📚 IB Mathematics: Step-by-Step Guide to Proof by Mathematical Induction | IB数学：数学归纳法解题步骤
Mathematical induction is a powerful proof technique used to show that a statement is true for all natural numbers, usually positive integers. It is a cornerstone of IB Mathematics Analysis and Approaches (AA) and is frequently tested in Paper 1 and Paper 2.
数学归纳法是一种强有力的证明技巧，用于证明某个命题对所有自然数（通常是正整数）都成立。它是IB数学分析与方法（AA）的核心内容，常在Paper 1和Paper 2中出现。
In essence, induction allows us to prove infinitely many cases by proving just two things: a starting case and a logical chain from one case to the next.
本质上，归纳法让我们只需证明两件事，就能证明无穷多种情形：一个起点，以及从前一个情形过渡到下一个情形的逻辑链条。
1. The Principle: Dominoes and Infinite Ladders | 原理：多米诺骨牌与无穷阶梯
Imagine an infinite row of dominoes. If you push the first domino, and every domino is arranged so that when one falls, the next one falls, then all dominoes will fall.
想象一排无穷长的多米诺骨牌。如果你推倒第一块，而且每一块骨牌倒下时都能碰倒下一块，那么所有骨牌都会倒下。
Mathematically, we must prove two conditions:
数学上，我们需要证明两个条件：
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The first domino falls: the statement is true for the first value, usually n = 1.
第一块骨牌倒下：命题对第一个值成立，通常 n = 1。
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in reaction: whenever the statement is true for n = k, it must also be true for n = k + 1.
连锁反应：每当命题在 n = k 时成立，它在 n = k + 1 时也必然成立。
Once both conditions hold, we conclude the statement is true for every positive integer n.
一旦两个条件都成立，我们就可以断定命题对所有正整数 n 都成立。
2. Knowing the Four Parts of an Induction Proof | 归纳法证明的四个部分
A complete induction proof in IB mathematics must include four clear parts.
在IB数学中，一个完整的归纳法证明必须包含四个清晰的部分。
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Base case: verify the statement for the smallest allowed value.
基础情形：验证命题对最小允许值成立。
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Inductive hypothesis: assume the statement is true for some arbitrary integer n = k.
归纳假设：假设命题对某个任意整数 n = k 成立。
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Inductive step: using the hypothesis, prove the statement for n = k + 1.
归纳递推：利用假设证明命题在 n = k + 1 时成立。
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Conclusion: state clearly that by the principle of mathematical induction, the statement is true for all allowed n.
结论：明确指出根据数学归纳法原理，命题对所有符合条件的 n 都成立。
3. Step 1: Prove the Base Case | 第一步：证明基础情形
, but sometimes the question specifies a different starting value, such as n = 0 or n = 4.
基础情形是证明的根基。对于大多数IB题目，这意味着检查 n = 1，但有时题目会指定不同的起始值，例如 n = 0 或 n = 4。
To prove the base case, substitute the starting value into both sides of the statement and show that they are equal, or that the required property holds.
证明基础情形时，将起始值代入命题的左右两边，证明它们相等，或说明所需性质成立。
Never skip this step. Without a valid base case, the whole induction collapses.
绝不能跳过这一步。没有有效的基础情形，整个归纳法证明就会崩塌。
4. Step 2: State the Inductive Hypothesis | 第二步：写出归纳假设
some arbitrary positive integer n = k. This assumption is called the inductive hypothesis.
假设命题对某个任意正整数 n = k 成立。这个假设称为归纳假设。
It is important to write this step explicitly in your exam solution. Many IB mark schemes award a mark just for stating the hypothesis correctly.
在考试解答中明确写出这一步非常重要。许多IB评分标准仅仅因为正确写出假设就给分。
Use wording such as: “Assume that the statement holds for n = k.”
可以使用这样的措辞：“假设命题在 n = k 时成立。”
5. Step 3: Perform the Inductive Step | 第三步：执行归纳递推
hypothesis, we manipulate the expression for n = k to reach the expression for n = k + 1.
这是证明的核心。从归纳假设出发，我们通过变形 n = k 时的表达式，推出 n = k + 1 时的表达式。
Common strategies include:
常用策略包括：
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Add the next term to both sides of a summation formula.
在求和公式两边同时加上下一项。
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Factor or rewrite an algebraic expression to use the hypothesis.
因式分解或重新改写代数表达式，以便利用归纳假设。
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Use an inequality chain for proofs involving inequalities.
对于不等式证明，使用不等式链。
Always show the key algebra clearly; do not simply state “by induction”.
务必清楚展示关键的代数步骤，不要只写一句“由归纳法可得”。
6. Step 4: Write the Conclusion | 第四步：写出结论
After the inductive step, you must formally conclude the proof.
完成归纳递推后，你必须正式地总结证明。
A standard conclusion is:
标准的结论表述是：
the principle of mathematical induction the statement is true for all positive integers n.
由于命题在 n = 1 时成立，且假设它在 n = k 时成立，则它在 n = k + 1 时也成立，因此根据数学归纳法原理，命题对所有正整数 n 都成立。
In some questions, replace “positive integers” with the required range, such as “integers n ≥ 4”.
在某些题目中，把“正整数”替换成题目要求的范围，例如“整数 n ≥ 4”。
7. Worked Example 1: Sum of the First n Natural Numbers | 示例1：前n个自然数求和
Problem: Prove that for all positive integers n,
题目：证明对所有正整数 n，
1 + 2 + 3 + ⋯ + n = n(n + 1) / 2
Solution:
解答：
Base case: 1, the left-hand side is 1, and the right-hand side is 1(2)/2 = 1. So the statement holds.
基础情形：当 n = 1 时，左边 = 1，右边 = 1(2)/2 = 1。因此命题成立。
Inductive hypothesis: Assume the formula holds for n = k, that is,
归纳假设：假设公式在 n = k 时成立，即
1 + 2 + ⋯ + k = k(k + 1
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