📚 Mastering Sequences: From Arithmetic Progressions to Infinite Series | 掌握数列：从等差数列到无穷级数
Sequences appear everywhere in the IB curriculum — from modelling interest rates to describing the patterns of nature. Mastering arithmetic, geometric, and recursive sequences gives you a powerful toolkit for both Paper 1 and Paper 2, as well as for internal assessments. This guide unpacks key concepts, formulas, and strategies to help you conquer sequences with confidence.
数列在 IB 课程中无处不在——从模拟利率到描述自然界的模式。掌握等差数列、等比数列和递推数列，将为你在试卷一和试卷二以及内部评估中提供强大的工具。本指南将解析关键概念、公式和策略，帮助你自信地征服数列题目。
1. Introduction to Sequences | 数列简介
A sequence is simply an ordered list of numbers that follow a particular rule. Each number in the list is called a term. Sequences can be finite (with a fixed number of terms) or infinite (continuing forever). In IB Mathematics, you will work with sequences that lead to series, calculus ideas, and financial models.
数列就是遵循特定规则的一列有序的数。列表中的每一个数都叫做一个项。数列可以是有限的（拥有固定数量的项）或无限的（永远继续下去）。在 IB 数学中，你将处理那些通向级数、微积分概念和金融模型的数列。
The notation is essential: we usually denote the first term by u₁, the second by u₂, and the nth term by uₙ. The position index is written as a subscript. For example, the sequence 3, 7, 11, 15,… can be described by a formula for uₙ. Recognising patterns and using the correct notation saves time and avoids errors in exams.
表示法至关重要：我们通常用 u₁ 表示第一项，u₂ 表示第二项，uₙ 表示第 n 项。位置索引写成下标形式。例如，数列 3, 7, 11, 15,… 可以用 uₙ 的公式来描述。识别模式并使用正确的表示法可以节省时间并避免考试中的错误。
- Finite sequence: 2, 5, 8, 11 (4 terms).
- Infinite sequence: 1, 3, 5, 7, … (continues).
- 有限数列：2, 5, 8, 11（4 项）。
- 无限数列：1, 3, 5, 7, …（持续下去）。
2. Arithmetic Sequences: Definition and nth Term | 等差数列：定义与第 n 项
An arithmetic sequence, or arithmetic progression (AP), is a sequence where the difference between any term and the previous term is constant. This constant is called the common difference, represented by d. The nth term formula is the cornerstone of all AP problems.
等差数列（或称算术数列）是指任意一项与它前一项之差为常数的数列。这个常数叫做公差，用 d 表示。第 n 项公式是所有 AP 问题的基石。
If the first term is u₁ and the common difference is d, then the nth term uₙ is obtained by starting at u₁ and adding (n−1) lots of d:
如果首项是 u₁，公差为 d，那么第 n 项 uₙ 可以通过从 u₁ 开始，加上 (n−1) 个 d 来求得：
uₙ = u₁ + (n − 1)d
For example, the sequence 5, 9, 13, 17,… has u₁ = 5 and d = 4. The 10th term is u₁₀ = 5 + 9×4 = 41. Pay close attention to the (n−1) factor — using n directly is one of the most common mistakes. Always double-check by listing a few terms mentally.
例如，数列 5, 9, 13, 17,… 有 u₁ = 5，d = 4。第 10 项 u₁₀ = 5 + 9×4 = 41。要特别注意 (n−1) 这个因子——直接使用 n 是最常见的错误之一。务必在脑海中列出几项来双重检查。
3. Sum of an Arithmetic Series | 等差数列的求和
When you add up the terms of an arithmetic sequence, you obtain an arithmetic series. The sum Sₙ of the first n terms can be remembered either in terms of the first and last terms or in terms of u₁ and d.
当你将等差数列的项相加，就得到了等差级数。前 n 项的和 Sₙ 既可以用首项和末项表示，也可以用 u₁ 和 d 表示。
Sₙ = n/2 (u₁ + uₙ) = n/2 [2u₁ + (n − 1)d]
The formula comes from a clever pairing idea: write the sum forward and backward, then add. Each column sums to (u₁+uₙ), and there are n columns. This reasoning often appears in IB “show that” questions, so it is worth knowing the derivation.
该公式源于一个巧妙的配对思想：将和式正着写和倒着写，然后相加。每一列的和都是 (u₁+uₙ)，共有 n 列。这个推理过程经常出现在 IB 的“证明”题中，因此值得了解推导过程。
Example: Find the sum of the first 20 terms of 3 + 7 + 11 + … . Here u₁=3, d=4, n=20, so S₂₀ = 20/2 [2×3 + 19×4] = 10 [6 + 76] = 820. Always write down what you know before substituting.
例题：求 3 + 7 + 11 + … 前 20 项的和。这里 u₁=3, d=4, n=20，所以 S₂₀ = 20/2 [2×3 + 19×4] = 10 [6 + 76] = 820。代入前务必先写下已知信息。
4. Geometric Sequences: Definition and nth Term | 等比数列：定义与第 n 项
A geometric sequence (geometric progression, GP) has a constant ratio between consecutive terms. This common ratio is denoted by r (r ≠ 0). Each term is found by multiplying the previous term by r.
等比数列（几何数列）相邻两项之间具有恒定的比值。这个公比用 r 表示（r ≠ 0）。每一项都是前一项乘以 r 得到的。
uₙ = u₁ · rⁿ⁻¹
Notice the exponent is n−1, not n. If r is positive, the terms either grow or decay. If r is negative, the signs alternate, creating an oscillating sequence. Both cases appear in IB exams.
注意指数是 n−1，而不是 n。如果 r 为正，各项会增长或衰减；如果 r 为负，符号会交替出现，形成摆动数列。两种情况都会在 IB 考试中出现。
For instance, 4, 12, 36, 108,… has u₁=4 and r=3; the 6th term is u₆ = 4·3⁵ = 972. Another example: 8, −4, 2, −1,… with u₁=8 and r=−½, giving u₅ = 8·(−½)⁴ = 8·(1/16) = ½.
例如，4, 12, 36, 108,… 有 u₁=4，r=3；第 6 项 u₆ = 4·3⁵ = 972。再比如 8, −4, 2, −1,… 有 u₁=8，r=−½，第 5 项 u₅ = 8·(−½)⁴ = 8·(1/16) = ½。
5. Sum of a Finite Geometric Series | 有限等比数列求和
Adding the first n terms of a geometric sequence gives a finite geometric series. The sum formula works for any r ≠ 1.
将等比数列的前 n 项相加得到有限几何级数。求和公式适用于任何 r ≠ 1 的情况。
Sₙ = u₁(1 − rⁿ) / (1 − r), r ≠ 1
The derivation uses a trick: write Sₙ, then multiply the whole equation by r, subtract, and solve for Sₙ. Many IB questions ask you to use this formula directly, but understanding the derivation helps with algebra skills and proof-type problems.
推导使用了一个技巧：写出 Sₙ，然后将整个等式乘以 r，相减，再解出 Sₙ。许多 IB 题目要求直接使用这个公式，但理解推导过程有助于提升代数技能和应对证明类问题。
Example: Sum the series 1 + 3 + 9 + 27 + 81. Here u₁=1, r=3, n=5, so S₅ = 1(1 − 3⁵)/(1 − 3) = (1 − 243)/(−2) = 121. Recall that rⁿ grows quickly, so always compute carefully.
例题：求级数 1 + 3 + 9 + 27 + 81 的和。这里 u₁=1，r=3，n=5，因此 S₅ = 1(1 − 3⁵)/(1 − 3) = (1 − 243)/(−2) = 121。记住 rⁿ 增长很快，因此计算时要格外小心。
6. Infinite Geometric Series | 无穷等比级数
If the absolute value of the common ratio is less than 1 (|r| < 1), the terms get progressively smaller and the infinite series converges to a finite sum.
如果公比的绝对值小于 1 (|r| < 1)，各项会逐渐变小，无穷级数收敛于一个有限的和。
S∞ = u₁ / (1 − r), for |r| < 1
This elegant formula emerges because rⁿ → 0 as n → ∞ when |r| < 1. If |r| ≥ 1, the series diverges — the sum either grows without bound or oscillates without settling. Always check the convergence condition first.
这个简洁的公式之所以成立，是因为当 |r| < 1 时，rⁿ → 0（n → ∞）。如果 |r| ≥ 1，级数发散——和要么无限增长，要么振荡而不定居。务必先检验收敛条件。
Classic example: 1 + ½ + ¼ + ⅛ + … has u₁=1, r=½, giving S∞ = 1 / (1 − ½) = 2. This type of question often appears alongside real-life contexts like bouncing balls or cooling temperatures.
经典例子：1 + ½ + ¼ + ⅛ + … 有 u₁=1，r=½
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