IGCSE Maths Binomial Expansion Revision | IGCSE 数学：二项式展开考点精讲

📚 IGCSE Maths Binomial Expansion Revision | IGCSE 数学：二项式展开考点精讲

Binomial expansion is a fundamental algebraic skill required for the IGCSE Mathematics syllabus. It allows us to expand expressions of the form (a+b)ⁿ efficiently, where n is a positive integer. Mastering this topic involves understanding Pascal’s Triangle, the binomial theorem, and the ability to find specific terms and coefficients without fully expanding large powers. This article breaks down every key concept, provides clear examples, and highlights common pitfalls to help you secure top marks.

二项式展开是IGCSE数学课程中的一项基本代数技能。它使我们能够高效地展开形如 (a+b)ⁿ 的表达式，其中 n 为正整数。掌握这一主题需要理解帕斯卡三角形、二项式定理，以及在不完全展开高次幂的情况下找到特定项和系数的能力。本文拆解了每一个关键概念，提供了清晰的示例，并指出常见错误，助你稳拿高分。

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

A binomial is an algebraic expression containing two terms, such as (x+y) or (2a-3b). Binomial expansion is the process of multiplying out a binomial raised to a power, writing it as a sum of terms. For small powers like (x+y)², we can simply use FOIL or the distributive law, but for higher powers like (x+y)⁵, systematic methods are essential.

二项式是包含两个项的代数式，例如 (x+y) 或 (2a-3b)。二项式展开就是将乘方后的二项式展开成多项之和的过程。对于 (x+y)² 这样的低次幂，我们可以直接使用乘法分配律；但对于 (x+y)⁵ 这样的高次幂，就需要系统的方法了。

The IGCSE syllabus focuses on expansions where the exponent is a positive integer. You will learn to use Pascal’s Triangle and the concept of combinations (nCr) to determine the coefficients of each term in the expansion.

IGCSE 考纲要求掌握指数为正整数的二项式展开。你将学习使用帕斯卡三角形和组合数 (nCr) 的概念来确定展开式中每一项的系数。

2. Pascal’s Triangle | 帕斯卡三角形

Pascal’s Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The triangle begins with a 1 at the top, and each row corresponds to the coefficients of the expansion of (a+b)ⁿ. For example, row 0 is just 1, row 1 is 1 1, row 2 is 1 2 1, row 3 is 1 3 3 1, and so on.

帕斯卡三角形是一个由数字组成的三角形阵列，其中每个数字是其正上方两个数字之和。三角形的顶端是 1，每一行对应 (a+b)ⁿ 展开式的系数。例如，第 0 行是 1，第 1 行是 1 1，第 2 行是 1 2 1，第 3 行是 1 3 3 1，依此类推。

To expand (a+b)⁴, we look at the fifth row (since n=4, row 4) which is 1, 4, 6, 4, 1. These coefficients multiply the terms a⁴, a³b, a²b², ab³, b⁴ respectively, with powers of a decreasing from n to 0 and powers of b increasing from 0 to n. It is an excellent tool for small values of n, typically up to n=7 or 8 in exam questions.

要展开 (a+b)⁴，我们查找第 4 行（因为 n=4）：1, 4, 6, 4, 1。这些系数分别乘以 a⁴, a³b, a²b², ab³, b⁴，其中 a 的指数从 n 降至 0，b 的指数从 0 升至 n。对于较小的 n 值（考试中通常最高到 n=7 或 8），帕斯卡三角形是一个非常实用的工具。

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

The coefficients in the binomial expansion can also be calculated using the combination formula: C(n,r) = n! / [r!(n-r)!], often read as “n choose r”. Here n is the power and r is the position of the term, starting from r=0 for the first term. This formula gives the number of ways to choose r items from n items, and it matches the entries in Pascal’s Triangle.

二项式展开中的系数也可以使用组合公式计算：C(n,r) = n! / [r!(n-r)!]，也常读作“n 选 r”。其中 n 是指数，r 是项的位置，第一项从 r=0 开始。这个公式给出了从 n 个物品中选择 r 个的组合数，结果与帕斯卡三角形中的数字一致。

IGCSE candidates are expected to be able to compute C(n,r) using a calculator or manually for small n. For instance, C(5,2) = 5! / (2!3!) = (5×4) / (2×1) = 10. Understanding the symmetry C(n,r) = C(n,n-r) is very helpful for verifying coefficients quickly.

IGCSE 考生应能使用计算器或手动计算较小的组合数，例如 C(5,2) = 5! / (2!3!) = (5×4) / (2×1) = 10。理解对称性 C(n,r) = C(n,n-r) 有助于快速核对系数。

4. The Binomial Theorem for Positive Integers | 正整数指数的二项式定理

The formal statement of the binomial theorem for a positive integer n is:

(a+b)ⁿ = C(n,0)aⁿ + C(n,1)aⁿ⁻¹b + C(n,2)aⁿ⁻²b² + … + C(n,n)bⁿ

This compact representation allows us to write any term directly without expanding the whole bracket. The term for a particular value of r (r = 0,1,2,…,n) is given by Tᵣ₊₁ = C(n,r) aⁿ⁻ʳ bʳ. Note that the first term corresponds to r=0, so the (r+1)th term uses this formula.

对于正整数 n，二项式定理的正式表述为：

(a+b)ⁿ = C(n,0)aⁿ + C(n,1)aⁿ⁻¹b + C(n,2)aⁿ⁻²b² + … + C(n,n)bⁿ

这一紧凑的表示法使我们能够直接写出任意一项，无需展开整个括号。对于特定的 r 值（r = 0,1,2,…,n），项 Tᵣ₊₁ = C(n,r) aⁿ⁻ʳ bʳ。注意第一项对应 r=0，因此第 (r+1) 项即可使用此公式。

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

Let’s illustrate the expansion process for (2x-3)⁴. First, identify a=2x, b=-3, and n=4. The general term is C(4,r) (2x)⁴⁻ʳ (-3)ʳ. For r=0: C(4,0)(2x)⁴(-3)⁰ = 1 × 16x⁴ × 1 = 16x⁴. For r=1: C(4,1)(2x)³(-3)¹ = 4 × 8x³ × (-3) = -96x³. For r=2: C(4,2)(2x)²(-3)² = 6 × 4x² × 9 = 216x². For r=3: C(4,3)(2x)¹(-3)³ = 4 × 2x × (-27) = -216x. For r=4: C(4,4)(2x)⁰(-3)⁴ = 1 × 1 × 81 = 81. So the full expansion is 16x⁴ – 96x³ + 216x² – 216x + 81.

我们以 (2x-3)⁴ 为例展示展开过程。首先确定 a=2x，b=-3，n=4。通项为 C(4,r) (2x)⁴⁻ʳ (-3)ʳ。当 r=0：C(4,0)(2x)⁴(-3)⁰ = 1×16x⁴×1 = 16x⁴。r=1：C(4,1)(2x)³(-3)¹ = 4×8x³×(-3) = -96x³。r=2：C(4,2)(2x)²(-3)² = 6×4x²×9 = 216x²。r=3：C(4,3)(2x)¹(-3)³ = 4×2x×(-27) = -216x。r=4：C(4,4)(2x)⁰(-3)⁴ = 1×1×81 = 81。因此，完整的展开式为 16x⁴ – 96x³ + 216x² – 216x + 81。

6. Finding a Specific Term | 求特定项

Exam questions often ask for a single term rather than the full expansion. To find the term containing a certain power, set up the general term formula and solve for r. For example, in the expansion of (x² + 1/x)⁸, find the term in x⁷. General term: C(8,r) (x²)⁸⁻ʳ (1/x)ʳ = C(8,r) x¹⁶⁻²ʳ x⁻ʳ = C(8,r) x¹⁶⁻³ʳ. Set exponent 16-3r = 7 → 3r = 9 → r=3. So the term is C(8,3) x⁷ = 56x⁷.

考试常要求找出单个项而非完整展开式。要找到包含特定幂次的项，先写出通项公式并解出 r。例如，在 (x² + 1/x)⁸ 的展开式中，求 x⁷ 项。通项：C(8,r) (x²)⁸⁻ʳ (1/x)ʳ = C(8,r) x¹⁶⁻²ʳ x⁻ʳ = C(8,r) x¹⁶⁻³ʳ。令指数 16-3r = 7 → 3r = 9 → r=3。因此，该项为 C(8,3) x⁷ = 56x⁷。

7. Finding the Coefficient of a Term | 求项的系数

Sometimes you are asked for the coefficient of a particular term, such as “find the coefficient of x⁵ in (3x – 2)⁷”. First locate the term using the general term approach: Tᵣ₊₁ = C(7,r) (3x)⁷⁻ʳ (-2)ʳ. The power of x is 7-r, so set 7-r = 5 → r=2. Then the coefficient is C(7,2) × 3⁵ × (-2)² = 21 × 243 × 4 = 20412. Note: the coefficient includes the sign.

有时题目要求找出特定项的系数，例如“求 (3x – 2)⁷ 的展开式中 x⁵ 的系数”。首先用通项法定位：Tᵣ₊₁ = C(7,r) (3x)⁷⁻ʳ (-2)ʳ。x 的指数为 7-r，令 7-r = 5 → r=2。此时系数为 C(7,2) × 3⁵ × (-2)² = 21 × 243 × 4 = 20412。注意系数包含符号。

8. Applying Binomial Expansion in Problem Solving | 二项式展开在解题中的应用

Binomial expansion skills are used in approximation problems. For example, to estimate (1.02)⁵ without a calculator, write it as (1 + 0.02)⁵ and expand: 1 + 5×0.02 + 10×0.0004 + 10×0.000008 + 5×0.00000016 + 0.0000000032. Summing the first few terms gives a good approximation quickly. Similarly, expanding (1+x)ⁿ can help approximate values of roots or powers.

二项式展开技巧可用于估值问题。例如，不借助计算器估计 (1.02)⁵，可将其写成 (1 + 0.02)⁵ 并展开：1 + 5×0.02 + 10×0.0004 + 10×0.000008 + 5×0.00000016 + 0.0000000032。求前几项的和即可迅速得到较好近似值。类似地，展开 (1+x)ⁿ 可帮助估算方根或幂次的值。

9. Handling Binomials with a Coefficient Not Equal to 1 | 处理系数不为 1 的二项式

When the binomial is of the form (kx + m)ⁿ, the exponent on the coefficient k must be carefully tracked. Every term includes k raised to the power (n-r). For instance, in (5x-2)³, the general term is C(3,r) (5x)³⁻ʳ (-2)ʳ. For r=1, the term is C(3,1)(5x)²(-2)¹ = 3 × 25x² × (-2) = -150x². A common mistake is forgetting to apply the exponent to the coefficient 5.

当二项式为 (kx + m)ⁿ 形式时，必须注意系数 k 的指数。每一项都包含 k 的 (n-r) 次幂。例如，在 (5x-2)³ 中，通项为 C(3,r) (5x)³⁻ʳ (-2)ʳ。当 r=1 时，项为 C(3,1)(5x)²(-2)¹ = 3 × 25x² × (-2) = -150x²。常见错误是忘记将指数应用于系数 5。

10. The Relationship Between Binomial Coefficients and Symmetry | 二项式系数的对称性

The coefficients in a binomial expansion are symmetric: the first coefficient equals the last, the second equals the second-last, and so on. This is because C(n,r) = C(n,n-r). For example, in (x+y)⁶, the coefficients are 1, 6, 15, 20, 15, 6, 1. This property can serve as a quick check of your work and reduces calculation if you are writing out the full expansion.

二项式展开的系数具有对称性：首项系数等于末项系数，第二项系数等于倒数第二项系数，依此类推。这是因为 C(n,r) = C(n,n-r)。例如，在 (x+y)⁶ 中，系数为 1, 6, 15, 20, 15, 6, 1。利用这一特性可以快速检查结果，并在书写完整展开式时减少计算量。

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

One frequent error is mishandling negative signs, especially when b is negative. Always use brackets: (-b)ʳ to maintain correct sign. Another mistake is confusing the power of a and b; remember the sum of exponents in each term is n. Also, when finding a specific term, ensure r is a whole number between 0 and n — if your calculated r is not an integer, the term does not exist.

一个常见错误是处理负号不当，特别是当 b 为负数时。务必使用括号：(-b)ʳ 以保持符号正确。另一个错误是混淆 a 和 b 的指数；记住每一项的指数之和为 n。此外，求特定项时，要确保 r 是 0 到 n 之间的整数——如果计算出的 r 不是整数，说明该项不存在。

Time-saving tip: use your calculator’s nCr function efficiently. Practice with past papers to get comfortable identifying terms quickly. In IGCSE, questions usually carry 2-4 marks each, so writing the general term earns partial credit even if arithmetic slips.

省时技巧：熟练使用计算器上的 nCr 功能。通过历年真题练习，提高快速定位项的能力。在 IGCSE 中，此类题目通常每题 2-4 分，因此写出通项即可获得部分分数，即使后续计算出现小失误。

12. Practice Examples with Solutions | 带解答的练习示例

Example 1: Expand (3x + y)⁴.
Solution: Using coefficients 1,4,6,4,1: = 1(3x)⁴ + 4(3x)³y + 6(3x)²y² + 4(3x)y³ + 1y⁴ = 81x⁴ + 108x³y + 54x²y² + 12xy³ + y⁴.

示例 1：展开 (3x + y)⁴。
解答：使用系数 1,4,6,4,1：= 1(3x)⁴ + 4(3x)³y + 6(3x)²y² + 4(3x)y³ + 1y⁴ = 81x⁴ + 108x³y + 54x²y² + 12xy³ + y⁴。

Example 2: Find the coefficient of x⁴ in the expansion of (2x² – 1/x)⁵.
Solution: General term C(5,r)(2x²)⁵⁻ʳ(-1/x)ʳ = C(5,r) 2⁵⁻ʳ x¹⁰⁻²ʳ (-1)ʳ x⁻ʳ = C(5,r) 2⁵⁻ʳ (-1)ʳ x¹⁰⁻³ʳ. Set 10-3r = 4 → r=2. Coefficient: C(5,2) × 2³ × (-1)² = 10 × 8 × 1 = 80.

示例 2：求 (2x² – 1/x)⁵ 展开式中 x⁴ 的系数。
解答：通项 C(5,r)(2x²)⁵⁻ʳ(-1/x)ʳ = C(5,r) 2⁵⁻ʳ x¹⁰⁻²ʳ (-1)ʳ x⁻ʳ = C(5,r) 2⁵⁻ʳ (-1)ʳ x¹⁰⁻³ʳ。令 10-3r = 4 → r=2。系数：C(5,2) × 2³ × (-1)² = 10 × 8 × 1 = 80。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IGCSE Maths Binomial Expansion Revision | IGCSE 数学：二项式展开 考点精讲