. Find the first term and the common ratio.
示例1：某几何级数的第3项为18，第6项为486。求首项和公比。
Solution: Let the first term be a and ratio r. Then ar² = 18 and ar⁵ = 486. Dividing gives r³ = 27, so r = 3. Then a × 9 = 18, so a = 2.
解：设首项为 a，公比为 r。则 ar² = 18，ar⁵ = 486。相除得 r³ = 27，所以 r = 3。于是 a × 9 = 18，所以 a = 2。
Example 2: A geometric series has first term 12 and sum to infinity 30. Find the common ratio.
示例2：一个几何级数的首项为12，无穷和为30。求公比。
Solution: Use S∞ = a/(1 − r) = 30. So 12/(1 − r) = 30, giving 1 − r = 0.4, hence r = 0.6.
解：利用 S∞ = a/(1 − r) = 30。所以 12/(1 − r) = 30，得 1 − r = 0.4，故 r = 0.6。
Practice by converting 0.181818… to a fraction. Answer: 2/11.
练习：将 0.181818… 化为分数。答案：2/11。
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