GCSE CIE Mathematics: Binomial Expansion – Key Points | GCSE CIE 数学：二项式展开考点精讲

📚 GCSE CIE Mathematics: Binomial Expansion – Key Points | GCSE CIE 数学：二项式展开考点精讲

Binomial expansion is a core algebraic skill in the GCSE CIE Mathematics syllabus. It involves raising an expression with two terms, such as (a + b), to a positive integer power and writing the result as a sum of terms. Mastering this topic requires a clear understanding of Pascal’s triangle, binomial coefficients (nCr), and the general term formula. This revision guide will walk you through every essential concept, common pitfalls, and exam techniques to help you secure full marks.

二项式展开是 GCSE CIE 数学教学大纲中的核心代数技能。它是指将一个包含两项的表达式（例如 (a + b)）进行正整数次乘方，并将结果写成若干项的和。掌握这个考点需要清晰理解帕斯卡三角、二项式系数 (nCr) 以及通项公式。本篇复习指南将带你梳理每一个关键概念、常见错误和考试技巧，帮助你稳拿满分。

1. What is a Binomial? | 什么是二项式？

A binomial is an algebraic expression that contains exactly two terms, connected by a plus or minus sign. Examples include (x + 2), (3a – 5b), and even (p² + 7). When we carry out a binomial expansion, we are multiplying this expression by itself several times and then simplifying the result into individual powers of the variables.

二项式是指恰好含有两个项的代数表达式，之间用加号或减号连接。例如 (x + 2)、(3a – 5b) 甚至 (p² + 7) 都是二项式。当我们进行二项式展开时，是将这个表达式自身相乘若干次，然后将结果化成各变量次幂的独立项之和。

2. Pascal’s Triangle Construction | 帕斯卡三角的构建

Pascal’s triangle is a triangular array of numbers that gives the coefficients for binomial expansions of the form (a + b)ⁿ. Each number is the sum of the two directly above it. To build the triangle, start with a ‘1’ at the top (row 0). Row 1: 1 1. Row 2: 1 2 1. Row 3: 1 3 3 1. Row 4: 1 4 6 4 1, and so on. For (a + b)⁴, the coefficients are exactly 1, 4, 6, 4, 1.

帕斯卡三角是一个三角形数字阵列，它给出了形如 (a + b)ⁿ 展开式的各项系数。每个数字是其上方两数之和。构建三角形时，最顶行（第0行）先写一个 ‘1’。第1行：1 1。第2行：1 2 1。第3行：1 3 3 1。第4行：1 4 6 4 1，以此类推。对于 (a + b)⁴，其系数恰好是 1, 4, 6, 4, 1。

3. Binomial Coefficients and nCr | 二项式系数与组合数

Instead of always drawing Pascal’s triangle, we can calculate coefficients directly using the combination formula nCr, also written as C(n, r) or (ⁿ_r). It is defined as nCr = n! / [r! (n – r)!], where n! means n factorial. For example, 5C2 = (5×4)/(2×1) = 10. The calculator button nCr speeds this up in exams. These values represent the number of ways to choose r items from n, matching the triangle entries.

除了每次都画出帕斯卡三角，我们还可以直接用组合数公式 nCr（也写作 C(n, r) 或 (ⁿ_r)）来计算系数。其定义为 nCr = n! / [r! (n – r)!]，其中 n! 表示 n 的阶乘。例如 5C2 = (5×4)/(2×1) = 10。考试中使用计算器上的 nCr 按钮可以快速得出结果。这些数值表示从 n 个物件中选取 r 个的组合数，恰好对应三角形中的数字。

4. The Binomial Theorem Formula | 二项式定理公式

For a positive integer n, the binomial theorem states:

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n–1 b + ⁿC₂ a^n–2 b² + … + ⁿC_n bⁿ

This compact form uses sigma notation: Σ_r=0ⁿ ⁿC_r a^n–r b^r. Notice that the powers of a decrease from n to 0, while the powers of b increase from 0 to n. The sum of the indices in each term is always n. You must be comfortable using this formula to write out full expansions or to find a specific term.

对于正整数 n，二项式定理如下：

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n–1 b + ⁿC₂ a^n–2 b² + … + ⁿC_n bⁿ

也可以用求和符号简洁地写作：Σ_r=0ⁿ ⁿC_r a^n–r b^r。注意 a 的指数从 n 递减至 0，而 b 的指数从 0 递增至 n。每一项中两个指数的和总等于 n。你必须能够熟练使用这个公式写出完整的展开式或找出特定的某一项。

5. Expanding (a + b)ⁿ Step by Step | 逐步展开 (a + b)ⁿ

Let’s expand (2x + 3)³ using the theorem. Here a = 2x, b = 3, n = 3. The coefficients from row 3 are 1, 3, 3, 1. Write the terms with decreasing powers of (2x):
Term 1: 1 × (2x)³ × 3⁰ = 8x³
Term 2: 3 × (2x)² × 3¹ = 3 × 4x² × 3 = 36x²
Term 3: 3 × (2x)¹ × 3² = 3 × 2x × 9 = 54x
Term 4: 1 × (2x)⁰ × 3³ = 27
Final expansion: 8x³ + 36x² + 54x + 27. Always simplify each term fully.

我们用这个定理展开 (2x + 3)³。这里 a = 2x, b = 3, n = 3。第3行的系数为 1, 3, 3, 1。按照 (2x) 的降幂写出各项：
第一项：1 × (2x)³ × 3⁰ = 8x³
第二项：3 × (2x)² × 3¹ = 3 × 4x² × 3 = 36x²
第三项：3 × (2x)¹ × 3² = 3 × 2x × 9 = 54x
第四项：1 × (2x)⁰ × 3³ = 27
最终展开式为：8x³ + 36x² + 54x + 27。务必完整化简每一项。

6. Special Case: (1 + x)ⁿ | 特殊形式 (1 + x)ⁿ

When the first term is 1, the expansion simplifies because powers of 1 remain 1. For example, (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴. More generally, (1 + x)ⁿ = 1 + nx + [n(n–1)/2!] x² + [n(n–1)(n–2)/3!] x³ + … . This form is especially useful for estimation because the coefficients are easier to compute mentally for small x.

当首项为 1 时，展开式会简化，因为 1 的任意次幂仍是 1。例如 (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴。更一般地，(1 + x)ⁿ = 1 + nx + [n(n–1)/2!] x² + [n(n–1)(n–2)/3!] x³ + …。这种形式在估算时特别有用，因为当 x 很小时，系数可以用心算快速得出。

7. Finding the r-th Term | 求展开式中的第 r 项

Exam questions often ask for a specific term without expanding everything. The general term, also known as the (r+1)-th term, is given by T_r+1 = ⁿC_r a^n–r b^r. Suppose we want the term in x⁵ from (2 + x)⁹. Here a = 2, b = x, n = 9. We need the power of x to be 5, so set r = 5. Then the term is ⁹C₅ × 2^9–5 × x⁵ = 126 × 2⁴ x⁵ = 126 × 16 x⁵ = 2016x⁵.

考试中常会要求找出某一特定的项，而不把整个式子展开。通项，也称为第 (r+1) 项，公式为 T_r+1 = ⁿC_r a^n–r b^r。例如，求 (2 + x)⁹ 展开式中含 x⁵ 的项。这里 a = 2, b = x, n = 9。我们需要 x 的指数为 5，因此设 r = 5。该项为 ⁹C₅ × 2^9–5 × x⁵ = 126 × 2⁴ x⁵ = 126 × 16 x⁵ = 2016 x⁵。

8. Estimating Values Using Binomial Expansion | 利用二项式展开进行估算

Because higher powers of small numbers become very small, we can use expansions to find approximate values. For example, to estimate (1.02)⁵, write it as (1 + 0.02)⁵. Expand using the (1 + x)ⁿ pattern: 1 + 5(0.02) + 10(0.02)² + 10(0.02)³ + … = 1 + 0.1 + 10×0.0004 + 10×0.000008 = 1 + 0.1 + 0.004 + 0.00008 = 1.10408 approximately. The exact value is 1.1040808032, so the first few terms already give excellent accuracy.

由于小数的较高次幂会变得非常小，我们可以利用展开式求近似值。例如，估算 (1.02)⁵，可写成 (1 + 0.02)⁵。按照 (1 + x)ⁿ 的模式展开：1 + 5(0.02) + 10(0.02)² + 10(0.02)³ + … = 1 + 0.1 + 10×0.0004 + 10×0.000008 = 1 + 0.1 + 0.004 + 0.00008 = 1.10408 约。精确值为 1.1040808032，只有前几项的近似结果已十分准确。

9. Common Pitfalls and How to Avoid Them | 常见错误及避免方法

Many mistakes come from mishandling coefficients, signs, or brackets. Here are key points to watch:

Forgetting the binomial coefficients: Never write (a + b)³ = a³ + a²b + ab² + b³. Always include the coefficients 1, 3, 3, 1.
Sign errors with negative terms: If b is negative, signs alternate. For (2x – 3)³, the signs become +, –, +, –. Raise the negative sign along with the number.
Not applying powers to the whole term: In (3y)², the coefficient must be squared: 9y², not 3y². Use brackets carefully: (3y)² vs 3y².
Using the wrong value of r: Remember the term with x^r corresponds to r in nCr, but the first term is r=0. For the 5th term, use r=4.

许多错误源于对系数、正负号或括号的处理不当。以下是需要留意的关键点：

忘记二项式系数：不要把 (a + b)³ 写成 a³ + a²b + ab² + b³。一定要加上系数 1, 3, 3, 1。
负项引起的正负号错误：如果 b 是负数，符号会交替变化。对于 (2x – 3)³，符号变为 +, –, +, –。计算时把负号同数字一起乘方。
没有对整项进行乘方：在 (3y)² 中，系数必须平方得出 9y²，而不是 3y²。小心使用括号：(3y)² 与 3y² 不同。
使用错误的 r 值：记住含 x^r 的项对应 nCr 中的 r，但第一项是 r=0。若要找第 5 项，须用 r=4。

10. Exam-style Questions and Solutions | 考试类型题与解答

Question: Find the coefficient of x³ in the expansion of (2 + 5x)⁶.
Solution: The general term is ⁶C_r × 2^6–r × (5x)^r. We need the term with x³, so r = 3. Coefficient = ⁶C₃ × 2³ × 5³ = 20 × 8 × 125 = 20000. Thus the term is 20000 x³.

题目：求 (2 + 5x)⁶ 展开式中 x³ 的系数。
解答： 通项为 ⁶C_r × 2^6–r × (5x)^r。我们需要 x³ 的项，故 r = 3。系数 = ⁶C₃ × 2³ × 5³ = 20 × 8 × 125 = 20000。因此该项为 20000 x³。

Question: Use a binomial expansion to estimate (0.98)⁴ to four decimal places.
Solution: Write 0.98 = 1 – 0.02. Then (1 – 0.02)⁴ = 1 + 4(–0.02) + 6(–0.02)² + 4(–0.02)³ + (–0.02)⁴ = 1 – 0.08 + 6(0.0004) – 4(0.000008) + 0.00000016 = 1 – 0.08 + 0.0024 – 0.000032 + 0.00000016 = 0.92236816. Rounded to four decimal places, the answer is 0.9224.

题目：利用二项式展开估算 (0.98)⁴ 至四位小数。
解答： 将 0.98 写成 1 – 0.02。则 (1 – 0.02)⁴ = 1 + 4(–0.02) + 6(–0.02)² + 4(–0.02)³ + (–0.02)⁴ = 1 – 0.08 + 6(0.0004) – 4(0.000008) + 0.00000016 = 1 – 0.08 + 0.0024 – 0.000032 + 0.00000016 = 0.92236816。四舍五入至四位小数，答案为 0.9224。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

GCSE CIE Mathematics: Binomial Expansion – Key Points | GCSE CIE 数学：二项式展开 考点精讲