📚 Binomial Expansion for GCSE Maths | GCSE 数学：二项式展开 考点精讲
1. Introduction to Binomial Expansion | 二项式展开简介
Binomial expansion is a fundamental algebraic technique that allows us to expand expressions raised to a power, such as (a + b)n, without having to multiply the brackets out term by term. This topic appears frequently in GCSE Higher Tier examinations and provides a crucial foundation for A-Level mathematics. Mastering binomial expansion saves time, reduces errors, and equips you with a powerful tool for tackling polynomial problems.
二项式展开是一种基础的代数技巧，它使我们能够展开带幂次的表达式，例如 (a + b)n，而无需逐项将括号相乘。这一主题经常出现在 GCSE 高阶考试中，并为 A-Level 数学奠定了重要基础。掌握二项式展开能够节省时间、减少错误，并为你提供处理多项式问题的强大工具。
2. What is a Binomial? | 什么是二项式？
A binomial is an algebraic expression that contains exactly two terms connected by either a plus or a minus sign. Examples include (x + 3), (2a − 5), and (p + q). The word ‘binomial’ comes from ‘bi-‘ meaning two, and ‘-nomial’ meaning term. When we talk about binomial expansion, we are referring to the process of raising a binomial to a positive integer power and writing out the result as a sum of individual terms.
二项式是恰好包含两个项的代数表达式，这两个项由加号或减号连接。例如 (x + 3)、(2a − 5) 和 (p + q)。”binomial” 一词中，”bi-” 表示”两个”，”-nomial” 表示”项”。当我们谈论二项式展开时，我们指的是将一个二项式进行正整数次幂运算，并将结果写成各项之和的过程。
The general form of a binomial raised to the power n is written as (a + b)n, where a and b can be numbers, variables, or a combination of both, and n is a positive integer. At GCSE level, n is typically 2, 3, 4, or occasionally 5, and you may also encounter expressions with coefficients, such as (2x + 3y)3 or (x − 4)4.
二项式的 n 次幂的一般形式写作 (a + b)n，其中 a 和 b 可以是数字、变量或两者的组合，n 为正整数。在 GCSE 级别，n 通常为 2、3、4，偶尔为 5，你还可能遇到带有系数的表达式，例如 (2x + 3y)3 或 (x − 4)4。
3. Expanding (a + b)2 | 展开 (a + b)2
Let us begin with the simplest non-trivial case: expanding (a + b)2. Using the distributive property (also known as the FOIL method for two binomials), we multiply (a + b) by itself. The step-by-step working is: (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2. Since ab and ba are identical, we combine them to obtain a2 + 2ab + b2. This result should be memorised as it forms the basis for all further expansions.
让我们从最简单的非平凡情况开始：展开 (a + b)2。利用分配律（对于两个二项式相乘也称为 FOIL 方法），我们将 (a + b) 乘以自身。逐步计算如下：(a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2。由于 ab 和 ba 完全相同，我们将它们合并得到 a2 + 2ab + b2。这个结果应当牢记，因为它是所有进一步展开的基础。
Notice the pattern in the coefficients: 1, 2, 1. The powers of a decrease from 2 down to 0, while the powers of b increase from 0 up to 2. This symmetrical pattern is not accidental — it is the first glimpse of a deeper structure that we will explore throughout this article.
请注意系数的规律：1、2、1。a 的幂次从 2 递减到 0，而 b 的幂次从 0 递增到 2。这种对称规律并非偶然——它是我们在本文中将深入探讨的更深层结构的初次展现。
4. Expanding (a + b)3 | 展开 (a + b)3
To expand (a + b)3, we can multiply the expansion of (a + b)2 by another factor of (a + b). Starting from (a + b)2 = a2 + 2ab + b2, we multiply this entire expression by (a + b). The working involves distributing each term: a2(a + b) = a3 + a2b, then 2ab(a + b) = 2a2b + 2ab2, and finally b2(a + b) = ab2 + b3. Summing these results and collecting like terms gives us: a3 + 3a2b + 3ab2 + b3.
要展开 (a + b)3，我们可以将 (a + b)2 的展开式再乘以一个 (a + b) 因式。从 (a + b)2 = a2 + 2ab + b2 出发，将整个表达式乘以 (a + b)。计算过程涉及逐项分配：a2(a + b) = a3 + a2b，然后 2ab(a + b) = 2a2b + 2ab2，最后 b2(a + b) = ab2 + b3。将这些结果相加并合并同类项，我们得到：a3 + 3a2b + 3ab2 + b3。
Again, examine the coefficient pattern: 1, 3, 3, 1. The powers of a descend as 3, 2, 1, 0, while the powers of b ascend as 0, 1, 2, 3. This consistent behaviour hints at a predictable system. In fact, these coefficients — 1, 3, 3, 1 — correspond to the fourth row of Pascal’s Triangle, a topic we shall now investigate.
再次观察系数规律：1、3、3、1。a 的幂次以 3、2、1、0 递减，而 b 的幂次以 0、1、2、3 递增。这种一致的行为暗示着一个可预测的系统。实际上，这些系数——1、3、3、1——对应帕斯卡三角形的第四行，这是我们接下来要探究的主题。
5. Pascal’s Triangle | 帕斯卡三角形
Pascal’s Triangle (known in China as Yang Hui’s Triangle, discovered by the Chinese mathematician Yang Hui in the 13th century) is a triangular array of numbers where each entry is the sum of the two numbers directly above it. The triangle begins with a single 1 at the top, and every row starts and ends with 1. To construct it, simply add adjacent numbers from the row above and place the result between them on the next row. Below are the first six rows (Row 0 to Row 5):
帕斯卡三角形（在中国称为杨辉三角，由13世纪中国数学家杨辉发现）是一个三角形的数字阵列，其中每个数字是正上方两个数字之和。三角形顶部以单个 1 开始，每行以 1 开头和结尾。要构建它，只需将上一行相邻的两个数字相加，并将结果放在下一行的两者之间。以下是前六行（第 0 行至第 5 行）：
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
The remarkable property of Pascal’s Triangle is that its rows provide the coefficients for binomial expansions. Row n (starting from n = 0) gives the coefficients for expanding (a + b)n. For instance, Row 2 gives 1, 2, 1 for (a + b)2; Row 3 gives 1, 3, 3, 1 for (a + b)3; and Row 4 gives 1, 4, 6, 4, 1 for (a + b)4. This connection transforms a tedious multiplication problem into a simple pattern-matching exercise.
帕斯卡三角形的一个显著特性是，它的每一行提供了二项式展开的系数。第 n 行（从 n = 0 开始）给出了展开 (a + b)n 的系数。例如，第 2 行给出 (a + b)2 的系数 1、2、1；第 3 行给出 (a + b)3 的系数 1、3、3、1；第 4 行给出 (a + b)4 的系数 1、4、6、4、1。这一联系将繁琐的乘法问题转化为简单的模式匹配练习。
6. Using Pascal’s Triangle for Expansion | 利用帕斯卡三角形进行展开
To expand a binomial using Pascal’s Triangle, follow a systematic procedure. First, identify the value of n — the exponent outside the brackets. Next, locate Row n in Pascal’s Triangle and write down the coefficients. Then, construct each term by taking the first variable a and decreasing its power from n down to 0, while taking the second variable b and increasing its power from 0 up to n. Multiply each pair of powers by the corresponding coefficient from the triangle. Finally, connect all terms with plus signs.
要利用帕斯卡三角形展开二项式，请遵循一个系统化的步骤。首先，确定 n 的值——即括号外的指数。接着，在帕斯卡三角形中找到第 n 行并写下系数。然后，构建每一项：将第一个变量 a 的幂次从 n 递减到 0，同时将第二个变量 b 的幂次从 0 递增到 n。将每对幂次乘以三角形中对应的系数。最后，用加号连接所有项。
Let us apply this method to expand (x + y)4. Row 4 of Pascal’s Triangle gives coefficients 1, 4, 6, 4, 1. The powers of x descend: x4, x3, x2, x1, x0, while powers of y ascend: y0, y1, y2, y3, y4. Multiplying coefficients by the corresponding variable powers produces:
让我们运用这个方法来展开 (x + y)4。帕斯卡三角形第 4 行给出系数 1、4、6、4、1。x 的幂次递减：x4、x3、x2、x1、x0，而 y 的幂次递增：y0、y1、y2、y3、y4。将系数与对应的变量幂次相乘得到：
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
Remember that x0 = 1 and y0 = 1, so the first and last terms simplify accordingly. This method is efficient and significantly reduces the risk of arithmetic mistakes compared to repeated multiplication. For GCSE exams, being able to write down the expansion directly from Pascal’s Triangle is a valuable skill that saves precious time.
请记住 x0 = 1 且 y0 = 1，因此首项和末项可以相应简化。与反复相乘相比，这种方法效率高，并显著降低了算术错误的风险。对于 GCSE 考试，能够直接从帕斯卡三角形写出展开式是一项宝贵的技能，可以节省宝贵的时间。
7. The Binomial Coefficient Notation | 二项式系数符号
While Pascal’s Triangle is excellent for small values of n, mathematicians use a more compact notation called binomial coefficients. The coefficient of the term containing an−rbr in the expansion of (a + b)n is written as nCr (read as ‘n choose r’), where nCr = n! / [r! (n − r)!]. Here, the exclamation mark denotes the factorial function: n! = n × (n − 1) × (n − 2) × … × 2 × 1. For example, 4! = 4 × 3 × 2 × 1 = 24.
虽然帕斯卡三角形对于较小的 n 值非常有效，但数学家使用一种更紧凑的符号，称为二项式系数。在 (a + b)n 的展开式中，包含 an−rbr 的项的系数写作 nCr（读作”n 选 r”），其中 nCr = n! / [r! (n − r)!]。这里，感叹号表示阶乘函数：n! = n × (n − 1) × (n − 2) × … × 2 × 1。例如，4! = 4 × 3 × 2 × 1 = 24。
At GCSE level, you are not required to use the factorial formula extensively; Pascal’s Triangle is usually sufficient. However, some exam boards introduce the nCr notation, and it is helpful to recognise it. The key relationship is that the rth number in Row n of Pascal’s Triangle (counting from r = 0) equals nCr. So for (a + b)4, the coefficients are 4C0 = 1, 4C1 = 4, 4C2 = 6, 4C3 = 4, and 4C4 = 1. Notice the symmetry: nCr = nC(n−r), which explains why binomial expansions are always symmetric.
在 GCSE 级别，你不需要广泛使用阶乘公式；帕斯卡三角形通常就足够了。然而，一些考试局引入了 nCr 符号，认识它是有帮助的。关键关系是，帕斯卡三角形第 n 行中第 r 个数字（从 r = 0 开始计数）等于 nCr。因此对于 (a + b)4，系数为 4C0 = 1、4C1 = 4、4C2 = 6、4C3 = 4 和 4C4 = 1。注意对称性：nCr = nC(n−r)，这解释了为什么二项式展开总是对称的。
8. General Formula for Binomial Expansion | 二项式展开的一般公式
Bringing together the ideas from Pascal’s Triangle and binomial coefficients, we can state the general binomial theorem for a positive integer n. The expansion of (a + b)n is given by:
将帕斯卡三角形和二项式系数的概念综合起来，我们可以陈述正整数 n 的一般二项式定理。(a + b)n 的展开式由以下公式给出：
(a + b)n = nC0·an + nC1·an−1b + nC2·an−2b2 + … + nCn·bn
Each term follows the pattern: coefficient nCr multiplied by a raised to the power (n − r) multiplied by b raised to the power r, where r runs from 0 to n. The total number of terms in the expansion is always n + 1. For example, (a + b)3 has 4 terms, and (a + b)5 has 6 terms. This formula is the backbone of binomial expansion theory, and although you may not need to quote it explicitly in a GCSE exam, understanding the underlying pattern is essential.
每一项遵循以下模式：系数 nCr 乘以 a 的 (n − r) 次幂再乘以 b 的 r 次幂，其中 r 从 0 到 n。展开式中的总项数始终为 n + 1。例如，(a + b)3 有 4 项，(a + b)5 有 6 项。这个公式是二项式展开理论的支柱，虽然你在 GCSE 考试中可能不需要明确引用它，但理解其内在规律至关重要。
9. Expanding (a + b)4 and Beyond | 展开 (a + b)4 及更高次幂
Let us consolidate our understanding by expanding (a + b)4 in full detail. Using Row 4 of Pascal’s Triangle (coefficients 1, 4, 6, 4, 1), we write each term systematically. For r = 0: the term is 1 × a4 × b0 = a4. For r = 1: 4 × a3 × b1 = 4a3b. For r = 2: 6 × a2 × b2 = 6a2b2. For r = 3: 4 × a1 × b3 = 4ab3. For r = 4: 1 × a0 × b4 = b4. The complete expansion is therefore a4 + 4a3b + 6a2b2 + 4ab3 + b4.
让我们通过详细展开 (a + b)4 来巩固理解。使用帕斯卡三角形第 4 行（系数 1、4、6、4、1），我们系统地写出每一项。当 r = 0：项为 1 × a4 × b0 = a4。当 r = 1：4 × a3 × b1 = 4a3b。当 r = 2：6 × a2 × b2 = 6a2b2。当 r = 3：4 × a1 × b3 = 4ab3。当 r = 4：1 × a0 × b4 = b4。因此完整展开式为 a4 + 4a3b + 6a2b2 + 4ab3 + b4。
Now consider a more challenging example: expand (2x + 3)3. Here, the binomial is not simply (a + b) but contains coefficients within the brackets. We treat a = 2x and b = 3, with n = 3
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