GCSE OCR Maths: Binomial Expansion | GCSE OCR 数学：二项式展开考点精讲

📚 GCSE OCR Maths: Binomial Expansion | GCSE OCR 数学：二项式展开考点精讲

The binomial expansion is a fundamental algebraic technique that allows us to expand expressions of the form (a + b)ⁿ, where n is a positive integer. In the OCR GCSE mathematics syllabus, mastering this topic is essential for handling higher‑tier problems involving polynomials, coefficients, and specific terms. This guide provides a thorough yet accessible breakdown of all the key concepts, reinforced with examples and exam‑style advice.

二项式展开是一个基础的代数技巧，让我们能够展开形如 (a + b)ⁿ（n 为正整数）的表达式。在 OCR GCSE 数学大纲中，掌握这一专题对于处理涉及多项式、系数和特定项的高阶问题至关重要。本指南将对所有关键概念进行透彻而通俗的拆解，并结合例题与考试风格的建议加以巩固。

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

A binomial is simply an algebraic expression containing two terms, such as (x + y) or (2a − 3b). Expanding a binomial raised to a positive integer power n means multiplying it out fully to remove the brackets and write the result as a sum of simpler terms. For small values of n, this can be done by repeated multiplication, but for larger n we need a systematic method.

二项式就是包含两个项的代数表达式，例如 (x + y) 或 (2a − 3b)。将二项式进行正整数次幂 n 的展开，意味着把它完全乘开以去掉括号，并把结果写为若干更简单的项之和。对于较小的 n，可以通过多次乘法完成，但对于较大的 n，我们需要一种系统的方法。

2. Pascal’s Triangle and Coefficients | 帕斯卡三角形与系数

One of the earliest tools for binomial expansion is Pascal’s triangle. Each row of the triangle gives the coefficients of the expansion of (a + b)ⁿ for n = 0, 1, 2, 3, … The first and last entry in each row is 1, and every other number is the sum of the two numbers directly above it. For example, the expansion of (a + b)⁴ has coefficients 1, 4, 6, 4, 1, which you can read straight from the 4th row (if the top row is counted as row 0).

二项式展开最早的工具之一是帕斯卡三角形。三角形的每一行给出了 (a + b)ⁿ 当 n = 0, 1, 2, 3, … 时的展开系数。每行的第一个和最后一个数都是 1，其他每一个数都是它正上方两个数之和。例如，(a + b)⁴ 的展开系数为 1, 4, 6, 4, 1，你可以直接从第 4 行（如果将第一行算作第 0 行）读出来。

Below is a quick look at the first few rows of Pascal’s triangle, which you might often recreate quickly in an exam:

以下是帕斯卡三角形前几行的快速展示，你在考试中可以迅速画出：

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

Remember that while Pascal’s triangle is handy for small powers, using it for n = 10 or higher becomes time‑consuming, which is why we introduce the nCr notation.

请记住，虽然帕斯卡三角形在处理小指数时很方便，但对于 n = 10 或以上时就会变得耗时，因此我们要引入 nCr 记号。

3. Introducing the Binomial Coefficient (nCr) | 引入二项式系数（nCr）

The binomial coefficient, often written as ⁿCᵣ or ‘n choose r’, gives the number of ways to choose r items from n without regard to order. In expansion, it represents the coefficient of aⁿ⁻ʳ bʳ. On your calculator, you find it using the nCr button. For example, ⁵C₂ = 10, which matches the third term’s coefficient in (a + b)⁵.

二项式系数，通常写作 ⁿCᵣ 或 “n 选 r”，表示从 n 个物品中不考虑顺序地选出 r 个的方法数。在展开式中，它代表 aⁿ⁻ʳ bʳ 的系数。在计算器上，你可以使用 nCr 按钮求得。例如，⁵C₂ = 10，这与 (a + b)⁵ 中第三项的系数一致。

The formula for ⁿCᵣ is given by ⁿCᵣ = n! / [r!(n − r)!], but at GCSE level you will mainly use the calculator function or Pascal’s triangle for small n. Understanding the link between ⁿCᵣ and the triangle is important: the r‑th entry (starting from r = 0) in row n of Pascal’s triangle is exactly ⁿCᵣ.

ⁿCᵣ 的公式为 ⁿCᵣ = n! / [r!(n − r)!]，但在 GCSE 阶段你主要使用计算器功能或针对较小的 n 使用帕斯卡三角形。理解 ⁿCᵣ 与三角形之间的联系很重要：帕斯卡三角形第 n 行中的第 r 个数（从 r = 0 开始）正是 ⁿCᵣ。

4. The Binomial Theorem | 二项式定理

The binomial theorem provides the full expansion formula for any positive integer n:

二项式定理给出了针对任何正整数 n 的全展开公式：

(a + b)ⁿ = ⁿC₀ aⁿ b⁰ + ⁿC₁ aⁿ⁻¹ b¹ + ⁿC₂ aⁿ⁻² b² + … + ⁿCᵣ aⁿ⁻ʳ bʳ + … + ⁿCₙ a⁰ bⁿ

Notice the pattern: the powers of a decrease from n to 0, the powers of b increase from 0 to n, and each term picks up a binomial coefficient ⁿCᵣ. The term with aⁿ⁻ʳ bʳ is often called the general term.

注意其中的规律：a 的指数从 n 递减至 0，b 的指数从 0 递增至 n，每一项都带有一个二项式系数 ⁿCᵣ。含 aⁿ⁻ʳ bʳ 的那一项通常被称为通项。

5. Expanding (a + b)ⁿ in Full | 完全展开 (a + b)ⁿ

Let’s see the theorem in action by expanding (x + 2)³. Here n = 3, and the coefficients from the 3rd row of Pascal’s triangle are 1, 3, 3, 1. Applying the pattern:

我们来实际运用这一定理展开 (x + 2)³。这里 n = 3，帕斯卡三角形第 3 行的系数为 1, 3, 3, 1。应用这个规律：

(x + 2)³ = ³C₀ x³·2⁰ + ³C₁ x²·2¹ + ³C₂ x¹·2² + ³C₃ x⁰·2³

Evaluate each coefficient and power: ³C₀ = 1, ³C₁ = 3, ³C₂ = 3, ³C₃ = 1. So we get:

计算每一个系数和幂：³C₀ = 1, ³C₁ = 3, ³C₂ = 3, ³C₃ = 1。于是我们得到：

(x + 2)³ = 1·x³·1 + 3·x²·2 + 3·x·4 + 1·1·8 = x³ + 6x² + 12x + 8

Always remember to apply the power to both the constant and the variable part when the second term is a number, like the 2² and 2³ above.

始终要记住，当第二项是一个数字时，要将幂同时作用于该常数，像上面的 2² 和 2³ 那样。

6. Expanding (1 + x)ⁿ | 展开 (1 + x)ⁿ

A very common special case is (1 + x)ⁿ. By setting a = 1 and b = x in the theorem, the expansion simplifies significantly because all powers of 1 are just 1. The result is:

一个很常见的特例是 (1 + x)ⁿ。在定理中令 a = 1、b = x，展开式会大大简化，因为 1 的任何次幂都是 1。结果是：

(1 + x)ⁿ = 1 + ⁿC₁ x + ⁿC₂ x² + ⁿC₃ x³ + … + xⁿ

For example, (1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵. This form appears regularly in questions that ask for the first few terms or the coefficient of a particular power of x.

例如，(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵。这种形式经常出现在要求前几项或求 x 某次幂系数的题目中。

7. Handling (a + bx)ⁿ Type | 处理 (a + bx)ⁿ 类型

When the binomial is of the form (a + bx)ⁿ, you can treat it exactly like (a + b)ⁿ by letting the second term be ‘bx’. The general term becomes ⁿCᵣ aⁿ⁻ʳ (bx)ʳ = ⁿCᵣ aⁿ⁻ʳ bʳ xʳ. Be particularly careful with the power on b when finding the coefficient of xʳ.

当二项式形如 (a + bx)ⁿ 时，你可以像处理 (a + b)ⁿ 那样处理它，只需将第二项视为 “bx”。通项变为 ⁿCᵣ aⁿ⁻ʳ (bx)ʳ = ⁿCᵣ aⁿ⁻ʳ bʳ xʳ。在求 xʳ 的系数时要特别留意 b 的幂次。

Worked example: Find the coefficient of x³ in the expansion of (2 + 3x)⁵.

解答示例：求 (2 + 3x)⁵ 的展开式中 x³ 的系数。

We need the term with x³, so the exponent of (3x) must be 3, i.e. r = 3. The term is ⁵C₃ · 2⁵⁻³ · (3x)³ = ⁵C₃ · 2² · 27x³ = 10 · 4 · 27 x³ = 1080 x³. Hence the coefficient is 1080.

我们需要含有 x³ 的项，因此 (3x) 的指数必须为 3，即 r = 3。该项为 ⁵C₃ · 2⁵⁻³ · (3x)³ = ⁵C₃ · 2² · 27x³ = 10 · 4 · 27 x³ = 1080 x³。因此系数是 1080。

8. Finding a Specific Term | 求特定项

Instead of writing out the full expansion, exam questions very often direct you to find a single term, such as the term independent of x (constant term) or the term in x². The key is to identify the correct value of r by looking at the power of x you require.

考试题目常常要求你找出某一项，而不是写出完整的展开式，比如与 x 无关的项（常数项）或含 x² 的项。关键在于通过你所需要的 x 指数来确定正确的 r 值。

For the expansion of (1 + 2x)⁷, to find the term in x⁴ we set r = 4 because the x⁴ arises from (2x)⁴. The term is ⁷C₄ · 1⁷⁻⁴ · (2x)⁴ = 35 · 1 · 16 x⁴ = 560 x⁴.

对于 (1 + 2x)⁷ 的展开式，要求 x⁴ 的项，我们设 r = 4，因为 x⁴ 来自 (2x)⁴。该项为 ⁷C₄ · 1⁷⁻⁴ · (2x)⁴ = 35 · 1 · 16 x⁴ = 560 x⁴。

When the question asks for the constant term (x⁰), you must determine the value of r that makes the exponent of x equal to zero. This sometimes happens when both parts of the binomial contain powers of x.

当题目要求常数项（x⁰）时，你必须确定使 x 指数为零的 r 值。这种情况有时发生在二项式的两部分都含有 x 的幂次时。

9. Using the General Term Formula | 使用通项公式

The general term (the (r+1)th term) of (a + b)ⁿ is given by:

(a + b)ⁿ 的通项（第 (r+1) 项）由下式给出：

Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ

In this formula, r starts at 0 for the first term. To use it efficiently, write down the given a and b carefully, substitute them, and then simplify. If you need the term in xᵏ, equate the exponent of x to k and solve for r. Then plug r back to find the coefficient.

在此公式中，r 从 0 开始对应第一项。为高效使用它，仔细写下给定的 a 和 b，代入，然后化简。如果你需要 xᵏ 的项，就让 x 的指数等于 k，解出 r，再将 r 代回求出系数。

For example, in (x² + 1/x)⁸, the general term is ⁸Cᵣ (x²)⁸⁻ʳ (x⁻¹)ʳ = ⁸Cᵣ x¹⁶⁻²ʳ · x⁻ʳ = ⁸Cᵣ x¹⁶⁻³ʳ. To find the term independent of x, set 16 − 3r = 0 ⇒ r = 16/3, which is not an integer, so there is no constant term. This type of reasoning is frequently tested.

例如，在 (x² + 1/x)⁸ 中，通项为 ⁸Cᵣ (x²)⁸⁻ʳ (x⁻¹)ʳ = ⁸Cᵣ x¹⁶⁻²ʳ · x⁻ʳ = ⁸Cᵣ x¹⁶⁻³ʳ。要找到与 x 无关的项，令 16 − 3r = 0 ⇒ r = 16/3，不是整数，所以没有常数项。这类推理经常被考查。

10. Common Exam Question Types | 常见考题类型

OCR GCSE papers typically feature binomial expansion in the following forms:

OCR GCSE 试题通常以下列形式考查二项式展开：

Straightforward expansion: “Expand (2x − 3)⁴ completely.”

直接展开：“将 (2x − 3)⁴ 完全展开。”
Find a specific coefficient: “Find the coefficient of x² in the expansion of (1 − 2x)⁶.”

求特定系数：“求 (1 − 2x)⁶ 的展开式中 x² 的系数。”
Evaluating binomial expressions: “Use the expansion of (1 + x)⁵ to estimate (1.02)⁵.”

利用二项式估值：“利用 (1 + x)⁵ 的展开式估算 (1.02)⁵。”
Combining with algebraic manipulation: “In the expansion of (1 + kx)(2 + x)⁶, the coefficient of x² is 0. Find k.”

与代数运算结合：“在 (1 + kx)(2 + x)⁶ 的展开式中，x² 的系数为 0。求 k。”

Practise each type to build speed and accuracy, and always check your nCr values with a calculator to avoid arithmetic slips.

练习每种题型以提高速度和准确性，并始终用计算器检验你的 nCr 值，以避免计算失误。

11. Pitfalls and How to Avoid Them | 陷阱与避免方法

Many marks are lost due to simple mistakes. Watch out for these common errors:

许多分数是由于简单的错误而丢掉的。提防以下常见错误：

Forgetting that the coefficient ⁿCᵣ multiplies everything, including any numbers already inside the bracket. In (3x)⁴, the coefficient is 3⁴, not just 3.

忘记系数 ⁿCᵣ 要乘以一切，包括括号内已有的数字。在 (3x)⁴ 中，系数是 3⁴，而不仅仅是 3。
Applying the power to only one factor: (2x)³ is 8x³, not 2x³.

只对一个因子施加幂次：(2x)³ 是 8x³，而不是 2x³。
Misaligning Pascal’s triangle with the term index: row n gives coefficients for (a + b)ⁿ, but the first coefficient is for aⁿ, not aⁿ⁻¹.

将与项序号对应的帕斯卡三角形行弄错：第 n 行给出的是 (a + b)ⁿ 的系数，但第一个系数对应的是 aⁿ，而不是 aⁿ⁻¹。

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GCSE OCR Maths: Binomial Expansion | GCSE OCR 数学：二项式展开 考点精讲