Summation of a Series by the Method of Induction | 归纳法求和级数

📚 Summation of a Series by the Method of Induction | 归纳法求和级数

Mathematical induction is one of the most elegant and powerful techniques in pure mathematics. For A-Level students, it is essential for proving formulas involving the sum of a series, especially those that depend on a positive integer n. This article explains the method step by step, with fully worked examples and common pitfalls to avoid.

数学归纳法是最优雅且强有力的数学证明技巧之一。对于 A-Level 学生来说，它在证明依赖于正整数 n 的级数求和公式时至关重要。本文将逐步解释这一方法，并给出完整的例题与常见的易错点。

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1. What is Mathematical Induction? | 什么是数学归纳法？

Mathematical induction is a proof technique used to establish that a statement is true for all positive integers. It is particularly useful for summation formulas, such as proving that a given expression for the sum of the first n terms is correct.

数学归纳法是一种用于证明命题对所有正整数都成立的证明技巧。它特别适用于求和公式，例如证明前 n 项和式的某个表达式是正确的。

Think of a row of dominoes. If you can knock over the first domino, and each falling domino knocks over the next one, then all dominoes will fall. Induction works in exactly the same way.

可以想象一排多米诺骨牌：如果你能推倒第一块骨牌，而且每一块倒下的骨牌都能推倒下一块，那么所有骨牌都会倒下。归纳法的原理正是如此。

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2. The Principle Behind Induction | 归纳法的基本原理

To prove a statement P(n) for all positive integers n, we need to verify two things. First, P(1) is true. Second, if P(k) is true for some positive integer k, then P(k+1) must also be true. If both conditions hold, we conclude that P(n) is true for every positive integer n.

为了证明命题 P(n) 对所有正整数 n 成立，我们需要验证两件事。第一，P(1) 为真。第二，若 P(k) 对某个正整数 k 为真，则 P(k+1) 也必定为真。若这两个条件都满足，我们就得出结论：P(n) 对每个正整数 n 都成立。

This second condition is called the inductive step. It connects one truth to the next, creating an infinite chain of truths. The first condition, the base case, starts the chain.

第二个条件称为归纳步骤。它把一个真命题连接到下一个真命题，形成一条无穷的真命题链条。第一个条件称为基础情形，它启动这条链条。

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3. The Three Key Steps | 三个关键步骤

When using induction to prove a summation formula, always follow these three steps:

用归纳法证明求和公式时，始终遵循以下三个步骤：

• Base case: Verify the statement is true for the smallest value, usually n = 1.

  基础情形：验证命题对最小取值成立，通常为 n = 1。

• Inductive hypothesis: Assume the statement is true for some arbitrary positive integer n = k. Do not prove anything here.

  归纳假设：假设命题对某个任意正整数 n = k 成立。此步骤无需证明任何东西。

• Inductive step: Using the assumption for n = k, prove that the statement is true for n = k+1. Then state the conclusion.

  归纳步骤：利用 n = k 的假设，证明命题对 n = k+1 成立。然后写出结论。

The inductive step usually involves algebraic manipulation: write the sum for k+1 depending on the sum for k, substitute the hypothesis, and simplify.

归纳步骤通常涉及代数变形：写出 k+1 的和式并依赖 k 的和式，代入假设并化简。

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4. Example 1: Sum of First n Positive Integers | 示例1：前 n 个正整数之和

We want to prove that for every positive integer n,

我们想证明：对每个正整数 n，

1 + 2 + 3 + … + n = n(n+1)/2

Base case: For n he left-hand side is 1, and the right-hand side is 1(1+1)/2 = 1. So the statement is true.

基础情形：当 n = 1 时，左边为 1，右边为 1(1+1)/2 = 1。因此命题成立。

Inductive hypothesis: Assume true for n = k:

归纳假设：假设对 n = k 成立：

1 + 2 + … + k = k(k+1)/2

Inductive step: Consider the sum up to k+1. Separate the last term:

归纳步骤：考虑加到 k+1 的和式。分离出最后一项：

1 + 2 + … + k + (k+1) = k(k+1)/2 + (k+1)

Put the right-hand side over a common denominator and factor:

右边通分并因式分解：

= (k+1)(k/2 + 1) = (k+1)(k+2)/2

This is exactly the formula for n = k+1. By induction, the formula holds for all positive integers n.

这正是 n = k+1 时的公式。由归纳法，公式对所有正整数 n 成立。

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5. Example 2: Sum of Squares | 示例2：平方和

Now let us prove a more challenging formula:

现在我们来证明一个更有挑战性的公式：

1² + 2² + 3² + … + n² = n(n+1)(2n+1)/6

Base case: For n = 1, left side is 1² = 1, right side is 1(1+1)(2×1+1)/6 = 1(2)(3)/6 = 1. True.

基础情形：当 n = 1 时，左边为 1² = 1，右边为 1(1+1)(2×1+1)/6 = 1(2)(3)/6 = 1。成立。

Inductive hypothesis: Assume true for n = k:

归纳假设：假设对 n = k 成立：

1² + 2² + … + k² = k(k+1)(2k+1)/6

Inductive step: Add (k+1)² to both sides:

归纳步骤：两边同时加上 (k+1)²：

1² + … + k² + (k+1)² = k(k+1)(2k+1)/6 + (k+1)²

Factor out (k+1)/6:

提取公因子 (k+1)/6：

= (k+1)/6 [ k(2k+1) + 6(k+1) ]

Simplify the bracket: 2k² + k + 6k + 6 = 2k² + 7k + 6 = (k+2)(2k+3).

化简括号内：2k² + k + 6k + 6 = 2k² + 7k + 6 = (k+2)(2k+3)。

= (k+1)(k+2)(2k+3)/6

Notice that (k+1)(k+2)(2k+3)/6 is exactly the formula with n replaced by k+1. Hence the statement is true for n = k+1, and by induction it is true for all positive integers.

注意 (k+1)(k+2)(2k+3)/6 正是把原公式中的 n 换成 k+1 的结果。因此命题对 n = k+1 成立，由归纳法它对所有正整数成立。

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6. Example 3: Sum of Cubes | 示例3：立方和

Another classic result is the sum of cubes:

另一个经典结论是立方和公式：

1³ + 2³ + 3³ + … + n³ = [n(n+1)/2]²

Base case: For n = 1, left side = 1³ = 1, right side = [1(2)/2]² = 1² = 1. True.

基础情形：当 n = 1 时，左边 = 1³ = 1，右边 = [1(2)/2]² = 1² = 1。成立。

Inductive hypothesis: Assume true for n = k:

归纳假设：假设对 n = k 成立：

1³ + 2³ + … + k³ = [k(k+1)/2]²

Inductive step: Add (k+1)³:

归纳步骤：加上 (k+1)³：

= [k(k+1)/2]² + (k+1)³

Factor out (k+1)²/4:

提取公因子 (k+1)²/4：

= (k+1)²/4 [ k² + 4(k+1) ]

Simplify: k² + 4k + 4 = (k+2)². Therefore:

化简：k² + 4k + 4 = (k+2)²。因此：

= (k+1)²(k+2)²/4 = [(k+1)(k+2)/2]²

This is exactly the formula for n = k+1. Thus the formula is proven.

这正是 n = k+1 时的公式。于是公式得证。

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7. Working with Sigma Notation | 西格玛记号的运用

Sigma notation provides a compact way to write a series. For example, the sum of squares can be written as Σ_r=1ⁿ r². In induction proofs, we often write the left-hand side using sigma notation.

西格玛记号提供了一种简洁的级数表示方式。例如，平方和可写作 Σ_r=1ⁿ r²。在归纳证明中，我们常常用西格玛记号表示左边。

When performing the inductive step, we split the sum:

在归纳步骤中，我们分解和式：

Σ_r=1^k+1 f(r) = Σ_r=1^k f(r) + f(k+1)

Then we substitute the inductive hypothesis for the first part and simplify.

然后我们代入第一部分的归纳假设并化简。

Remember that the index r is a dummy variable; it does not matter what letter you use. This makes sigma notation very flexible in algebraic manipulations.

记住下标 r 是一个哑变量，使用什么字母并不重要。这使得西格玛记号在代数运算中非常灵活。

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8. Common Pitfalls | 常见错误

Many students make avoidable mistakes when using induction. Here are the most frequent ones:

许多学生会在使用归纳法时犯一些可以避免的错误。以下是最常见的几种:

• Missing the base case: Without verifying n=1, the chain never starts.

  遗漏基础情形：不验证 n=1，链条便无法启动。

• Assuming what you need to prove: In the inductive step, you cannot start with the formula for k+1. You must derive it from the formula for k.

  假设了需要证明的东西：在归纳步骤中，你不能从一开始就假设 k+1 的公式成立。你必须从 k 的公式推导出来。

• Algebraic errors: When factoring or combining fractions, take your time. A single small error can invalidate the proof.

  代数错误：在因式分解或合并分数时，要耐心细致。一个很小的错误都可能导致证明无效。

• Ending too early: A proof by induction is complete only after you explicitly state, “By the principle of mathematical induction, the statement is true for all positive integers n.”

  过早结束：归纳证明只有在明确写出“根据数学归纳法原理，该命题对所有正整数 n 成立”之后才算完整。

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9. Practice Problem | 练习

Try to prove the following formula on your own using induction:

请尝试用归纳法自行证明以下公式：

1 × 2 + 2 × 3 + 3 × 4 + … + n(n+1) = n(n+1)(n+2)/3

Hint: In the inductive step, add (k+1)(k+2) to both sides and factorise.

提示：在归纳步骤中，两边同时加上 (k+1)(k+2)，然后因式分解。

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10. Conclusion | 结论

Induction is a reliable and satisfying method for proving summation formulas. With a solid understanding of the base case, the inductive hypothesis, and the inductive step, you can handle any A-Level question on this topic. The key is to practise writing proofs clearly and logically.

归纳法是证明求和公式的可靠而令人满意的方法。只要牢固掌握基础情形、归纳假设和归纳步骤，你就能应对 A-Level 中关于这一主题的任何问题。关键在于练习清晰而有逻辑地书写证明。

Keep in mind that the structure of the proof matters as much as the algebra. Examiners award method marks for the correct structure, even if you make a small arithmetic slip. So always lay out your proof with clear labels: base case, hypothesis, inductive step, and conclusion.

记住，证明的结构和代数同样重要。即使计算有小的失误，考官也会为正确的结构给出方法分。因此，始终用清晰的标签组织你的证明：基础情形、归纳假设、归纳步骤和结论。

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