Exercise 1G: Mastering Mathematical Induction | 练习1G：精通数学归纳法

📚 Exercise 1G: Mastering Mathematical Induction | 练习1G：精通数学归纳法

In many IB Mathematics textbooks, Exercise 1G is devoted to proof by mathematical induction – a cornerstone technique for establishing the truth of infinitely many statements about positive integers. Whether you are working on summation formulas, divisibility results, inequalities or properties of recursively defined sequences, Exercise 1G challenges you to master the rigorous three‑step structure of induction. This revision guide walks you through all the key question types you are likely to encounter, gives worked examples written in clear logical steps, and highlights common mistakes that can cost you marks. Use it as your go‑to companion when tackling the problems in Exercise 1G, and you will soon find induction both intuitive and satisfying.

在许多 IB 数学课本中，练习 1G 专门针对数学归纳法证明——这是确定无穷多个关于正整数的命题是否成立的基础技巧。无论你处理的是求和公式、整除性结论、不等式还是递推数列的性质，练习 1G 都要求你掌握归纳法的严谨三步结构。这篇复习指南将带你梳理所有你可能遇到的典型题型，配有步骤清晰的工作示例，并强调了可能让你丢分的常见错误。把它当作你攻克练习 1G 时的随身指南，你很快就会发现归纳法既直观又令人满足。

---

1. The Principle of Induction | 归纳法原理

Mathematical induction rests on a simple but powerful idea: if a statement P(n) is true for n = 1, and if the truth of P(k) always forces the truth of P(k + 1), then P(n) must be true for all positive integers n. This is often compared to an infinite row of dominoes – knocking over the first domino and ensuring each domino knocks over the next one guarantees that every domino will fall.

数学归纳法基于一个简单而强大的思想：如果命题 P(n) 对 n = 1 成立，并且 P(k) 成立总能推出 P(k + 1) 成立，那么 P(n) 必然对所有正整数 n 都成立。常把它比作无限长的一排骨牌——推倒第一块骨牌并确保每一块骨牌都能推倒下一块，就可以保证所有骨牌都会倒下。

To write a complete induction proof you must clearly present three stages: the base case (usually n = 1, but sometimes n = 0 or a small starting value), the induction hypothesis (assume P(k) is true for some arbitrary integer k ≥ 1), and the induction step (prove that P(k) ⇒ P(k + 1)). In IB exams you are expected to label these parts explicitly and to justify algebraic manipulations at each stage.

要写出完整的归纳法证明，你必须清楚地展示三个阶段：基础情形（通常取 n = 1，但有时取 n = 0 或其他较小的起始值），归纳假设（假设对某个任意的整数 k ≥ 1，P(k) 成立），以及归纳递推（证明 P(k) ⇒ P(k + 1)）。在 IB 考试中，你需要明确标出这几个步骤，并在每一步对代数变形给出充分理由。

---

2. Proving Summation Formulas | 证明求和公式

One of the most common Exercise 1G tasks is to prove a closed‑form expression for a sum of a series. For example, you might be asked to show that 1 + 2 + 3 + … + n = n(n + 1)/2, or that the sum of the first n odd numbers equals n². The induction step typically begins by writing the sum to k + 1 terms as the sum to k terms plus the (k + 1)‑th term, then applying the induction hypothesis.

练习 1G 中最常见的任务之一是证明级数求和的闭式表达式。比如，你可能需要证明 1 + 2 + 3 + … + n = n(n + 1)/2，或者证明前 n 个奇数的和等于 n²。归纳递推步骤通常先把前 k + 1 项的和写成前 k 项的和再加上第 (k + 1) 项，然后代入归纳假设进行变形。

Worked example: Prove that 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1)/6 for all n ∈ ℤ⁺.

示例：证明对所有正整数 n，有 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1)/6。

Base case n = 1: LHS = 1² = 1; RHS = 1×2×3/6 = 1. True.
Induction hypothesis: Assume ∑ᵢ₌₁ᵏ i² = k(k + 1)(2k + 1)/6.
Induction step: For n = k + 1, LHS = ∑ᵢ₌₁ᵏ i² + (k + 1)² = k(k + 1)(2k + 1)/6 + (k + 1)². Factor (k + 1) to obtain (k + 1)[k(2k + 1) + 6(k + 1)]/6 = (k + 1)(2k² + 7k + 6)/6 = (k + 1)(k + 2)(2k + 3)/6, which matches RHS with n = k + 1. Hence proven.

基础情形 n = 1：左边 = 1² = 1；右边 = 1×2×3/6 = 1。成立。
归纳假设：假设 ∑ᵢ₌₁ᵏ i² = k(k + 1)(2k + 1)/6。
归纳递推：对 n = k + 1，左边 = ∑ᵢ₌₁ᵏ i² + (k + 1)² = k(k + 1)(2k + 1)/6 + (k + 1)²。提取因子 (k + 1) 得到 (k + 1)[k(2k + 1) + 6(k + 1)]/6 = (k + 1)(2k² + 7k + 6)/6 = (k + 1)(k + 2)(2k + 3)/6，与 n = k + 1 时的右边一致。证毕。

---

3. Divisibility Proofs | 整除性证明

Divisibility questions ask you to show that a given expression is divisible by a fixed integer for all positive integers n. The key to the induction step is to express f(k + 1) in terms of f(k) and a multiple of the divisor. A typical approach is to write f(k + 1) = a·f(k) + M, where M is clearly a multiple of the divisor. This structure allows you to apply the induction hypothesis immediately.

整除性问题要求你证明某个表达式对所有正整数 n 都能被一个固定的整数整除。归纳递推的关键在于将 f(k + 1) 用 f(k) 和除数的倍数表示出来。常见思路是写成 f(k + 1) = a·f(k) + M，其中 M 显然能被该除数整除。这样就可以直接套用归纳假设。

Worked example: Prove that n³ – n is divisible by 6 for all n ∈ ℤ⁺.

示例：证明对所有正整数 n，n³ – n 能被 6 整除。

Base case n = 1: 1³ – 1 = 0, divisible by 6.
Induction hypothesis: Assume k³ – k = 6m for some integer m.
Induction step: (k + 1)³ – (k + 1) = k³ + 3k² + 3k + 1 – k – 1 = (k³ – k) + 3k(k + 1). The first bracket is 6m by hypothesis. The second term 3k(k + 1) is a multiple of 6 because k(k + 1) is always even, so 3 times an even number yields a multiple of 6. Therefore the whole expression is a multiple of 6.

基础情形 n = 1：1³ – 1 = 0，能被 6 整除。
归纳假设：假设 k³ – k = 6m，其中 m 为某整数。
归纳递推：(k + 1)³ – (k + 1) = k³ + 3k² + 3k + 1 – k – 1 = (k³ – k) + 3k(k + 1)。由归纳假设，第一部分等于 6m。第二项 3k(k + 1) 是 6 的倍数，因为 k(k + 1) 总是偶数，而 3 乘以偶数得到的是 6 的倍数。因此整个表达式是 6 的倍数。

---

4. Induction with Inequalities | 含不等式的归纳法

Inequality induction problems require you to manipulate both sides of an inequality while preserving the direction. The core of the induction step is to start from the assumed inequality P(k), perform algebraic or logical operations to reach P(k + 1), and then justify that these operations are legitimate. Often you will need to use the fact that k ≥ some value, or combine the induction hypothesis with a separate simple estimate.

含不等式的归纳法问题要求你在保持不等号方向的前提下对不等式的两边进行变形。归纳递推的核心是从假设的不等式 P(k) 出发，经过代数或逻辑运算推出 P(k + 1)，同时要说明这些运算是合理的。通常你需要利用 k ≥ 某个值这个条件，或者将归纳假设与另外一个简单估计结合起来。

Worked example: Prove that 2ⁿ > n² for all integers n ≥ 5.

示例：证明对所有整数 n ≥ 5，有 2ⁿ > n²。

Base case n = 5: 2⁵ = 32 > 25 = 5². True.
Induction hypothesis: Assume 2ᵏ > k² for some k ≥ 5.
Induction step: Consider 2ᵏ⁺¹ = 2·2ᵏ > 2·k² (by hypothesis). We need to show 2·k² ≥ (k + 1)² for k ≥ 5. Expand: 2k² – (k + 1)² = 2k² – k² – 2k – 1 = k² – 2k – 1 = (k – 1)² – 2. For k ≥ 5, (k – 1)² – 2 ≥ 16 – 2 = 14 > 0, so 2·k² > (k + 1)². Thus 2ᵏ⁺¹ > (k + 1)².

基础情形 n = 5：2⁵ = 32 > 25 = 5²。成立。
归纳假设：假设对某个 k ≥ 5 有 2ᵏ > k²。
归纳递推：考虑 2ᵏ⁺¹ = 2·2ᵏ > 2·k²（根据假设）。需证明当 k ≥ 5 时 2·k² ≥ (k + 1)²。展开：2k² – (k + 1)² = 2k² – k² – 2k – 1 = k² – 2k – 1 = (k – 1)² – 2。当 k ≥ 5 时，(k – 1)² – 2 ≥ 16 – 2 = 14 > 0，所以 2·k² > (k + 1)²。因此 2ᵏ⁺¹ > (k + 1)²。

---

5. Induction with Recursively Defined Sequences | 递推定义数列的归纳法

IB problems frequently present a sequence defined by a recurrence relation, such as uₙ₊₁ = 2uₙ + 1, and ask you to prove a closed‑form formula like uₙ = 2ⁿ – 1. The base case verifies the formula for the initial term, and the induction step uses the recurrence to express uₖ₊₁ in terms of uₖ. Substituting the induction hypothesis for uₖ yields the desired closed form for index k + 1.

IB 试题中经常出现用递推关系定义的数列，例如 uₙ₊₁ = 2uₙ + 1，然后要求你证明形如 uₙ = 2ⁿ – 1 的通项公式。基础情形要验证首项满足公式，而归纳递推步骤则利用递推关系把 uₖ₊₁ 用 uₖ 表示出来。代入关于 uₖ 的归纳假设即可得到指标 k + 1 时的目标闭式。

Worked example: A sequence satisfies u₁ = 3 and uₙ₊₁ = 2uₙ – 1. Prove that uₙ = 2ⁿ + 1.

示例：数列满足 u₁ = 3，uₙ₊₁ = 2uₙ – 1。证明 uₙ = 2ⁿ + 1。

Base case n = 1: u₁ = 2¹ + 1 = 3, matches given value.
Induction hypothesis: Assume uₖ = 2ᵏ + 1 for some k ≥ 1.
Induction step: uₖ₊₁ = 2uₖ – 1 = 2(2ᵏ + 1) – 1 = 2·2ᵏ + 2 – 1 = 2ᵏ⁺¹ + 1, which is exactly the formula for n = k + 1.

基础情形 n = 1：u₁ = 2¹ + 1 = 3，与给定值相符。
归纳假设：假设对某个 k ≥ 1 有 uₖ = 2ᵏ + 1。
归纳递推：uₖ₊₁ = 2uₖ – 1 = 2(2ᵏ + 1) – 1 = 2·2ᵏ + 2 – 1 = 2ᵏ⁺¹ + 1，恰好是 n = k + 1 时的公式。

---

6. Strong Induction | 强归纳法

Sometimes the truth of P(k + 1) depends not only on P(k) but on several earlier statements. In such cases you use strong induction: assume that P(1), P(2), …, P(k) are all true, and then deduce P(k + 1). The base case may also require checking more than one initial value. This technique is particularly useful for sequences defined by higher‑order recurrences, like Fibonacci‑style relations.

有时 P(k + 1) 的真实性不仅依赖于 P(k)，还依赖于前面若干个命题。这时你可以使用强归纳法：假设 P(1), P(2), …, P(k) 全部成立，然后推出 P(k + 1)。基础情形可能也需要检验不止一个初始值。对于像斐波那契型递推这样的高阶递推关系，这一技巧尤为有用。

Example structure: To prove that every term of the Fibonacci sequence Fₙ (with F₁ = 1, F₂ = 1) satisfies some property, check n = 1 and n = 2, then assume the property holds for all indices up to k, and use the recurrence Fₖ₊₁ = Fₖ + Fₖ₋₁ to prove it for k + 1.

结构示例：要证明斐波那契数列 Fₙ（满足 F₁ = 1，F₂ = 1）的每一项都具有某种性质，需先检验 n = 1 和 n = 2，然后假设该性质对所有不超过 k 的指标都成立，再利用递推关系 Fₖ₊₁ = Fₖ + Fₖ₋₁ 证明它对 k + 1 也成立。

---

7. Common Pitfalls and How to Avoid Them | 常见陷阱与避免方法

Even well‑prepared students make predictable errors in induction proofs. One frequent mistake is forgetting to verify the base case thoroughly – if the base case fails, the entire argument collapses. Another is assuming P(k) without stating it explicitly as the induction hypothesis. In the induction step, some candidates manipulate both sides of the target equation simultaneously, which can mask logical gaps. Always work from one side, apply the hypothesis, and simplify to the other side.

即使是准备充分的学生也会在归纳证明中犯一些可预见的错误。一个常见的错误就是忘记充分验证基础情形——如果基础情形不成立，整个论证都会崩塌。另一个错误是使用了 P(k) 却没有明确声明那是归纳假设。在归纳递推步骤中，有些考生会同时从目标等式的两边入手进行变形，这样可能会掩盖逻辑漏洞。正确的做法是始终从一边出发，代入归纳假设，再化简到另一边。

For inequalities, a typical pitfall is multiplying or dividing by a quantity whose sign could be negative without checking it. Always verify that k is large enough to keep every operation valid. With divisibility proofs, students sometimes introduce expressions that are not obviously multiples of the divisor; always break the k + 1 case into a sum of a multiple of f(k) and a clearly identifiable multiple of the divisor.

处理不等式时，一个典型陷阱是在未检查符号的情况下就乘以或除以一个可能为负数的量。要始终确保 k 足够大，使得每一步运算都是合法的。在整除性证明中，学生有时会引入一些并非明显是除数倍数的表达式；务必把 k + 1 情形拆分为 f(k) 的倍数与一个明显可被整除的项之和。

---

8. Structuring Your Answer for Maximum Marks | 为获取满分而组织答案

IB examiners expect your induction proof to be clearly signposted. Use phrases like ‘Base case (n = 1)’, ‘Induction hypothesis: Assume true for n = k’, and ‘Induction step: Prove true for n = k + 1’. Write the statement P(n) explicitly at the beginning. When you use the induction hypothesis, show exactly where you substitute. For algebraic steps, a short justification – such as ‘by the induction hypothesis’ or ‘factorising’ – can make your reasoning easy to follow and earn method marks even if a slip occurs.

IB 阅卷老师希望你的归纳证明有清晰的标志。请使用诸如“基础情形（n = 1）”“归纳假设：假设 n = k 时命题成立”“归纳递推：证明 n = k + 1 时命题成立”这样的短语。在证明开头明确写出 P(n) 是什么。当使用归纳假设时，要明确标出你是如何代入的。对于代数步骤，简短的说明——如“根据归纳假设”或“因式分解”——可以让你的推理易于理解，即使出现小失误也能拿到方法分。

---

9. Practice Questions Mirroring Exercise 1G | 模拟练习 1G 的练习题

Here is a selection of problems that closely resemble those found in typical IB Exercise 1G sets. Try to prove each statement using a fully written‑out induction argument.

以下选编了一些与典型 IB 练习 1G 题目极为相似的习题。请尝试用完整写出的归纳法论证来证明每一个命题。

• Prove that 1 + 3 + 5 + … + (2n – 1) = n². / 证明 1 + 3 + 5 + … + (2n – 1) = n²。
• Show that 7ⁿ – 1 is divisible by 6 for all n ∈ ℤ⁺. / 证明对所有正整数 n，7ⁿ – 1 能被 6 整除。
• Prove that n! > 2ⁿ for all n ≥ 4. / 证明对所有 n ≥ 4，有 n! > 2ⁿ。
• A sequence is given by a₁ = 2 and aₙ₊₁ = √(2aₙ + 3). Prove by induction that aₙ < 3 for all n. / 数列满足 a₁ = 2，aₙ₊₁ = √(2aₙ + 3)。用归纳法证明对所有 n 有 aₙ < 3。
• Prove that the sum of the cubes of the first n positive integers equals [n(n+1)/2]². / 证明前 n 个正整数的立方和等于 [n(n+1)/2]²。

---

10. Final Tips and Review Strategy | 最后提示与复习策略

Mastering Exercise 1G is less about memorising dozens of proofs and more about internalising the pattern: state P(n), verify the base case, assume P(k), connect P(k) to P(k + 1) using algebraic manipulation or logical deduction, and then write a concluding sentence. As you practise, vary the types of statements – summation, divisibility, inequality, recurrence – so that no exam question takes you by surprise. Time yourself while writing full solutions, because a carefully structured proof can be completed in under ten minutes and will earn you substantial marks.

掌握练习 1G 不在于死记硬背几十个证明，而在于内化其套路：写出 P(n)，验证基础情形，假设 P(k)，通过代数变形或逻辑推理将 P(k) 与 P(k + 1) 连接起来，最后写上结论句。练习时要变换命题的类型——求和、整除、不等式、递推——这样考场上就不会遇到意料之外的题目。在写完整解答时要计时，因为一个结构清晰的证明完全可以在十分钟内完成，并能为你赢得可观的分数。

In your revision, create a summary card listing the three compulsory steps and the most common algebraic tricks: factorising, adding and subtracting the same quantity, using the induction hypothesis early, and checking the direction of inequalities carefully. This compact reference will be invaluable in the days before the exam.

复习时可以做一张总结卡片，列出三步必写项和最常见的代数技巧：因式分解、同时加减同一个量、尽早使用归纳假设、仔细检查不等号方向。这张简明的参考卡片将在考试前几天发挥巨大作用。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）