📚 Number Sequences | 数列
In IB Mathematics, mastering number sequences is essential for building a strong foundation in algebra and discrete mathematics. Sequences appear in a wide range of contexts—from financial models and population growth to geometric patterns and computer algorithms. This article provides a thorough yet exam-focused revision of arithmetic and geometric sequences, covering key formulae, summation, sigma notation, and exam-style problem solving.
在 IB 数学中，掌握数列是打好代数与离散数学基础的关键。数列广泛出现在金融模型、人口增长、几何图案和计算机算法等众多背景中。本文以备考为导向，全面复习等差数列和等比数列，涵盖核心公式、求和、求和符号以及考试风格的问题求解。
1. What is a Number Sequence? | 什么是数列？
A number sequence is an ordered list of numbers following a particular pattern or rule. Each number in the sequence is called a term, typically denoted by u1, u2, u3, …, un, where n is the term number. The rule defining the relationship between consecutive terms determines the type of sequence.
数列是按照特定模式或规则排列的一列有序数字。数列中的每个数字称为项，通常记作 u1, u2, u3, …, un，其中 n 是项数。定义相邻项之间关系的规则决定了数列的类型。
In IB Mathematics: Analysis and Approaches (AA) and Applications and Interpretation (AI), you will primarily work with arithmetic and geometric sequences, but understanding the general concept of recursive and explicit definitions is equally important.
在 IB 数学：分析与方法 (AA) 以及应用与解释 (AI) 中，你主要处理等差数列和等比数列，但理解递推定义和显式定义的一般概念同样重要。
2. Arithmetic Sequences | 等差数列
An arithmetic sequence is a sequence in which the difference between any term and the previous term is constant. This constant is called the common difference and is usually denoted by d. For instance, the sequence 2, 5, 8, 11, 14, … has a common difference of 3 because each term is obtained by adding 3 to the preceding term.
等差数列是任意一项与前一项之差保持不变的数列。这个常数称为公差，通常记作 d。例如，数列 2, 5, 8, 11, 14, … 的公差为 3，因为每一项都是在前一项上加 3 得到的。
The general recursive relationship can be written as un+1 = un + d. To fully determine the sequence, you also need the first term, often labelled u1 or a. A negative common difference produces a decreasing arithmetic sequence.
一般的递推关系可以写作 un+1 = un + d。要完全确定数列，你还需要首项，通常记作 u1 或 a。负的公差会产生递减的等差数列。
3. nth Term of an Arithmetic Sequence | 等差数列的通项
The explicit formula for the nth term of an arithmetic sequence is derived directly from the definition. Starting from u1, each step adds d, so after (n−1) steps you reach the nth term:
等差数列的第 n 项显式公式直接由定义推出。从 u1 出发，每一步加 d，因此经过 (n−1) 步后即得第 n 项：
un = u1 + (n − 1) × d
This formula is extremely powerful for finding any term without having to list all preceding ones. For example, the 50th term of 3, 7, 11, 15, … with u1 = 3, d = 4 is u50 = 3 + (50 − 1) × 4 = 3 + 196 = 199.
这个公式非常强大，可以让你无需列出前面所有项就能找到任意一项。例如，数列 3, 7, 11, 15, … 中 u1 = 3, d = 4，其第 50 项为 u50 = 3 + (50 − 1) × 4 = 3 + 196 = 199。
IB questions often ask you to find the number of terms n when a term value is given, or to determine the first term and common difference from two known terms. Rearranging the formula algebraically is a key skill.
IB 试题经常要求你在已知某项值时求项数 n，或者根据两个已知项求首项和公差。对公式进行代数重组是一项关键技能。
4. Sum of an Arithmetic Series | 等差数列求和
When you add the terms of an arithmetic sequence together, you obtain an arithmetic series. The sum of the first n terms, denoted by Sn, can be calculated using a clever pairing argument or by averaging the first and last term:
当你把等差数列的项相加时，就得到了一个等差级数。前 n 项的和记作 Sn，可以利用巧妙的配对论证或首末项的平均值来计算：
Sn = n/2 × [ 2u1 + (n − 1)d ]
Alternatively, Sn = n/2 × (u1 + un) is often more convenient when the last term is known. This formula reflects the idea that the average of the first and last term is the same as the average of the second and second-to-last, and so on.
另一种形式是 Sn = n/2 × (u1 + un)，当已知末项时使用起来往往更方便。这个公式体现了一个思路：首项和末项的平均值等于第二项和倒数第二项的平均值，依此类推。
Suppose u1 = 5, d = 2, and n = 20. Then S20 0. Understanding both forms is crucial because exam problems may provide different givens.
假设 u1 = 5, d = 2, n = 20。那么 S20 = 20/2 × [ 2×5 + (20−1)×2 ] = 10 × [10 + 38] = 480。理解两种形式至关重要，因为考试题目可能给出不同的已知条件。
5. Geometric Sequences | 等比数列
A geometric sequence is defined by a constant common ratio ach term is obtained by multiplying the previous term by r. For example, 3, 6, 12, 24, 48, … has r = 2, while 8, 4, 2, 1, 0.5, … has r = ½.
等比数列由相邻项之间的恒定公比定义，记作 r。每一项都是前一项乘以 r 得到的。例如，3, 6, 12, 24, 48, … 的公比 r = 2，而 8, 4, 2, 1, 0.5, … 的公比 r = ½。
Geometric sequences model exponential growth or decay, making them highly relevant in finance (compound interest), biology (population growth), and physics (radioactive decay). In IB exams, you must be able to distinguish arithmetic from geometric sequences by checking the pattern of differences versus ratios.
等比数列模拟指数增长或衰减，这使得它们在金融（复利）、生物学（人口增长）和物理学（放射性衰变）中非常实用。在 IB 考试中，你必须能够通过检查差的模式与比的模式来区分等差数列和等比数列。
6. nth Term of a Geometric Sequence | 等比数列的通项
The nth term of a geometric sequence is found by repeatedly multiplying the first term by r. The explicit formula is:
等比数列的第 n 项是通过首项不断乘以 r 得到的。显式公式为：
un = u1 × rn−1
Notice the exponent is (n−1), not n, because the first term corresponds to r0 = 1. For a sequence with u1 = 4 and r = 3, the 6th term is u6 = 4 × 35 = 4 × 243 = 972. If r is negative, the terms alternate in sign, e.g., 5, −10, 20, −40, … with r = −2.
注意指数为 (n−1) 而非 n，因为首项对应 r0 = 1。对于首项 u1 = 4、公比 r = 3 的数列，第 6 项为 u6 = 4 × 35 = 4 × 243 = 972。若 r 为负，则各项正负交替，如 5, −10, 20, −40, … 公比 r = −2。
When |r| < 1, the terms get closer to zero; this behaviour is critical for sum to infinity. You may also be required to write the explicit rule from a given recursive definition and vice versa.
当 |r| < 1 时，各项趋近于零；这种行为对无穷求和至关重要。你也可能被要求根据给定的递推定义写出显式公式，或者反过来。
7. Sum of a Finite Geometric Series | 有限等比数列求和
The sum of the first n terms of a geometric sequence is found using a formula that arises from subtracting two matched expressions. The result is:
等比数列前 n 项和的公式通过两式相减得出，结果为：
Sn = u1 × (1 − rn) / (1 − r) , for r ≠ 1
This form is convenient when |r| < 1. An equivalent version is Sn = u1 × (rn − 1) / (r − 1), often used when r > 1 to avoid negative denominators. Both give the same numerical result.
当 |r| < 1 时，这种形式很方便。另一种等价形式为 Sn = u1 × (rn − 1) / (r − 1)，常在 r > 1 时使用以避免负分母。两者给出的数值结果相同。
For example, sum the first 10 terms of 1, 2, 4, 8, … Here u1 = 1, r = 2, n = 10, so S10 = 1 × (210 / (2 − 1) = 1024 − 1 = 1023. Practise choosing the right version to simplify calculations.
例如，求数列 1, 2, 4, 8, … 的前 10 项和。这里 u1 = 1, r = 2, n = 10，所以 S10 = 1 × (210 − 1) / (2 − 1) = 1024 − 1 = 1023。练习选择合适的版本以简化计算。
8. Sum to Infinity | 无穷等比数列求和
When the common ratio satisfies −1 < r < 1 (i.e., |r| < 1), the terms of the geometric sequence become vanishingly small as n increases. The series converges to a finite sum, called the sum to infinity:
当公比满足 −1 < r < 1（即 |r| < 1）时，随着 n 增大，等比数列的项变得极小。该级数收敛于一个有限的无穷和：
S∞ = u1 / (1 − r)
elegant result is derived by taking the limit as n → ∞ in the finite sum formula, where rn → 0. For instance, the series 8 + 4 + 2 + 1 + ½ + … has u1 = 8, r = ½, so S∞ = 8 / (1−½) = 16. Any series with |r| ≥ 1 diverges and has no finite sum.
这个简洁的结果是在有限求和公式中令 n → ∞、rn → 0 取极限得到的。例如，级数 8 + 4 + 2 + 1 + ½ + … 中 u1 = 8, r = ½，所以 S∞ = 8 / (1−½) = 16。任何满足 |r| ≥ 1 的级数均发散，没有有限和。
IB questions often ask you to find the sum to infinity, determine the value of r given the sum, or interpret infinite geometric series in real-life contexts like bouncing balls or recurring decimals.
IB 题目经常要求你求无穷和、根据已知无穷和求公比 r，或者在实际情境中解释无穷等比级数，例如弹跳球或循环小数。
9. Sigma Notation | 求和符号 Σ
n provides a compact way to represent series. The general form is Σ (from k = 1 to n) of ak, where k is the index. For arithmetic and geometric series, you can write:
求和符号 Σ 提供了一种表示级数的紧凑方式。一般形式为 Σ (k=1 到 n) ak，其中 k 为指标。对于等差和等比级数，可以写作：
Σ (k=1 to n) (u1 + (k−1)d) or Σ (k=1 to n) u1 rk−1
ower and upper bounds is essential when the index does not start at 1, e.g., Σ (k=3 to 8).
你需要能够展开求和符号以显示各项，相反地，也要能将展开的级数写成求和符号形式。当指标不从 1 开始时（例如 Σ (k=3 到 8)），理解下限和上限至关重要。
ion when evaluating sums with technology, and to recognise the properties such as Σ (c × ak) = c × Σ ak. This notation simplifies handling large data sets and series proofs.
此外，IB 要求你在使用技术求值时运用求和符号，并识别其性质，例如 Σ (c × ak) = c × Σ ak。这种符号简化了处理大型数据集和级数证明的过程。
10. Problem-Solving Tips for Sequences | 数列解题技巧
When tackling IB sequence problems, always begin by identifying the type of sequence. Check if differences are constant (arithmetic) or if ratios are constant (geometric). Write down the known values: u1, d or r, n, un, Sn, and select the appropriate formula.
处理 IB 数列问题时，始终从识别数列类型开始。检查差是否恒定（等差）或比是否恒定（等比）。写下已知值：u1、d 或 r、n、un、Sn，并选择合适的公式。
Many problems involve forming equations from given information. For example, if you know u3 and u7 in an arithmetic sequence, treat u3 = u1 + 2d and u7 = u1 + 6d, then solve simultaneously. A similar approach works for geometric sequences by setting up a pair of exponential equations.
许多问题涉及根据给定信息建立方程。例如，若已知等差数列中的 u3 和 u7，可设 u3 = u1 + 2d 和 u7 = u1 + 6d，然后联立求解。类似方法也适用于等比数列，需建立指数方程组。
When modelling real-world situations, clearly define your parameters and decide whether an arithmetic or geometric model is more appropriate. Arithmetic models imply linear growth; geometric models imply exponential growth or decay. Always check for convergence constraints when dealing with sum to infinity.
在对现实情境建模时，要清晰地定义参数，并判断是等差数列模型还是等比数列模型更合适。等差模型意味着线性增长；等比模型意味着指数增长或衰减。在处理无穷和时，务必检查收敛条件。
Finally, use your calculator’s sequence or spreadsheet functions to verify results, especially for iterative series. IB encourages a balanced approach between analytical work and technology use.
最后，利用计算器的数列或电子表格功能验证结果，特别是迭代级数。IB 鼓励分析工作与技术使用之间的平衡。
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