Binomial Expansion for IGCSE CCEA Mathematics | IGCSE CCEA 数学：二项式展开考点精讲

📚 Binomial Expansion for IGCSE CCEA Mathematics | IGCSE CCEA 数学：二项式展开考点精讲

Binomial expansion is a core algebraic skill in the IGCSE CCEA Mathematics syllabus. It allows you to expand expressions of the form (a + b)ⁿ without having to multiply the brackets repeatedly. Understanding the pattern of coefficients, the role of Pascal’s triangle and the nCr formula, and being able to find any specific term are essential for exam success. This revision guide breaks down every key concept with worked examples, common mistakes and practical tips.

二项式展开是 IGCSE CCEA 数学考试大纲中的核心代数技能。它能让你不用反复乘法就能展开形如 (a + b)ⁿ 的表达式。理解系数的规律、帕斯卡三角形与 nCr 公式的作用，并能求出任意指定项，是取得考试成功的关键。这份复习指南通过详细示例、常见错误和实用技巧，分解每一个重要概念。

1. What is Binomial Expansion? | 什么是二项式展开？

A binomial is an algebraic expression that contains exactly two terms, such as (x + 3) or (2a – 5b). Binomial expansion is the process of raising a binomial to a positive integer power n and writing the result as a sum of terms. Instead of multiplying out (x + 2)³ as (x+2)(x+2)(x+2), expansion gives the polynomial directly: x³ + 6x² + 12x + 8.

二项式是恰好包含两项的代数表达式，例如 (x + 3) 或 (2a – 5b)。二项式展开是指将一个二项式提升到正整数 n 次幂，并将结果写成若干项的和。例如，不用将 (x+2)³ 乘开为 (x+2)(x+2)(x+2)，展开式直接给出多项式：x³ + 6x² + 12x + 8。

In the IGCSE CCEA examination, you will often be asked to expand binomials like (1 + 2x)⁵ or (3 – y)⁴, or to find a particular coefficient. The power n is usually a small positive integer, but the method generalises to any n using the binomial theorem.

在 IGCSE CCEA 考试中，你常常会被要求展开如 (1 + 2x)⁵ 或 (3 – y)⁴ 的二项式，或者求出某一特定项的系数。幂指数 n 通常是一个较小的正整数，但利用二项式定理，这一方法可以推广到任意 n。

2. Pascal’s Triangle | 帕斯卡三角形

Pascal’s triangle is a simple and visual way to find the coefficients of a binomial expansion. Each row corresponds to the power n, starting with n = 0 at the top. Row n gives the coefficients for (a + b)ⁿ. The triangle is constructed by adding the two numbers directly above to obtain the number below.

帕斯卡三角形是一种简单直观的寻找二项式展开系数的方法。每一行对应幂次 n，顶部从 n = 0 开始。第 n 行给出 (a + b)ⁿ 的系数。三角形的构造方法是将正上方的两个数相加，得到下方数字。

For example, the first few rows are:
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

例如，前几行如下所示：
第 0 行：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
第 5 行：1 5 10 10 5 1

To use the triangle for expansion, you take the coefficients from row n and attach them to descending powers of the first term and ascending powers of the second term. This method works neatly for small values of n, such as n ≤ 5, and is often the fastest approach in a non-calculator paper.

利用三角形进行展开时，从第 n 行取出系数，并将它们与第一项的降幂和第二项的升幂组合在一起。对于较小的 n 值（例如 n ≤ 5），这种方法十分整洁，而且往往是非计算器试卷中最快的解题方式。

3. Binomial Coefficients and the nCr Formula | 二项式系数与 nCr 公式

When n becomes larger, writing out Pascal’s triangle is impractical. Instead, we use the combination formula nCr, also written as C(n, r) or ⁿCᵣ. This tells you the coefficient of the term that contains bʳ. The formula is: nCr = n! / [r! (n – r)!], where ‘!’ denotes the factorial function.

当 n 较大时，写出帕斯卡三角形就不切实际了。我们转而使用组合公式 nCr，也写作 C(n, r) 或 ⁿCᵣ。它告诉你含有 bʳ 的那一项的系数。公式为：nCr = n! / [r! (n – r)!]，其中 ‘!’ 表示阶乘函数。

For example, ⁵C₂ = 5! / (2! × 3!) = (5×4×3×2×1) / (2×1 × 3×2×1) = 10. This matches the third entry in row 5 of Pascal’s triangle. Your scientific calculator will have an nCr button, but you must also know how to compute it manually for non-calculator papers.

例如，⁵C₂ = 5! / (2! × 3!) = (5×4×3×2×1) / (2×1 × 3×2×1) = 10。这与帕斯卡三角形第 5 行的第三个数字相吻合。你的科学计算器上会有 nCr 键，但在不允许使用计算器的试卷中，你必须掌握手算的方法。

The role of r: in the expansion of (a + b)ⁿ, the general term is nCr × aⁿ⁻ʳ × bʳ, where r starts at 0 (giving the first term aⁿ) and runs to n (giving the last term bⁿ).

r 的角色：在 (a + b)ⁿ 的展开式中，通项为 nCr × aⁿ⁻ʳ × bʳ，其中 r 从 0 开始（给出首项 aⁿ），一直取到 n（给出末项 bⁿ）。

4. The Binomial Theorem Statement | 二项式定理的陈述

The binomial theorem provides a compact way to write the full expansion of (a + b)ⁿ for any positive integer n:

(a + b)ⁿ = Σ_{r=0}ⁿ nCr aⁿ⁻ʳ bʳ

二项式定理为任意正整数 n 的 (a + b)ⁿ 展开式提供了一种简洁的写法：

(a + b)ⁿ = Σ_{r=0}ⁿ nCr aⁿ⁻ʳ bʳ

Writing this out in full gives:

(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹ b + nC2 aⁿ⁻² b² + … + nCn bⁿ

把它完整写出就是：

(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹ b + nC2 aⁿ⁻² b² + … + nCn bⁿ

Remember that nC0 = 1 and nCn = 1. The powers of a decrease from n to 0, while the powers of b increase from 0 to n. The sum of the exponents in each term is always n.

记住 nC0 = 1 且 nCn = 1。a 的幂从 n 递减到 0，而 b 的幂从 0 递增到 n。每一项中指数的和恒为 n。

5. Step-by-step Expansion of (a + b)ⁿ | 逐步展开 (a + b)ⁿ

Let’s expand (2x + 3)⁴ using the binomial theorem. We identify a = 2x, b = 3 and n = 4. We then compute the five terms (since r = 0 to 4) step by step.

让我们用二项式定理来展开 (2x + 3)⁴。我们确定 a = 2x，b = 3，n = 4。然后逐步计算出五项（因为 r 从 0 到 4）。

r = 0: ⁴C₀ (2x)⁴ (3)⁰ = 1 × 16x⁴ × 1 = 16x⁴
r = 1: ⁴C₁ (2x)³ (3)¹ = 4 × 8x³ × 3 = 96x³
r = 2: ⁴C₂ (2x)² (3)² = 6 × 4x² × 9 = 216x²
r = 3: ⁴C₃ (2x)¹ (3)³ = 4 × 2x × 27 = 216x
r = 4: ⁴C₄ (2x)⁰ (3)⁴ = 1 × 1 × 81 = 81

The final expansion is 16x⁴ + 96x³ + 216x² + 216x + 81. Notice how the powers of x decrease and the powers of 3 increase.

最终展开式为 16x⁴ + 96x³ + 216x² + 216x + 81。注意 x 的幂次如何递减，而 3 的幂次如何递增。

Always double-check that the number of terms is n+1 and that the coefficients follow a symmetric pattern when the original a and b are symmetric – although here a = 2x and b = 3 are not symmetric, so the coefficients are not palindromic.

务必再次检查：项数应为 n+1 个；当原来的 a 与 b 对称时，系数呈现对称模式——虽然此处 a = 2x，b = 3 并不对称，因此系数并不具有回文对称性。

6. Handling Negative Terms and Subtraction | 处理负项与减法

When the binomial involves subtraction, such as (x – 2)⁵, treat it as (x + (-2))⁵. This means b = -2. The alternating signs will automatically appear because odd powers of a negative number remain negative, while even powers become positive.

当二项式涉及减法时，例如 (x – 2)⁵，将其视为 (x + (-2))⁵。这意味着 b = -2。符号会自动交替出现，因为负数的奇次幂仍为负，偶次幂则变为正。

For example, expanding (2y – 3)³:
a = 2y, b = -3, n = 3.
Term 1: ³C₀ (2y)³ (-3)⁰ = 8y³
Term 2: ³C₁ (2y)² (-3)¹ = 3 × 4y² × (-3) = -36y²
Term 3: ³C₂ (2y)¹ (-3)² = 3 × 2y × 9 = 54y
Term 4: ³C₃ (2y)⁰ (-3)³ = 1 × 1 × (-27) = -27
Thus (2y – 3)³ = 8y³ – 36y² + 54y – 27.

例如，展开 (2y – 3)³：
a = 2y，b = -3，n = 3。
第 1 项：³C₀ (2y)³ (-3)⁰ = 8y³
第 2 项：³C₁ (2y)² (-3)¹ = 3 × 4y² × (-3) = -36y²
第 3 项：³C₂ (2y)¹ (-3)² = 3 × 2y × 9 = 54y
第 4 项：³C₃ (2y)⁰ (-3)³ = 1 × 1 × (-27) = -27
因此 (2y – 3)³ = 8y³ – 36y² + 54y – 27。

Never ignore the negative sign – it is one of the most common mistakes. Write the binomial as a sum first, then apply the theorem systematically.

千万不要忽略负号——这是最常见的错误之一。先将二项式写成求和形式，再有条理地运用定理。

7. Finding a Specific Term without Full Expansion | 无需全部展开即可找到特定项

A very common exam question asks for ‘the term in x⁵’ or ‘the coefficient of x³’ without requiring the whole expansion. You use the general term formula: T_{r+1} = nCr × aⁿ⁻ʳ × bʳ. The subscript r+1 simply indicates that the first term corresponds to r = 0.

一个非常常见的考试题型是要求给出 ‘含有 x⁵ 的项’ 或 ‘x³ 的系数’，而不必写出整个展开式。此时使用通项公式：T_{r+1} = nCr × aⁿ⁻ʳ × bʳ。下标 r+1 仅仅表示首项对应 r = 0。

Example: find the term in x⁴ in the expansion of (2 + x)⁷.
Here a = 2, b = x, n = 7. The general term is ⁷Cᵣ × 2⁷⁻ʳ × xʳ. We need the power of x to be 4, so set r = 4. Then the term is ⁷C₄ × 2⁷⁻⁴ × x⁴ = 35 × 2³ × x⁴ = 35 × 8 × x⁴ = 280x⁴. The coefficient is 280.

示例：在 (2 + x)⁷ 的展开式中找出含有 x⁴ 的项。
这里 a = 2，b = x，n = 7。通项为 ⁷Cᵣ × 2⁷⁻ʳ × xʳ。我们需要 x 的幂次为 4，因此设 r = 4。那么该项为 ⁷C₄ × 2⁷⁻⁴ × x⁴ = 35 × 2³ × x⁴ = 35 × 8 × x⁴ = 280x⁴。系数为 280。

Now consider a trickier case: find the coefficient of x⁵ in (3x – 1/x²)⁸. First identify a = 3x, b = -1/x², n = 8. The general term is ⁸Cᵣ (3x)⁸⁻ʳ (-1/x²)ʳ. Simplify the x-part: (x)⁸⁻ʳ × (x⁻²)ʳ = x⁸⁻ʳ⁻²ʳ = x⁸⁻³ʳ. We need the exponent to be 5, so 8 – 3r = 5 → 3r = 3 → r = 1. Substitute r = 1: ⁸C₁ × (3x)⁷ × (-1/x²)¹ = 8 × 3⁷ x⁷ × (-1) x⁻² = 8 × 2187 × (-1) × x⁵ = -17496x⁵. The coefficient is -17496.

再来看一道更复杂的题：求 (3x – 1/x²)⁸ 展开式中 x⁵ 的系数。首先确定 a = 3x，b = -1/x²，n = 8。通项为 ⁸Cᵣ (3x)⁸⁻ʳ (-1/x²)ʳ。化简 x 的部分：(x)⁸⁻ʳ × (x⁻²)ʳ = x⁸⁻ʳ⁻²ʳ = x⁸⁻³ʳ。我们需要指数为 5，因此 8 – 3r = 5 → 3r = 3 → r = 1。代入 r = 1：⁸C₁ × (3x)⁷ × (-1/x²)¹ = 8 × 3⁷ x⁷ × (-1) x⁻² = 8 × 2187 × (-1) × x⁵ = -17496x⁵。系数为 -17496。

8. Finding the Constant Term | 求常数项

The constant term is the term that does not contain any variable, i.e. where the exponent of x becomes 0. To find it, set the exponent of x in the general term equal to 0 and solve for r. Then substitute back to find the coefficient.

常数项是不含任何变量的项，即 x 的指数变为 0 的那一项。要求常数项，就令通项中 x 的指数等于 0，解得 r，再代回求系数。

Example: find the constant term in the expansion of (x² + 2/x)⁹.
Here a = x², b = 2/x, n = 9. General term = ⁹Cᵣ (x²)⁹⁻ʳ (2/x)ʳ = ⁹Cᵣ × 2ʳ × x^{18 – 2r – r} = ⁹Cᵣ × 2ʳ × x^{18 – 3r}.
Set 18 – 3r = 0 → r = 6. The constant term is ⁹C₆ × 2⁶ x⁰ = 84 × 64 = 5376.

示例：求 (x² + 2/x)⁹ 展开式中的常数项。
这里 a = x²，b = 2/x，n = 9。通项 = ⁹Cᵣ (x²)⁹⁻ʳ (2/x)ʳ = ⁹Cᵣ × 2ʳ × x^{18 – 2r – r} = ⁹Cᵣ × 2ʳ × x^{18 – 3r}。
令 18 – 3r = 0 → r = 6。常数项为 ⁹C₆ × 2⁶ x⁰ = 84 × 64 = 5376。

This technique is highly examined. Always isolate the power of the variable, form a simple linear equation, and check that the resulting r is an integer between 0 and n.

这种方法在考试中出现频率很高。务必将变量的指数分离出来，建立一个简单的一次方程，并验证得到的 r 是介于 0 到 n 之间的整数。

9. Using the Expansion for Approximation | 利用展开式进行近似计算

When x is small, certain binomial expansions can be used to estimate values quickly. For a binomial of the form (1 + x)ⁿ where |x| < 1, the terms decrease rapidly, so truncating after the first few terms gives a good approximation.

当 x 很小时，某些二项式展开式可用来快速估算数值。对于形如 (1 + x)ⁿ 且 |x| < 1 的二项式，各项迅速减小，因此只取前几项就能得到很好的近似值。

Example: approximate (1.02)⁵ using the expansion of (1 + 2x)⁵, with x = 0.01. Actually, rewrite 1.02 = 1 + 0.02. Then (1 + 0.02)⁵ ≈ 1 + 5(0.02) + 10(0.02)² = 1 + 0.1 + 10(0.0004) = 1 + 0.1 + 0.004 = 1.104. The exact value is about 1.10408, so the approximation is excellent.

示例：利用 (1 + 2x)⁵ 当 x = 0.01 时的展开来估算 (1.02)⁵。实际上，将 1.02 改写为 1 + 0.02。那么 (1 + 0.02)⁵ ≈ 1 + 5(0.02) + 10(0.02)² = 1 + 0.1 + 10(0.0004) = 1 + 0.1 + 0.004 = 1.104。精确值大约是 1.10408，因此近似效果极佳。

You may also be asked to estimate square roots, cubes, or reciprocals by choosing a suitable x. Always identify the connection between the given binomial and the number to approximate.

你可能还会被要求通过选择合适的 x 来估算平方根、立方或倒数。务必找出给定二项式与待近似数值之间的联系。

10. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Mistake 1: Forgetting to apply the power to the coefficient inside the bracket. In (2x)³, many students write 2x³ instead of 8x³. Always apply the exponent to both the number and the variable.

错误 1：忘记对括号内的系数进行乘方。在 (2x)³ 中，许多学生写成 2x³ 而不是 8x³。要把指数同时作用于数字和变量。

Mistake 2: Misidentifying a and b. In (3 – 2x)⁴, a = 3, b = -2x, not 2x. Writing b as 2x and then manually alternating signs often leads to errors. Let the theorem handle the signs by using b = -2x.

错误 2：错误识别 a 与 b。在 (3 – 2x)⁴ 中，a = 3，b = -2x，而不是 2x。将 b 写成 2x 然后手动交替符号常常导致出错。应使用 b = -2x，让定理来处理符号。

Mistake 3: Getting the nCr values wrong under pressure. Practise using both the calculator nCr button and the factorial formula. Remember that nCr = nC(n-r), which can save time (e.g., ¹⁰C₈ = ¹⁰C₂ = 45).

错误 3：在紧张时算错 nCr 的值。要练习使用计算器上的 nCr 键以及阶乘公式。记住 nCr = nC(n-r)，这可以节省时间（例如 ¹⁰C₈ = ¹⁰C₂ = 45）。

Mistake 4: Forgetting that the first term corresponds to r = 0. When asked for the third term in the expansion, use r = 2, not r = 3. Always check whether the question means the term number or the value of r.

错误 4：忘记首项对应 r = 0。当被问到展开式中的第三项时，要用 r = 2，而不是 r = 3。一定要弄清楚题目指的是项序号还是 r 的值。

11. Exam Techniques and Summary | 应考技巧与总结

Read the question carefully: does it ask for the full expansion or just one term? If only one term, use the general term formula immediately – it saves time. If the full expansion is required, check the power n – for n ≤ 4, Pascal’s triangle is quick; for n > 4, use the nCr method.

仔细读题：它要求的是完整的展开式还是仅仅某一项？如果只求一项，立即使用通项公式——这能节省时间。如果要求完整展开，检查幂次 n——若 n ≤ 4，帕斯卡三角形很快；若 n > 4，使用 nCr 方法。

When writing the final answer, present terms in descending or ascending powers as requested. Simplify coefficients fully. If the question specifies ‘in ascending powers of x’, start with the constant term.

在书写最终答案时，按要求的降幂或升幂排列各项。系数要完全化简。如果题目指定 ‘按 x 的升幂排列’，就从常数项开始。

Finally, always check your expansion by substituting a small value, such as x = 1 or x = 0. If (1 + 1)ⁿ = 2ⁿ does not equal the sum of your coefficients, you have made an error. This quick validation can catch sign or coefficient mistakes before you finish the exam.

最后，总是通过代入一个简单的值来检验你的展开式，比如 x = 1 或 x = 0。如果 (1 + 1)ⁿ = 2ⁿ 不等于你各项系数的总和，那就说明出错了。这种快速验证能在考试结束前帮你揪出符号或系数上的错误。

Binomial expansion is a predictable and highly structured topic. Mastery comes from understanding the pattern, practising the nCr formula, and training your eye to spot the required term. With consistent practice, you can secure full marks on every expansion question.

二项式展开是一个可预测且结构严谨的课题。掌握它在于理解规律、练习 nCr 公式，并训练自己去发现题目所要求的项。通过持续的练习，你就有把握在每一个展开题上拿到满分。

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Binomial Expansion for IGCSE CCEA Mathematics | IGCSE CCEA 数学：二项式展开 考点精讲