📚 IB Mathematics: Finding the General Term of a Sequence | 数列通项公式的求法
In IB Mathematics, finding the general term (often denoted as aₙ) of a sequence is a fundamental skill that appears in both Analysis and Approaches (AA) and Applications and Interpretation (AI). A clear grasp of the methods for determining aₙ not only helps with direct questions but also underpins problems involving series, limits, and mathematical induction. This article provides a structured guide to the most common techniques.
在 IB 数学中，求数列的通项公式（通常记作 aₙ）是一项基本技能，无论是在分析与方法（AA）还是应用与解释（AI）课程中都会出现。掌握求 aₙ 的常见方法，不仅有助于直接解题，也是处理级数、极限和数学归纳法问题的基础。本文将系统梳理最常用的几类方法。
1. Arithmetic Sequences | 等差数列
An arithmetic sequence has a constant difference between consecutive terms. If the first term is a₁ and the common difference is d, then the general term is given by:
等差数列的相邻两项之差恒定。设首项为 a₁，公差为 d，则通项公式为：
aₙ = a₁ + (n − 1)d
This formula is derived by observing that a₂ = a₁ + d, a₃ = a₁ + 2d, and so on. It is often paired with the sum formula in IB exams, so make sure you can switch between the two quickly.
该公式由观察 a₂ = a₁ + d、a₃ = a₁ + 2d 等规律而得。IB 考试中经常将其与求和公式结合考查，因此务必能够快速切换使用两种公式。
2. Geometric Sequences | 等比数列
A geometric sequence has a constant ratio between consecutive terms. With first term a₁ and common ratio r, the general term is:
等比数列的相邻两项之比恒定。设首项为 a₁，公比为 r，则通项公式为：
aₙ = a₁ ⋅ rⁿ⁻¹
Take care with the exponent (n − 1), especially when solving for r or n using logarithms. In IB, you may also be asked to determine whether a sequence is geometric by checking the ratio aₙ₊₁ / aₙ.
指数为 (n − 1)，使用对数求解 r 或 n 时需格外小心。IB 考试中也可能要求通过检验比值 aₙ₊₁ / aₙ 来判断一个数列是否为等比数列。
3. Recurrence Relations: Iteration | 递推关系：迭代法
Some sequences are defined by a recurrence relation such as aₙ₊₁ = aₙ + f(n) or aₙ₊₁ = r ⋅ aₙ. For simple recurrences, expanding the first few terms reveals a pattern that can be generalised.
有些数列由递推关系定义，例如 aₙ₊₁ = aₙ + f(n) 或 aₙ₊₁ = r ⋅ aₙ。对于简单的递推式，展开前几项即可发现规律并归纳出通项。
Example | 例： Given a₁ = 1 and aₙ₊₁ = aₙ + 2n, write out a₂ = 1 + 2, a₃ = 1 + 2 + 4, suggesting that aₙ = 1 + 2(1 + 2 + … + (n−1)). This simplifies to aₙ = n² − n + 1.
例： 已知 a₁ = 1，且 aₙ₊₁ = aₙ + 2n，可写出 a₂ = 1 + 2、a₃ = 1 + 2 + 4，从而推测 aₙ = 1 + 2(1 + 2 + … + (n−1))，化简得 aₙ = n² − n + 1。
Iteration is best suited to recurrences where the change is linear or simple. The key is to recognise the emerging pattern and confirm it by substitution or induction.
迭代法最适合变化为线性或形式简单的递推式。关键在于识别显现出的规律，并通过代入或归纳法加以验证。
4. First-Order Linear Recurrences | 一阶线性递推
A recurrence of the form aₙ₊₁ = p aₙ + q (with p ≠ 1 and q ≠ 0) cannot be solved by simple iteration alone. The standard trick is to find a fixed point α such that α = pα + q, giving α = q/(1−p). Then rewrite the recurrence as:
形如 aₙ₊₁ = p aₙ + q 的递推式（其中 p ≠ 1 且 q ≠ 0）无法仅靠直接迭代求解。标准技巧是求不动点 α，令 α = pα + q，得 α = q/(1−p)。然后将递推式改写为：
aₙ₊₁ − α = p (aₙ − α)
This shows that the sequence bₙ = aₙ − α is geometric with common ratio p, so bₙ = b₁ pⁿ⁻¹, and hence aₙ = α + (a₁ − α) pⁿ⁻¹.
这表明新数列 bₙ = aₙ − α 是以 p 为公比的等比数列，因此 bₙ = b₁ pⁿ⁻¹，从而 aₙ = α + (a₁ − α) pⁿ⁻¹。
Example | 例： For aₙ₊₁ = 2aₙ + 3, a₁ = 1, we find α = −3, so aₙ = −3 + 4 ⋅ 2ⁿ⁻¹.
例： 对 aₙ₊₁ = 2aₙ + 3、a₁ = 1，可得 α = −3，故 aₙ = −3 + 4 ⋅ 2ⁿ⁻¹。
5. Second-Order Linear Recurrences | 二阶线性递推
For recurrences of the form aₙ₊₂ = P aₙ₊₁ + Q aₙ, we use the characteristic equation t² = P t + Q. If it has distinct real roots r₁ and r₂, the general term is:
对于形如 aₙ₊₂ = P aₙ₊₁ + Q aₙ 的递推式，可使用特征方程 t² = P t + Q。若特征方程有两个不相等的实根 r₁ 和 r₂，则通项为：
aₙ = A r₁ⁿ⁻¹ + B r₂ⁿ⁻¹
where A and B are found from the initial conditions a₁ and a₂. If the roots are equal (r₁ = r₂ = r), the general term takes the form:
其中常数 A 和 B 由初始条件 a₁ 和 a₂ 确定。若两根相等（r₁ = r₂ = r），通项形式变为：
aₙ = (A + Bn) rⁿ⁻¹
This method is particularly relevant in IB HL courses, where questions may also involve complex roots — in that case the solution can be expressed using sine and cosine functions.
这一方法在 IB HL 课程中尤为重要，有时还会出现复根的情形——此时解可用正弦和余弦函数表示。
6. Fractional Recurrences: Fixed-Point Method | 分式递推：不动点法
For a recurrence of the form aₙ₊₁ = (u aₙ + v)/(w aₙ + t), a direct iteration is rarely productive. Instead, solve the fixed-point equation x = (u x + v)/(w x + t), which is a quadratic in x. If it has two distinct fixed points p and q, then the transformation bₙ = (aₙ − p)/(aₙ − q) often turns the recurrence into a geometric one.
对于形如 aₙ₊₁ = (u aₙ + v)/(w aₙ + t) 的递推式，直接迭代往往收效甚微。此时应求解不动点方程 x = (u x + v)/(w x + t)，即关于 x 的二次方程。若存在两个不同的不动点 p 和 q，则通常可以通过变换 bₙ = (aₙ − p)/(aₙ − q) 将递推式化为等比形式。
Example | 例： For aₙ₊₁ = (aₙ + 2)/(aₙ + 1), solving x = (x + 2)/(x + 1) gives x = ±√2. Then bₙ = (aₙ − √2)/(aₙ + √2) satisfies bₙ₊₁ = k bₙ with k = (1 − √2)/(1 + √2), yielding a closed form for aₙ.
例： 对 aₙ₊₁ = (aₙ + 2)/(aₙ + 1)，解 x = (x + 2)/(x + 1) 得 x = ±√2。令 bₙ = (aₙ − √2)/(aₙ + √2)，可知 bₙ₊₁ = k bₙ，其中 k = (1 − √2)/(1 + √2)，由此可得 aₙ 的闭式表达式。
If the equation has a double root, a different substitution such as bₙ = 1/(aₙ − p) is used. IB questions usually point you toward the correct substitution, so read carefully.
若方程有重根，则需使用 bₙ = 1/(aₙ − p) 等不同的变换。IB 题目通常会在题干中暗示正确的代换方式，请仔细读题。
7. Using the Sum Formula: aₙ from Sₙ | 利用求和公式：由 Sₙ 求 aₙ
When the sum of the first n terms (denoted Sₙ) is given, the general term can be recovered using the identity:
当已知前 n 项和（记作 Sₙ）时，可由以下恒等式还原通项：
aₙ = Sₙ − Sₙ₋₁ (for n ≥ 2)
This follows because Sₙ = a₁ + a₂ + … + aₙ and Sₙ₋₁ = a₁ + a₂ + … + aₙ₋₁. It is essential to check the case n = 1 separately, since S₁ = a₁.
这是因为 Sₙ = a₁ + a₂ + … + aₙ，而 Sₙ₋₁ = a₁ + a₂ + … + aₙ₋₁。必须单独验证 n = 1 的情况，因为 S₁ = a₁。
Example | 例： If Sₙ = 3n² + 2n, then for n ≥ 2, aₙ = (3n² + 2n) − [3(n−1)² + 2(n−1)] = 6n − 1. Checking n = 1 gives a₁ = 5, which also satisfies 6(1) − 1 = 5.
例： 若 Sₙ = 3n² + 2n，则 n ≥ 2 时，aₙ = (3n² + 2n) − [3(n−1)² + 2(n−1)] = 6n − 1。检验 n = 1 得 a₁ = 5，同样满足 6(1) − 1 = 5。
This method is particularly common in IB Paper 1 and Paper 2 questions involving polynomial expressions for Sₙ. Remember: never assume that a formula valid for n ≥ 2 also holds for n = 1 — always verify.
此法在 IB Paper 1 和 Paper 2 中经常出现，尤其是 Sₙ 为多项式形式的问题。切记：不要假定 n ≥ 2 时成立的公式也适用于 n = 1——务必单独验证。
8. Mathematical Induction | 数学归纳法
When a candidate formula for aₙ is proposed (perhaps from iteration or pattern recognition), mathematical induction provides a rigorous proof. The induction process in the context of sequences has three steps:
当已通过迭代或规律观察提出 aₙ 的候选公式时，数学归纳法可为其提供严谨证明。在数列情境中，归纳法分为三步：
- Base case: verify the formula for n = 1 (or the first relevant value).
- Inductive step: assume the formula holds for n = k, and use the recurrence relation to show it holds for n = k + 1.
- Conclusion: state that by induction, the formula is true for all positive integers n.
- 基础情形：验证 n = 1（或第一个相关取值）时公式成立。
- 归纳步骤：假设公式对 n = k 成立，利用递推关系证明 n = k + 1 时也成立。
- 结论：由数学归纳法可知，该公式对所有正整数 n 成立。
Induction is not a method for finding aₙ, but rather for verifying a guess. In IB exams, the statement to be proved is often given explicitly, so practising careful algebraic manipulation in the inductive step is critical.
归纳法并非用来“发现”aₙ 的方法，而是用来“验证”猜想的方法。在 IB 考试中，待证明的公式通常已直接给出，因此熟练掌握归纳步骤中的代数变形至关重要。
Mastering the various techniques for finding the general term of a sequence is essential for success in IB Mathematics. Start by identifying whether the sequence is arithmetic, geometric, or recursive, then choose the appropriate method. For recursive forms, the fixed-point and characteristic equation methods are particularly powerful. Make sure to verify your results with initial terms — this simple check can save you from costly mistakes in an exam.
掌握求数列通项公式的各种方法是 IB 数学备考的核心环节。首先判断数列属于等差、等比还是递推类型，再选择对应方法。对于递推形式，不动点法和特征方程法尤其有效。最后务必将初始几项代入验证——这一简单检查能帮助你在考试中避免不必要的失分。
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