Binomial Theorem and Its Expansions | 二项式定理及其展开式

📚 Binomial Theorem and Its Expansions | 二项式定理及其展开式

The Binomial Theorem is one of the most fundamental tools in algebra and calculus. It provides a systematic method for expanding expressions of the form (a + b)ⁿ without carrying out repeated multiplication. This theorem appears in nearly every major examination board’s syllabus, from GCSE Further Mathematics to A-Level Pure Mathematics and IB HL.

二项式定理是代数和微积分中最基础的工具之一。它为展开形如 (a + b)ⁿ 的表达式提供了一种系统方法，无需进行反复乘法运算。该定理几乎出现在各大考试局的教学大纲中，从 GCSE 进阶数学到 A-Level 纯数学以及 IB 高级课程均有涉及。

1. The General Statement | 一般表述

For any positive integer n, the binomial expansion of (a + b)ⁿ can be written as a sum of terms involving binomial coefficients. The standard form is:

对于任意正整数 n，(a + b)ⁿ 的二项式展开可以写成包含二项式系数的和式。其标准形式为：

(a + b)ⁿ = ∑ₖ₌₀ⁿ C(n, k) aⁿ⁻ᵏ bᵏ

where C(n, k), also written as ⁿCₖ or (ₖⁿ), represents the number of ways to choose k items from n items. Each term in the expansion has the form C(n, k) aⁿ⁻ᵏ bᵏ, and there are exactly n + 1 terms in total.

其中 C(n, k)（也写作 ⁿCₖ 或 (ₖⁿ)）表示从 n 个元素中选取 k 个元素的方法数。展开式中的每一项都具有 C(n, k) aⁿ⁻ᵏ bᵏ 的形式，总计恰好有 n + 1 项。

2. Factorials and Binomial Coefficients | 阶乘与二项式系数

The binomial coefficients are defined using factorials. Recall that n! (read as “n factorial”) is the product of all positive integers from 1 to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.

二项式系数通过阶乘来定义。回顾一下，n!（读作“n 的阶乘”）是从 1 到 n 的所有正整数的乘积。例如，5! = 5 × 4 × 3 × 2 × 1 = 120。

C(n, k) = n! / [k! (n − k)!]

This formula allows us to compute any binomial coefficient directly. For instance, C(6, 2) = 6! / (2! × 4!) = 720 / (2 × 24) = 15. Notice that C(n, 0) = C(n, n) = 1 for every positive integer n.

这个公式允许我们直接计算任何二项式系数。例如，C(6, 2) = 6! / (2! × 4!) = 720 / (2 × 24) = 15。注意，对于每个正整数 n，C(n, 0) = C(n, n) = 1。

3. Pascal’s Triangle | 帕斯卡三角形

Pascal’s triangle is a triangular array of numbers in which each entry is the sum of the two entries directly above it. The rows correspond to increasing values of n, and the entries in row n are exactly the coefficients C(n, 0), C(n, 1), …, C(n, n).

帕斯卡三角形是一个数字组成的三角形阵列，其中每个数字是它正上方两个数字之和。各行对应逐渐增大的 n 值，第 n 行的数字恰好是系数 C(n, 0), C(n, 1), …, C(n, n)。

The first few rows are:

前几行如下：

Row 0: 1

第 0 行：1
Row 1: 1 1

第 1 行：1 1
Row 2: 1 2 1

第 2 行：1 2 1
Row 3: 1 3 3 1

第 3 行：1 3 3 1
Row 4: 1 4 6 4 1

第 4 行：1 4 6 4 1

For example, (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴. The coefficients 1, 4, 6, 4, 1 match row 4 exactly.

例如，(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴。系数 1、4、6、4、1 与第 4 行完全吻合。

4. Properties of the Expansion | 展开式的性质

The binomial expansion possesses several important structural properties. First, the number of terms is always n + 1 when n is a positive integer. Second, the powers of a decrease from n to 0, while the powers of b increase from 0 to n.

二项式展开具有几个重要的结构性质。首先，当 n 为正整数时，项数总是 n + 1。其次，a 的幂从 n 递减到 0，而 b 的幂从 0 递增到 n。

Third, the coefficients are symmetric: the coefficient of the k-th term from the beginning equals the coefficient of the k-th term from the end. This symmetry reflects the identity C(n, k) = C(n, n − k). Fourth, the sum of all coefficients in the expansion of (a + b)ⁿ equals 2ⁿ, which is obtained by setting a = b = 1.

第三，系数是对称的：从开头数第 k 项的系数等于从末尾数第 k 项的系数。这种对称性反映了恒等式 C(n, k) = C(n, n − k)。第四，在 (a + b)ⁿ 的展开式中所有系数之和等于 2ⁿ，这可以通过令 a = b = 1 得到。

5. Finding a Specific Term | 求指定项

A common examination question asks for the coefficient of a particular power of x in the expansion of a binomial expression. The general term in the expansion of (a + b)ⁿ is Tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ, which is the (k + 1)-th term.

一个常见的考试问题要求求二项式展开式中 x 的特定幂的系数。(a + b)ⁿ 的展开式中的通项是 Tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ，即第 (k + 1) 项。

For example, to find the coefficient of x³ in (2 + 3x)⁵, we set k = 3: the term is C(5, 3) × 2² × (3x)³ = 10 × 4 × 27x³ = 1080x³. The required coefficient is therefore 1080.

例如，要求 (2 + 3x)⁵ 中 x³ 的系数，我们令 k = 3：该项为 C(5, 3) × 2² × (3x)³ = 10 × 4 × 27x³ = 1080x³。因此所需系数为 1080。

6. The Binomial Theorem for Negative and Fractional Indices | 负指数与分数指数的二项式定理

When n is not a positive integer, the binomial expansion becomes an infinite series. This extension is valid only when |x| < 1 for the expression (1 + x)ⁿ. The formula is:

当 n 不是正整数时，二项式展开变成无穷级数。这种推广仅当 |x| < 1 时对表达式 (1 + x)ⁿ 有效。其公式为：

(1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …

For example, (1 + x)⁻¹ = 1 − x + x² − x³ + … for |x| < 1. Similarly, √(1 + x) = (1 + x)^(1/2) = 1 + (1/2)x − (1/8)x² + (1/16)x³ − ... for |x| < 1.

例如，当 |x| < 1 时，(1 + x)⁻¹ = 1 − x + x² − x³ + ...。类似地，当 |x| < 1 时，√(1 + x) = (1 + x)^(1/2) = 1 + (1/2)x − (1/8)x² + (1/16)x³ − ...。

7. Validity Conditions and Convergence | 有效条件与收敛性

Understanding when a binomial series converges is crucial. For positive integer n, the expansion terminates after n + 1 terms and is valid for all real x. For negative or fractional n, the series does not terminate and converges only under the condition |x| < 1.

理解二项级数何时收敛至关重要。对于正整数 n，展开式在 n + 1 项后终止，并且对所有实数 x 都有效。对于负指数或分数指数 n，级数不会终止，并且仅在 |x| < 1 的条件下收敛。

When dealing with (a + bx)ⁿ where n is not a positive integer, we first factor out aⁿ to rewrite it as aⁿ(1 + (b/a)x)ⁿ. The series then converges when |(b/a)x| < 1, i.e., |x| < |a/b|. Examiners frequently test both the algebraic manipulation and the statement of the validity condition.

当处理 n 不是正整数的 (a + bx)ⁿ 时，我们首先提取 aⁿ，将其改写为 aⁿ(1 + (b/a)x)ⁿ。然后当 |(b/a)x| < 1 时，即 |x| < |a/b| 时，级数收敛。考官经常同时考察代数变形和有效条件的陈述。

8. Approximations Using the Binomial Theorem | 利用二项式定理进行近似计算

One of the most practical applications of the binomial theorem is in numerical approximation. By truncating the infinite series after a few terms, we can obtain approximate values of roots and powers with remarkable accuracy, provided x is small.

二项式定理最实际的应用之一是数值近似。通过在几项之后截断无穷级数，只要 x 很小，我们就能以惊人的精度获得根和幂的近似值。

For instance, to approximate √1.04, we write √(1 + 0.04) = (1 + 0.04)^(1/2). Using the first three terms: 1 + (1/2)(0.04) − (1/8)(0.04)² = 1 + 0.02 − 0.0002 = 1.0198. The actual value is approximately 1.01980, showing excellent agreement.

例如，为了近似计算 √1.04，我们写成 √(1 + 0.04) = (1 + 0.04)^(1/2)。使用前三项：1 + (1/2)(0.04) − (1/8)(0.04)² = 1 + 0.02 − 0.0002 = 1.0198。实际值约为 1.01980，表明两者高度吻合。

9. Connection with Combinatorics | 与组合数学的联系

The binomial coefficients C(n, k) count the number of ways to select k objects from a set of n distinct objects. This combinatorial interpretation explains why the coefficients appear in the expansion: each term aⁿ⁻ᵏ bᵏ arises from choosing b from exactly k of the n factors (a + b).

二项式系数 C(n, k) 计算从 n 个不同对象中选取 k 个对象的方法数。这种组合解释说明了为什么这些系数出现在展开式中：每一项 aⁿ⁻ᵏ bᵏ 都源于从 n 个因子 (a + b) 中恰好选择 k 个因子取 b。

This connection leads to many useful identities. For example, the sum ∑ₖ₌₀ⁿ C(n, k)² = C(2n, n) follows from a combinatorial argument about choosing n objects from 2n objects. Such identities occasionally appear in extension questions on competitive examinations.

这种联系引出了许多有用的恒等式。例如，求和 ∑ₖ₌₀ⁿ C(n, k)² = C(2n, n) 可以通过一个关于从 2n 个对象中选取 n 个对象的组合论证得出。这类恒等式偶尔出现在竞赛型考试的拓展题中。

10. Worked Examples | 典型例题

Let us work through some classic questions step by step. First, expand (1 + 2x)⁶ up to the term in x³.

让我们逐步演算一些经典问题。首先，展开 (1 + 2x)⁶ 至 x³ 项。

(1 + 2x)⁶ = 1 + 6(2x) + 15(2x)² + 20(2x)³ + …

This simplifies to 1 + 12x + 60x² + 160x³ + …. The coefficient of x² is 60 and the coefficient of x³ is 160.

这化简为 1 + 12x + 60x² + 160x³ + …。x² 的系数是 60，x³ 的系数是 160。

Second, find the constant term in the expansion of (2x + 1/x)⁴. The general term is C(4, k)(2x)⁴⁻ᵏ(1/x)ᵏ = C(4, k)2⁴⁻ᵏx⁴⁻²ᵏ. Setting 4 − 2k = 0 gives k = 2. The constant term is therefore C(4, 2) × 2² = 6 × 4 = 24.

其次，求 (2x + 1/x)⁴ 展开式中的常数项。通项为 C(4, k)(2x)⁴⁻ᵏ(1/x)ᵏ = C(4, k)2⁴⁻ᵏx⁴⁻²ᵏ。令 4 − 2k = 0 得 k = 2。因此常数项为 C(4, 2) × 2² = 6 × 4 = 24。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students frequently make errors in three areas: forgetting the binomial coefficients, mishandling the powers of negative terms, and ignoring the validity condition for non-integer n. Always write out the general term formula first and double-check each exponent.

学生常在三个地方出错：忘记二项式系数、处理负项的幂时出错，以及忽略非整数 n 的有效条件。务必先写出通项公式，并仔细检查每一个指数。

When the second term inside the bracket has a coefficient, remember to raise that entire term to the appropriate power. For example, in (3 + 2x)⁵, the term in x² is C(5, 2) × 3³ × (2x)² = 10 × 27 × 4x² = 1080x², not 10 × 27 × 2x².

当括号内第二项带有系数时，记住要将整个项提升到相应的幂。例如，在 (3 + 2x)⁵ 中，x² 项是 C(5, 2) × 3³ × (2x)² = 10 × 27 × 4x² = 1080x²，而不是 10 × 27 × 2x²。

12. Summary and Revision Checklist | 总结与复习清单

The binomial theorem is a powerful and versatile result that every mathematics student must master. It connects algebra, combinatorics, and calculus, and it appears in a wide range of examination contexts. Memorize the standard formula, understand the conditions for infinite series, and practise finding specific terms efficiently.

二项式定理是一个强大而通用的结论，每位数学学生都必须掌握。它连接了代数、组合数学和微积分，并广泛出现在各种考试情境中。牢记标准公式，理解无穷级数的条件，并练习高效地求指定项。

Memorise C(n, k) = n! / [k!(n − k)!] and the general term formula.

牢记 C(n, k) = n! / [k!(n − k)!] 以及通项公式。
For positive integer n, the expansion has n + 1 terms and no restrictions on x.

对于正整数 n，展开式有 n + 1 项，对 x 没有限制。
For negative or fractional n, the series is infinite and requires |x| < 1.

对于负指数或分数指数 n，级数是无穷的，需要 |x| < 1。
Always factorise expressions into the form (1 + u)ⁿ before applying the infinite series formula.

在应用无穷级数公式之前，始终将表达式化为 (1 + u)ⁿ 的形式。
Check the validity condition whenever the index is not a positive integer.

当指数不是正整数时，务必检查有效条件。

With consistent practice and attention to detail, the binomial theorem becomes one of the most reliable tools in your mathematical toolkit.

通过持续练习和对细节的关注，二项式定理将成为你数学工具箱中最可靠的工具之一。

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Binomial Theorem and Its Expansions | 二项式定理及其展开式