Proof by Induction | 数学归纳法

📚 Proof by Induction | 数学归纳法

Proof by induction is a powerful technique used across A-Level Edexcel Mathematics to prove statements involving natural numbers. It is particularly common in sequences, series, divisibility, and inequalities. Mastering the three-step structure – base case, inductive hypothesis, and inductive step – is essential for scoring full marks on Pure Mathematics proof questions. This article breaks down each component with clear examples and common pitfalls, helping you build a solid understanding of the method.

数学归纳法是Edexcel A-Level数学中用于证明涉及自然数命题的强大方法，常见于数列、级数、整除性及不等式证明中。掌握其三步结构——基础情形、归纳假设与归纳步骤——对于在纯数证明题中取得满分至关重要。本文通过清晰的示例和常见误区，逐步拆解各个环节，助你扎实掌握这一方法。

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1. Introduction to Induction | 归纳法简介

r the first domino, and ensure that each domino knocks over the next, then all dominoes will fall. In mathematics, we first verify the statement for n = 1, then show that if it is true for n = k, it must also be true for n = k + 1.

数学归纳法是一种证明某个命题对所有自然数 n 成立的方法。可以把它想象成一排多米诺骨牌：如果你能推倒第一块骨牌，并保证每块骨牌倒下时都能推倒下一块，那么所有骨牌都会倒下。在数学中，我们先验证 n = 1 时命题成立，然后证明如果 n = k 时命题成立，则 n = k + 1 时命题也必然成立。

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2. The Principle of Mathematical Induction | 数学归纳法原理

The principle rests on the axiom that if P(1) is true and P(k) ⇒ P(k + 1) for every positive integer k, then P(n) is true for all n ∈ ℕ. This is a formal way of expressing the domino effect. Edexcel questions will often ask you to prove statements for all positive integers, and you must structure your answer around these two logical steps.

归纳法原理基于这样一条公理：如果 P(1) 为真，并且对每一个正整数 k 都有 P(k) ⇒ P(k + 1)，那么 P(n) 对所有自然数 n 都成立。这实际上是“多米诺效应”的形式化表述。在 Edexcel 考题中，常会要求你对所有正整数证明某个命题，你必须围绕这两个逻辑步骤组织答案。

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3. Base Case | 基础情形

ality and show they are equal (or the relation holds). Never skip this step, as without it the entire proof collapses. Write clearly: “When n = 1, LHS = ... , RHS = ... , so the statement is true for n = 1.”

基础情形用于验证命题对最小的取值（通常是 n = 1）成立。你需要将 n = 1 代入等式或不等式的两边，并说明它们相等（或关系成立）。切勿跳过这一步，否则整个证明将失去根基。书写时应清晰表达：“当 n = 1 时，左式 = ……，右式 = ……，因此 n = 1 时命题成立。”

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4. Inductive Hypothesis | 归纳假设

k. This is the inductive hypothesis: P(k) holds. You must state this clearly, for example: “Assume for n = k that 1 + 2 + ... + k = k(k+1)/2.” assumption is the bridge that allows you to prove the next case. It is essential to write “Assume true for n = k” rather than “for n = n”.

归纳假设是假定命题对某个任意的正整数 n = k 成立。你必须清楚地陈述这一假设，例如：“假设当 n = k 时，1 + 2 + ... + k = k(k+1)/2。”这一假设是证明下一情形的桥梁。必须写明“Assume true for n = k”，而不能写成“for n = n”。

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5. Inductive Step | 归纳步骤

. For a summation, add the (k+1)-th term to the assumed sum. For divisibility, express f(k+1) in terms of f(k). The goal is to arrive at the same formula form with k replaced by k+1. Conclude: “Thus the statement holds for n = k + 1.”

现在你需要证明若 P(k) 成立，则 P(k + 1) 也成立。从 n = k + 1 的表达式出发，利用归纳假设进行改写。对于求和问题，将第 (k+1) 项加到假设的和式上。对于整除性问题，将 f(k+1) 用 f(k) 表示。目标是得出与 k 换成 k+1 后形式一致的公式。最后作结：“因此 n = k + 1 时命题成立。”

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6. Summation Proofs | 求和证明

A typical summation proof asks you to show that ∑_r=1ⁿ r = n(n+1)/2 or ∑_r=1ⁿ r² = n(n+1)(2n+1)/6. In the inductive step, write LHS for n = k+1 as (∑_r=1^k r) + (k+1). Then replace the first part with the assumed formula and simplify to obtain (k+1)(k+2)/2, confirming the pattern. Always factorise the result to match the required closed form.

典型的求和证明题要求你证明 ∑_r=1ⁿ r = n(n+1)/2 或 ∑_r=1ⁿ r² = n(n+1)(2n+1)/6。在归纳步骤中，将 n = k+1 时的左式写作 (∑_r=1^k r) + (k+1)，然后用归纳假设代入第一部分，化简后得到 (k+1)(k+2)/2，从而验证规律。务必对结果进行因式分解，使其与要求的封闭形式一致。

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7. Divisibility Proofs | 整除性证明

d to show an expression like f(n) = 7ⁿ − 1 is divisible by 6. The inductive step: assume 7^k − 1 = 6m. Then write 7^k+1 − 1 = 7·7^k h is clearly a multiple of 6. This technique of expressing f(k+1) as a combination of f(k) and a multiple of the divisor is classic. Always state “divisible by” clearly.

对于整除性问题，你需要证明类似于 f(n) = 7ⁿ − 1 能被 6 整除的表达式。归纳步骤：假设 7^k − 1 = 6m，那么 7^k+1 − 1 = 7·7^k − 1 = 7(6m + 1) − 1 = 42m + 6 = 6(7m + 1)，显然它是 6 的倍数。这种将 f(k+1) 表示为 f(k) 与除数的倍数之和的方法是经典套路。务必清晰地写出“可被……整除”。

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8. Inequalities Proofs | 不等式证明

When proving inequalities such as 2ⁿ > n for all positive integers, proceed from the assumed 2^k > k. Then multiply both sides by 2 to get 2^k+1 > 2k. Now you need to show 2k ≥ k + 1 for k ≥ 1, which is true. Thus 2^k+1 > k + 1. Inequality induction often requires an extra justification linking the intermediate inequality to the target one – don't skip this reasoning.

在证明诸如 2ⁿ > n（对所有正整数）的不等式时，从假设 2^k > k 出发。两边同乘 2 得到 2^k+1 > 2k。此时需证明 2k ≥ k + 1 对 k ≥ 1 成立，这显然是正确的，因此 2^k+1 > k + 1。不等式归纳往往需要额外的理由来连接中间不等式与目标不等式，切勿省略这一推理过程。

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9. Common Mistakes | 常见错误

isuse the hypothesis by substituting k+1 into the assumption directly. Also, failing to simplify correctly in summation proofs, or not showing factorisation steps, loses marks. Always present a clear narrative: “ is true for n = k + 1. Since it is true for n = 1, it is true for all positive integers n.”

许多学生忘记写出“由归纳法知命题对所有 n 成立”的结论。还有人未经证明就错误地使用 P(k+1)，或将 k+1 直接代入假设而误用。在求和证明中未能正确化简，或未展示因式分解步骤，都会导致失分。务必呈现清晰的叙述：“因此，若 n = k 时命题成立，则 n = k + 1 时也成立。又因 n = 1 时成立，故对所有正整数 n 成立。”

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10. Worked Example: Summation | 例题：求和

Prove that for all n ∈ ℕ, ∑_r=1ⁿ (3r − 2) = (n/2)(3n − 1).

证明对所有自然数 n，∑_r=1ⁿ (3r − 2) = (n/2)(3n − 1)。

Base case n = 1: LHS = 3(1) − 2 = 1, RHS = (1/2)(3·1 − 1) = 1. ✓

基础情形 n = 1：左式 = 1，右式 = 1，成立。

Assume true for n = k: ∑_r=1^k (3r − 2) = (k/2)(3k − 1).

假设 n = k 时成立：∑_r=1^k (3r − 2) = (k/2)(3k − 1)。

Inductive step n = k+1: LHS = ∑_r=1^k +1) − 1). This matches the RHS form. Hence true for n = k+1. By induction, true for all n.

归纳步骤 n = k+1：左式 = 原和式 + 第 (k+1) 项 = (k/2)(3k − 1) + (3k+1) = (3k² +5k +2)/2，因式分解得 ((k+1)/2)(3(k+1) − 1)，与原式形式一致。故 n=k+1 时成立。由归纳法，对所有 n 成立。

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11. Worked Example: Divisibility | 例题：整除

Prove that f(n) = 3²ⁿ − 1 is divisible by 8 for all positive integers n.

证明对所有正整数 n，f(n) = 3²ⁿ − 1 可被 8 整除。

Base case n = 1: f(1) = 3² − 1 = 8, divisible by 8. ✓

基础情形 n = 1：f(1) = 8，可被 8 整除。

Assume for n = k: 3^2k − 1 = 8m, for some integer m.

假设 n = k 时：3^2k − 1 = 8m，m 为整数。

Inductive step n = k+1: f(k+1) = 3^2(k+1) − 1 = 3²·3^2k − 1 = 9·3^2k + 1). Clearly a multiple of 8. Thus f(k+1) is divisible by 8. By induction, f(n) is divisible by 8 for all positive integers n.

归纳步骤 n = k+1： f(k+1) = 9·3^2k − 1 = 9(8m+1) − 1 = 72m+8 = 8(9m+1)，显然是 8 的倍数。故 f(k+1) 可被 8 整除。由归纳法，对所有正整数 n 成立。

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12. Summary and Exam Tips | 总结与备考建议

Proof by induction is a highly structured topic — once you know the template, you can secure full marks. Always label each part: Base case, Inductive hypothesis, Inductive step, Conclusion. Show all algebraic simplification explicitly. In divisibility proofs, factor out the divisor cleanly. For inequalities, justify each inequality sign carefully. Practice a variety of question types, including summation, divisibility, matrix multiplication, and recurrence relations. Remember that induction is not just a mechanical process; it is a logical argument that builds truth from one integer to the next.

数学归纳法是结构极为清晰的主题，一旦掌握了模板，就能稳拿满分。请务必标明每个部分：基础情形、归纳假设、归纳步骤、结论。清晰展示所有代数化简过程。在整除性证明中，要将除数明确提取出来。对于不等式，要仔细说明每个不等号的理由。多练习不同类型的题目，包括求和、整除、矩阵乘法以及递推关系证明。记住，归纳法不仅是一套机械流程，更是一种从某一整数到下一整数逐步建立真理的逻辑论证。

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