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

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

Binomial expansion is a core algebraic skill in the IGCSE OCR Mathematics syllabus. It allows you to multiply out expressions of the form (a + b)ⁿ quickly without tedious repeated multiplication. Mastering this topic not only boosts your algebra marks but also lays the foundation for more advanced work in series and probability. This guide walks you through every key concept, common pitfalls and exam-style strategies you need to succeed.

二项式展开是 IGCSE OCR 数学大纲中的核心代数技能。它能帮助你快速展开 (a + b)ⁿ 这类表达式，无需重复进行繁琐的乘法运算。掌握这一知识点不仅能提高你在代数部分的得分，也为后续的级数和概率学习打下基础。本指南将带你梳理每一个关键概念、常见错误以及应对考试真题的策略。

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 + y) or (2a − 3b). Binomial expansion is the process of raising a binomial to a positive integer power and writing the result as a sum of terms. For instance, (x + 3)² expands to x² + 6x + 9. The expansion follows a predictable pattern, which saves time and reduces errors compared to multiplying out bracket by bracket.

二项式是指恰好包含两项的代数表达式，例如 (x + y) 或 (2a − 3b)。二项式展开是指将一个二项式进行正整数次幂运算，并将结果写成若干项之和的过程。比如 (x + 3)² 展开后得到 x² + 6x + 9。展开遵循一种可预测的规律，相比逐项相乘能节省时间并减少错误。

In the IGCSE OCR exam, you will be expected to expand binomials up to powers like 4, 5 or even higher using efficient methods. You must also be able to find a specific term within an expansion without writing out the entire polynomial. This relies heavily on understanding patterns, Pascal’s triangle and the binomial coefficients.

在 IGCSE OCR 考试中，你将被要求使用高效的方法将二项式展开到 4 次、5 次甚至更高的幂。你还必须能够在展开式中找到某一特定项，而无需写出整个多项式。这主要依赖于对规律、帕斯卡三角形以及二项式系数的理解。

2. Pascal’s Triangle | 帕斯卡三角形

Pascal’s triangle is a triangular array of numbers where each entry is the sum of the two numbers directly above it. The rows give the coefficients of the expanded binomial. Row 0 corresponds to (a + b)⁰ = 1, row 1 gives coefficients 1, 1 for (a + b)¹, row 2 gives 1, 2, 1 for (a + b)², and so on. This pattern continues indefinitely, making it easy to expand low-power binomials by hand.

帕斯卡三角形是一个数字三角形阵列，其中每个数字等于它正上方两个数字之和。每一行给出了二项式展开的系数。第 0 行对应 (a + b)⁰ = 1；第 1 行给出 (a + b)¹ 的系数 1, 1；第 2 行给出 (a + b)² 的系数 1, 2, 1；依此类推。这个规律无限延伸，使得手动展开低次幂的二项式变得非常容易。

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

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

While Pascal’s triangle works nicely for n ≤ 5, it becomes impractical for larger powers. That is where the binomial theorem using combinations (nCr) takes over. OCR questions often ask you to use the nCr method directly, especially when finding a single term.

虽然帕斯卡三角形对 n ≤ 5 时非常方便，但对于更大的指数就不太实用了。这时就需要使用组合数 (nCr) 的二项式定理来解决。OCR 题目常常要求你直接使用 nCr 方法，特别是在求单项时。

3. Combinations and Binomial Coefficients | 组合数与二项式系数

The binomial coefficient C(n, r), often pronounced “n choose r”, tells you how many ways you can choose r items from a set of n without regard to order. In binomial expansion, it represents the coefficient of the term containing bʳ. The formula is C(n, r) = n! / [r! (n − r)!], but on your OCR calculator you can use the nCr button directly.

二项式系数 C(n, r)，常读作 “n 选 r”，表示从 n 个不同元素中选取 r 个元素而不考虑顺序的方法数。在二项式展开中，它代表含有 bʳ 项的系数。计算公式为 C(n, r) = n! / [r! (n − r)!]，但在你的 OCR 计算器上可以直接使用 nCr 按钮。

For example, C(5, 2) = 10, which matches the third entry in row 5 of Pascal’s triangle. In an expansion of (a + b)⁵, the coefficient of a³b² is C(5, 2) = 10. Note the symmetry: C(n, r) = C(n, n − r), so coefficients read the same backwards and forwards.

例如，C(5, 2) = 10，正好对应帕斯卡三角形第 5 行的第三个数字。在 (a + b)⁵ 的展开中，a³b² 项的系数就是 C(5, 2) = 10。请注意对称性：C(n, r) = C(n, n − r)，因此系数从前往后读和从后往前读是一样的。

Understanding this connection between Pascal’s triangle and combinations is essential. It allows you to generate any coefficient without drawing out the triangle, and it underpins the general binomial formula used in all IGCSE OCR questions.

理解帕斯卡三角形与组合数之间的联系至关重要。它让你无需画出三角形就能得出任意系数，也为所有 IGCSE OCR 题目中通用的二项式展开公式打下了基础。

4. The Expansion Formula for (a + b)ⁿ | (a + b)ⁿ 的展开公式

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

This formula states that you start with aⁿ and finish with bⁿ. The powers of a decrease from n to 0 while powers of b increase from 0 to n. The coefficient of each term is the binomial coefficient C(n, r) where r is the power of b in that term. The general term, counting from r = 0, is C(n, r) aⁿ⁻ʳ bʳ.

这个公式表明展开式从 aⁿ 开始，到 bⁿ 结束。a 的指数从 n 递减至 0，b 的指数从 0 递增至 n。每一项的系数是二项式系数 C(n, r)，其中 r 就是该项中 b 的指数。一般项（从 r = 0 开始计数）为 C(n, r) aⁿ⁻ʳ bʳ。

In IGCSE OCR, you will often see expansions of the form (1 + x)ⁿ or (2x − 3)ⁿ. The same formula applies, but careful attention must be paid to signs and to the coefficients of the terms inside the brackets. A negative sign in the binomial introduces alternating signs in the expansion.

在 IGCSE OCR 考试中，你常常会遇到 (1 + x)ⁿ 或 (2x − 3)ⁿ 这类展开式。同一公式依然适用，但必须格外注意括号内各项的符号和系数。二项式中的负号会使展开式出现正负号交替的情况。

5. Step-by-Step Expansion | 展开步骤分解

To expand (x + 2)⁴ using the formula, first identify a = x, b = 2 and n = 4. Write down the six terms from r = 0 to r = 4 using the general form: C(4,0)x⁴(2)⁰ + C(4,1)x³(2)¹ + C(4,2)x²(2)² + C(4,3)x¹(2)³ + C(4,4)x⁰(2)⁴. Then replace each binomial coefficient with its value: 1x⁴ + 4x³·2 + 6x²·4 + 4x·8 + 1·16. Finally simplify the arithmetic: x⁴ + 8x³ + 24x² + 32x + 16.

要展开 (x + 2)⁴，首先确定 a = x，b = 2，n = 4。利用一般形式写出从 r = 0 到 r = 4 的六项：C(4,0)x⁴(2)⁰ + C(4,1)x³(2)¹ + C(4,2)x²(2)² + C(4,3)x¹(2)³ + C(4,4)x⁰(2)⁴。然后将每个二项式系数替换为具体数值：1x⁴ + 4x³·2 + 6x²·4 + 4x·8 + 1·16。最后化简算术部分：x⁴ + 8x³ + 24x² + 32x + 16。

This systematic approach prevents missing terms or misplacing exponents. When the binomial is a subtraction, for example (2x − 1)³, treat b as (−1). The terms become C(3,0)(2x)³(−1)⁰ + C(3,1)(2x)²(−1)¹ + …, resulting in alternating signs: 8x³ − 12x² + 6x − 1.

这种系统化的方法可以防止遗漏项或放错指数位置。当二项式是减法时，例如 (2x − 1)³，将 b 视为 (−1)。各项变为 C(3,0)(2x)³(−1)⁰ + C(3,1)(2x)²(−1)¹ + …，从而得到正负号交替的展开式：8x³ − 12x² + 6x − 1。

Always remember that the exponent of a and b must sum to n for each term. This is a quick-check mechanism: in (x + 2)⁴, the powers of x and 2 in the term 4x³·2¹ sum to 3 + 1 = 4, which matches n. If the sum isn’t n, you have made an index error.

请始终记住，每一项中 a 和 b 的指数之和必须等于 n。这是一个快速检验机制：在 (x + 2)⁴ 中，项 4x³·2¹ 的指数和为 3 + 1 = 4，与 n 吻合。如果和不等于 n，就说明你在指数上出了错。

6. Finding a Specific Term | 求特定项

A classic IGCSE OCR exam question asks, “Find the coefficient of x⁶ in the expansion of (2x + 3)⁸.” You do not need to write the whole expansion. Instead, set up the general term C(8, r) (2x)⁸⁻ʳ (3)ʳ = C(8, r) 2⁸⁻ʳ x⁸⁻ʳ 3ʳ. The power of x is 8 − r. Set 8 − r = 6 to get r = 2. Substitute r = 2 to find the coefficient: C(8, 2) × 2⁶ × 3² = 28 × 64 × 9 = 16128.

一道典型的 IGCSE OCR 考题是：“求 (2x + 3)⁸ 展开式中 x⁶ 的系数。”你不需要写出整个展开式。只需先写出一般项 C(8, r)(2x)⁸⁻ʳ(3)ʳ = C(8, r) 2⁸⁻ʳ x⁸⁻ʳ 3ʳ。x 的指数为 8 − r。令 8 − r = 6 可得 r = 2。代入 r = 2 即可求出系数：C(8, 2) × 2⁶ × 3² = 28 × 64 × 9 = 16128。

This technique is extremely powerful and time-saving. It works for any power of x, for constant terms (set exponent of x to 0) and for finding the term independent of x. Always double-check that you used the correct r and simplified all arithmetic without losing factors from the original binomial.

这种方法十分强大且省时。它适用于任意的 x 指数，也适用于常数项（令 x 的指数为 0），以及求与 x 无关的项。务必仔细核对你是否使用了正确的 r，并且化简了所有算术部分，没有丢掉原二项式中的任何因数。

7. Coefficient and Constant Term Problems | 系数与常数项问题

Sometimes the question asks for the term independent of x, which is the constant term. For example, in (x + 1/x²)⁹, the general term is C(9, r) x⁹⁻ʳ (1/x²)ʳ = C(9, r) x⁹⁻ʳ x⁻²ʳ = C(9, r) x⁹⁻³ʳ. For a constant term, the exponent of x must be 0, so 9 − 3r = 0 ⇒ r = 3. The constant term is then C(9, 3) = 84.

有时题目要求找出与 x 无关的项，也就是常数项。例如，在 (x + 1/x²)⁹ 中，一般项为 C(9, r) x⁹⁻ʳ (1/x²)ʳ = C(9, r) x⁹⁻ʳ x⁻²ʳ = C(9, r) x⁹⁻³ʳ。欲得常数项，x 的指数必须为 0，所以 9 − 3r = 0 ⇒ r = 3。于是常数项为 C(9, 3) = 84。

Always combine the powers of x carefully using index laws. When the binomial includes fractional or negative exponents, the same principle applies, but such cases are rare at IGCSE level. The main challenge is solving the simple linear equation to find r, then correctly evaluating the binomial coefficient and any remaining constants.

一定要运用指数运算法则仔细整理 x 的幂。虽然二项式中含有分式或负指数的情况在 IGCSE 阶段较少出现，但基本原理相同。主要难点在于求解简单的一次方程以确定 r，然后正确计算二项式系数以及任何剩余常数。

8. Expressions with Coefficients | 含系数变量的表达式

When the binomial contains coefficients other than 1, such as (3x − 2y)⁵, you must apply the power to both the variable and the coefficient within the bracket. In the general term C(5, r) (3x)⁵⁻ʳ (−2y)ʳ, the coefficient involves 3⁵⁻ʳ and (−2)ʳ multiplied by C(5, r). Simplify these constants carefully before multiplying. This is where many students lose marks by forgetting to raise the numerical coefficient to the power.

当二项式中含有 1 以外的系数时，例如 (3x − 2y)⁵，你必须对括号内的变量和系数同时进行幂运算。在一般项 C(5, r)(3x)⁵⁻ʳ(−2y)ʳ 中，系数部分包括 3⁵⁻ʳ 和 (−2)ʳ 再乘以 C(5, r)。在相乘之前要仔细化简这些常数。许多学生正是因为忘记对数字系数进行幂运算而丢分。

For instance, to find the coefficient of x³y² in (3x − 2y)⁵, set r = 2 (since y-power is 2). The term is C(5,2) (3x)³ (−2y)² = 10 × 27x³ × 4y² = 1080 x³y². The coefficient is 1080. Take your time to write every step; exam markers allocate method marks for correct setup even if the final arithmetic contains a slip.

例如，要求 (3x − 2y)⁵ 中 x³y² 项的系数，令 r = 2（因为 y 的指数为 2）。该项为 C(5,2) (3x)³ (−2y)² = 10 × 27x³ × 4y² = 1080 x³y²。系数为 1080。请逐步写下每一步；即使最终算术出现小差错，阅卷老师也会对设置正确的式子给步骤分。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

One frequent error is forgetting that the binomial coefficient C(n, r) and the powers of the bracketed numbers are separate factors. Always write the general term as C(n, r) × (first term)ⁿ⁻ʳ × (second term)ʳ. Another mistake is using r incorrectly: r corresponds to the power of the second term b, not to the power of the first term a. Mixing these up leads to wrong coefficients and wrong powers.

一个常见错误是忘记二项式系数 C(n, r) 与括号内数字的幂是相互独立的因子。一定要将一般项写成 C(n, r) × (第一项)ⁿ⁻ʳ × (第二项)ʳ 的形式。另一个错误是误用 r：r 对应第二项 b 的指数，而不是第一项 a 的指数。混淆两者会导致系数和指数都出错。

Sign errors also plague many answers. When the second term is negative, (−b)ʳ will be positive for even r and negative for odd r. A quick sign check is to see whether the final expansion has alternating signs. You can also test your expansion by substituting a simple value (like x = 1) into both the original binomial and your expansion to check they match.

符号错误也困扰着许多答案。当第二项为负数时，(−b)ʳ 在 r 为偶数时为正，奇数时为负。快速检验符号的一种方法是观察最终展开式的符号是否交替出现。你还可以代入一个简单数值（如 x = 1）到原二项式和你得出的展开式中，检查两者是否相等。

Finally, be careful with brackets: (2x)³ = 8x³, not 2x³. A slip here changes the entire coefficient. Underline or highlight the bracket to remind yourself to apply the power to both the number and the variable. Using the calculator’s nCr function correctly is also vital – practise finding, say, C(8,3) quickly during revision.

最后，要当心括号：(2x)³ = 8x³，而不是 2x³。这里一旦出错会改变整个系数。可以在括号下划线或高亮显示，以提醒自己要对数字和变量同时进行幂运算。正确使用计算器的 nCr 功能也至关重要——复习时就要练习快速求出如 C(8,3) 这样的值。

10. Exam-Style Question Examples | 考试真题示例

Example 1: Write down the expansion of (1 + 2x)⁴ in ascending powers of x. Solution: Use a=1, b=2x, n=4. Terms: C(4,0)·1⁴·(2x)⁰ = 1; C(4,1)·1³·(2x)¹ = 8x; C(4,2)·1²·(2x)² = 24x²; C(4,3)·1¹·(2x)³ = 32x³; C(4,4)·1⁰·(2x)⁴ = 16x⁴. Answer: 1 + 8x + 24x² + 32x³ + 16x⁴.

示例 1：将 (1 + 2x)⁴ 按 x 的升幂展开。解：令 a=1, b=2x, n=4。各项为：C(4,0)·1⁴·(2x)⁰ = 1；C(4,1)·1³·(2x)¹ = 8x；C(4,2)·1²·(2x)² = 24x²；C(4,3)·1¹·(2x)³ = 32x³；C(4,4)·1⁰·(2x)⁴ = 16x⁴。答案：1 + 8x + 24x² + 32x³ + 16x⁴。

Example 2 (Targeted term): Find the coefficient of x³ in the expansion of (2 − 3x)⁷. General term: C(7, r) 2⁷⁻ʳ (−3x)ʳ = C(7, r) 2⁷⁻ʳ (−3)ʳ xʳ. For x³, r = 3. Coefficient = C(7,3) × 2⁴ × (−3)³ = 35 × 16 × (−27) = −15120. Note the negative sign. Common mistake: forgetting the minus sign inside the bracket.

示例 2（特定项）：求 (2 − 3x)⁷ 展开式中 x³ 的系数。一般项：C(7, r) 2⁷⁻ʳ (−3x)ʳ = C(7, r) 2⁷⁻ʳ (−3)ʳ xʳ。令 x³ 可得 r = 3。系数 = C(7,3) × 2⁴ × (−3)³ = 35 × 16 × (−27) = −15120。注意负号。常见错误：忘记括号内的负号。

Example 3 (Constant term): In the expansion of (x² + 1/x)⁹, find the term independent of x. General term: C(9, r) (x²)⁹⁻ʳ (1/x)ʳ = C(9, r) x¹⁸⁻²ʳ x⁻ʳ = C(9, r) x¹⁸⁻³ʳ. Set 18 − 3r = 0 ⇒ r = 6. Term = C(9, 6) = C(9,3) = 84. Because r is within 0 to 9, solution is valid.

示例 3（常数项）：在 (x² + 1/x)⁹ 的展开式中，求与 x 无关的项。一般项：C(9, r) (x²)⁹⁻ʳ (1/x)ʳ = C(9, r) x¹⁸⁻²ʳ x⁻ʳ = C(9, r) x¹⁸⁻³ʳ。令 18 − 3r = 0 ⇒ r = 6。项 = C(9, 6) = C(9,3) = 84。由于 r 在 0 到 9 之间，解有效。

11. Summary and Exam Tips | 总结与备考建议

Binomial expansion can be broken into three core skills: using Pascal’s triangle for small powers, applying the nCr formula for any power, and finding specific terms without full expansion. Make sure your calculator’s nCr function is second nature, and practise setting up the general term with the exact notation used in the exam.

二项式展开可以分解为三项核心技能：对小指数使用帕斯卡三角形；对任意指数应用 nCr 公式；以及在不进行完整展开的情况下找出特定项。确保你能熟练使用计算器的 nCr 功能，并练习用考试要求的准确符号写出一般项。

Always read the question carefully: does it ask for the ‘coefficient of x⁴’ or the ‘term in x⁴’? The term includes the x⁴ part and its sign; the coefficient is only the numerical factor. Check how many marks the question carries – a full expansion of a higher power might be quicker with direct nCr rather than Pascal’s triangle, but a 2-mark question might only need a single term extraction.

一定要仔细审题：题目要求的是 “x⁴ 的系数” 还是 “含有 x⁴ 的项”？项包含 x⁴ 部分及其符号；系数仅仅是数字因子。看清题目分值——高次幂的完整展开用 nCr 直接计算可能比帕斯卡三角形更快，但一道 2 分的题目可能只需要提取单项。

Finally, manage your time wisely during the exam. If a full expansion is requested, write down the general formula line first, then systematically plug in r = 0, 1, 2,… Keep your work tidy so you can re-trace steps if an answer check reveals a mistake. With targeted practice, binomial expansion can become one of your highest-scoring topics.

最后，考试中要合理安排时间。如果要完整展开，先写出一般公式行，然后系统地代入 r = 0, 1, 2,……保持卷面整洁，这样一旦发现答案有误，可以回溯步骤。通过有针对性的练习，二项式展开完全可以成为你得分率最高的专题之一。

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