📚 Binomial Expansion: CIE A-Level Maths Key Points | CIE A-Level 数学二项式展开考点精讲
The binomial expansion is a fundamental topic in CIE A-Level Mathematics (both Pure 1 and Pure 3), appearing frequently in algebraic manipulation, series approximations and real-world modelling. Mastering the binomial theorem for positive integer powers and its extension to rational indices is essential for scoring highly on Paper 1 and Paper 3.
二项式展开是 CIE A-Level 数学（包括纯数 1 和纯数 3）的基础主题，经常出现于代数化简、级数近似和实际建模中。掌握正整数幂的二项式定理以及向有理指数的推广，对于在试卷 1 和试卷 3 取得高分至关重要。
1. Why Binomial Expansion Matters in CIE A-Level | 为什么二项式展开在 CIE A-Level 中重要
In syllabus 9709, binomial expansion is examined in Paper 1 for positive integer powers only, while Paper 3 extends the same idea to negative and fractional indices and links it to approximation and convergence. Because the formula is printed in the formula booklet, examiners test technique rather than memory: correct structure, correct binomial coefficients and correct signs.
在 9709 考纲中，试卷 1 只考正整数幂的二项式展开，而试卷 3 把同一思想推广到负指数与分数指数，并将其与近似计算、收敛性联系起来。由于公式印在公式册中，考官考的是技巧而非记忆：结构是否正确、二项式系数是否正确、符号是否正确。
- Paper 1: full expansion, single coefficient, unknown constants | 试卷 1：完整展开、单项系数、求未知常数
- Paper 3: (1 + x)ⁿ with rational n, validity range, numerical approximation | 试卷 3：有理指数下的 (1 + x)ⁿ、有效性范围、数值近似
- Later use: binomial probability, series in calculus, modelling compound growth | 后续用途：二项分布概率、微积分中的级数、复利增长建模
Most marks are lost through careless arithmetic rather than through misunderstanding, so this topic rewards disciplined, step-by-step writing.
大多数失分来自粗心计算而不是概念误解，因此这个主题特别奖励按步骤书写、条理清晰的作答习惯。
2. Binomial Theorem for Positive Integer Powers | 正整数幂的二项式定理
expansion of (a + b)ⁿ is given by the sum: (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCᵣ aⁿ⁻ʳ bʳ + … + ⁿCₙ bⁿ, where the binomial coefficient is ⁿCᵣ = n! / (r!(n − r)!).
对于正整数 n，(a + b)ⁿ 的展开由以下求和给出：(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCᵣ aⁿ⁻ʳ bʳ + … + ⁿCₙ bⁿ，其中二项式系数 ⁿCᵣ = n! / (r!(n − r)!)。
(a + b)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ aⁿ⁻ʳ bʳ
The expansion contains n + 1 terms, the powers of a decrease from n to 0, and the powers of b increase from 0 to n. The two powers always add up to n in every single term, which is the quickest way to check that your expansion is internally consistent.
展开式共有 n + 1 项，a 的幂从 n 递减到 0，b 的幂从 0 递增到 n。每一项中两个幂次之和恒等于 n，这是检查展开式是否自洽最快的方法。
Eg: (1 + x)³ = 1 + 3x + 3x² + x³
Notice that the coefficient never appears without the accompanying powers of a and b, which is the single most common written error in examination scripts.
注意系数永远不能脱离 a 与 b 的幂次单独出现，这是试卷书写中最常见的一类错误。
3. Binomial Coefficients and Factorial Notation | 二项式系数与阶乘记号
t ⁿCᵣ counts the number of ways of choosing r objects from n, and it is defined by ⁿCᵣ = n! / (r!(n − r)!), where n! = n × (n − 1) × (n − 2) × … × 2 × 1 and, by convention, 0! = 1.
二项式系数 ⁿCᵣ 表示从 n 个对象中取出 r 个的组合数，其定义为 ⁿCᵣ = n! / (r!(n − r)!)，其中 n! = n × (n − 1) × (n − 2) × … × 2 × 1，并按约定 0! = 1。
ⁿC₀ = 1, ⁿC₁ = n, ⁿC₂ = n(n − 1) / 2, ⁿCₙ = 1
In the examination you may use the calculator function nCr directly, but you should be able to quote the first three coefficients instantly, because most questions only need ⁿC₁, ⁿC₂ and ⁿC₃.
考试中你可以直接使用计算器的 nCr 功能，但应能瞬间写出前三个系数，因为大多数题目只用到 ⁿC₁、ⁿC₂ 和 ⁿC₃。
- ⁸C₂ = 8 × 7 / 2 = 28 | ⁸C₂ = 8 × 7 / 2 = 28
- ⁸C₃ = 8 × 7 × 6 / 6 = 56 | ⁸C₃ = 8 × 7 × 6 / 6 = 56
- ⁸C₅ = ⁸C₃ = 56 (symmetry) | ⁸C₅ = ⁸C₃ = 56（对称性）
When a question involves a large n with a small r, always cancel factorials by hand instead of computing enormous numbers first; this reduces both time and error.
当题目中 n 很大而 r 很小时，应先用阶乘约分而不是先算出巨大的数字，这样既省时间又减少错误。
4. Symmetry and Pascal's Triangle | 系数对称性与杨辉三角
The coefficients are symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ, which is why the numbers in any row of Pascal' ymmetry is extremely useful for checking answers and for reading off coefficients quickly.
系数具有对称性：ⁿCᵣ = ⁿCₙ₋ᵣ，这就是杨辉三角（帕斯卡三角）每一行的数字从左读到右与从右读到左完全相同的原因。这一对称性在检验答案和快速读出系数时极为有用。
1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1
Pascal's triangle is built by adding two adjacent entries to obtain the entry below, and the row indexed by n gives the coefficients of (a + b)ⁿ. For small values of n this is faster than using the nCr formula.
杨辉三角的构造方法是把相邻两个数相加得到下一行的数，第 n 行给出的正是 (a + b)ⁿ 的系数。当 n 较小时，用它比用 nCr 公式更快。
| n | Coefficients | 系数 | Sum | 和 |
| 2 | 1, 2, 1 | 4 |
| 3 | 1, 3, 3, 1 | 8 |
| 4 | 1, 4, 6, 4, 1 | 16 |
| 5 | 1, 5, 10, 10, 5, 1 | 32 |
which gives a famous one-line proof check: substitute a = b = 1 into the binomial theorem.
第 n 行系数之和等于 2ⁿ，这给出一个著名的一行验证方法：在二项式定理中代入 a = b = 1。
5. The General Term and Specific Coefficients | 通项与特定系数
Any single term of the expansion can be written directly as Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ to jump straight to the x⁵ term of a large expansion without writing every preceding term.
展开式中任意一项可直接写为通项 Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ，其中 r 从 0 开始计数。利用通项，你可以直接跳到很大的展开式中的 x⁵ 项，而无须写出前面所有项。
Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
Strategy: identify a and b, write the general term including all signs and constants inside the bracket, form an equation for the required power, solve for r, then substitute back.
策略：先确定 a 与 b，写出含括号内所有符号与常数的通项，再用所需幂次建立方程，解出 r，最后代回求值。
oefficient of x³ in the expansion of (2 + 3x)⁶. Here a = 2, b = 3x, so the term is ⁶Cᵣ 2⁶⁻ʳ (3x)ʳ. We need r = 3, giving ⁶C₃ × 2³ × 3³ = 20 × 8 × 27 = 4320.
例：求 (2 + 3x)⁶ 展开式中 x³ 的系数。此处 a = 2，b = 3x，故该项为 ⁶Cᵣ 2⁶⁻ʳ (3x)ʳ。需要 r = 3，于是得 ⁶C₃ × 2³ × 3³ = 20 × 8 × 27 = 4320。
Do not forget that the constant 3 inside b must be raised to the power r as well as the x; forgetting this cube is a classic examination error.
别忘了 b 中的常数 3 也要取 r 次幂，而不只是 x 取幂；漏掉这个三次方是经典考试错误。
6. Worked Example: Expanding Fully | 完整展开的例题
Expand (3 − 2x)⁵ completely. Write the terms in ascending powers of x, showing the binomial coefficients explicitly before evaluating them.
完整展开 (3 − 2x)⁵。按 x 的升幂写出各项，先明确写出二项式系数再求值。
(3 − 2x)⁵ = 243 − 810x + 1080x² − 720x³ + 240x⁴ − 32x⁵
(−8)x³ = −720x³; term 4 is ⁵C₄ 3 (−2x)⁴ = 5 × 3 × 16x⁴ = 240x⁴; term 5 is (−2x)⁵ = −32x⁵.
分步过程：第 0 项为 3⁵ = 243；第 1 项为 ⁵C₁ 3⁴ (−2x) = 5 × 81 × (−2)x = −810x；第 2 项为 ⁵C₂ 3³ (−2x)² = 10 × 27 × 4x² = 1080x²；第 3 项为 ⁵C₃ 3² (−2x)³ = 10 × 9 × (−8)x³ = −720x³；第 4 项为 ⁵C₄ 3 (−2x)⁴ = 5 × 3 × 16x⁴ = 240x⁴；第 5 项为 (−2x)⁵ = −32x⁵。
Signs alternate because the negative sign inside the bracket is raised to an odd power on odd-numbered terms. Always track this separately to avoid sign slips.
由于括号内的负号在奇数项上取奇次幂，所以符号交替变化。务必单独跟踪符号，避免正负号出错。
7. Finding Unknown Constants | 由已知系数求未知常数
Many Paper 1 questions give one coefficient and ask for an unknown constant. The method is always the same: write the general term, isolate the required power, and equate the resulting expression to the given value.
许多试卷 1 的题目给出一个系数并要求求未知常数。方法始终一致：写出通项，分离出所需幂次，再把得到的表达式与给定值相等。
² term is ⁶C₂ (ax)² = 15a²x², so 15a² = 240, giving a² = 16 and therefore a = 4 or a = −4.
例：在 (1 + ax)⁶ 的展开式中，x² 的系数为 240，求 a 的所有可能值。x² 项为 ⁶C₂ (ax)² = 15a²x²，所以 15a² = 240，得 a² = 16，因此 a = 4 或 a = −4。
Note that the question asks for values, not a value, precisely because a squared unknown produces two answers. Reading the command word carefully is worth easy marks.
注意题目问的是"值"（复数）而不是"一个值"，正是因为未知量以平方形式出现会给出两个答案。仔细读题指令词就能轻松拿分。
A harder variant may involve two unknown constants with two simultaneous conditions, for example the coefficients of x and x² both given; solve the linear equation first and substitute into the quadratic one.
更难的一种变式涉及两个未知常数与两个联立条件，例如同时给出 x 与 x² 的系数；应先解一次方程，再代入二次方程求解。
8. The Standard Form (1 + x)ⁿ | 标准形式 (1 + x)ⁿ
most examinable version of the theorem is the standard form (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …, valid for any real n when |x| < 1.
最常考的版本是标准形式 (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …，当 |x| < 1 时对任意实数 n 成立。
(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …
x)ⁿ, then apply the standard expansion to the bracket. This step makes the constant and the variable roles explicit.
对于 (a + bx)ⁿ 形式的括号，先提取 a：(a + bx)ⁿ = aⁿ (1 + (b/a)x)ⁿ，再对括号应用标准展开式
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