📚 AS Mathematics: Binomial Expansion – Key Points | AS 数学：二项式展开 考点精讲
Binomial expansion is a fundamental technique in AS Mathematics that transforms powers of two-term expressions into a sum of terms. Mastering the binomial theorem allows you to expand expressions efficiently, isolate specific coefficients, and solve problems that appear regularly in exam papers. This article breaks down the essential concepts, from Pascal’s triangle and combination notation to finding constant terms and handling rational powers, with clear bilingual explanations.
二项展开是 AS 数学中的基础技巧，可将两项式的幂展开为多项式之和。掌握二项式定理不仅能高效地展开表达式，还能提取特定项的系数，解决考试中反复出现的典型题目。本文从帕斯卡三角和组合记号讲起，直到求常数项与处理有理指数展开，用中英双语清晰讲解每一个核心考点。
1. What is a Binomial Expansion? | 什么是二项展开？
A binomial is an algebraic expression containing two terms, such as (a + b) or (3x – 2). When we raise a binomial to a power, say (a + b)n, we need to multiply it by itself n times. Doing this manually for higher powers is tedious, so we use systematic rules known as binomial expansion. The result is a polynomial with (n+1) terms, where each term has a coefficient derived from combinatorial choices.
二项式是含有两项的代数表达式，例如 (a + b) 或 (3x – 2)。当我们计算 (a + b)n 时，需要将其自身相乘 n 次，对于高次幂手工展开非常繁琐。因此我们使用一套系统性的规则——二项展开——将幂展开为一个包含 (n+1) 项的多项式，每一项的系数由组合数决定。
2. Pascal’s Triangle and Combinatorics | 帕斯卡三角与组合数
Before learning the binomial theorem, many students use Pascal’s triangle to find coefficients. In this triangle, each number is the sum of the two directly above it. The row corresponding to (a+b)n gives the binomial coefficients when n is a positive integer. For example, row 3 is 1, 3, 3, 1, which matches (a+b)3 = a3 + 3a2b + 3ab2 + b3. These coefficients are also given by the combination formula nCr = n! / [r!(n−r)!], read as ‘n choose r’.
在学习二项式定理之前，很多同学用帕斯卡三角来寻找系数。三角形的每一行数字都是上方两数之和，对应正整数 n 的 (a+b)n 展开系数。例如第 3 行是 1, 3, 3, 1，正好对应 (a+b)3 = a3 + 3a2b + 3ab2 + b3。这些系数也可以用组合公式 nCr = n! / [r!(n−r)!] 来表示，意为 ‘n 选 r’。
3. The Binomial Theorem for Positive Integer n | 正整数指数的二项式定理
For a positive integer n, the binomial theorem gives a compact formula: (a + b)n = Σr=0n nCr an−r br. In expanded form: an + nC1 an−1b + nC2 an−2b2 + … + bn. This theorem works for any term order; often we rewrite (a + bx)n by treating a and bx as the two parts. The power of a decreases as the power of b increases, while the sum of the exponents in each term is always n.
对于正整数 n，二项式定理给出简洁的求和形式：(a + b)n = Σr=0n nCr an−r br。展开后为 an + nC1 an−1b + nC2 an−2b2 + … + bn。无论变量顺序如何，这个定理都适用；我们通常把 (a + bx)n 中的 a 和 bx 视为两个部分。每一项中 a 的指数递减，b 的指数递增，且指数之和恒为 n。
(a + b)n = nC0 an + nC1 an−1b + nC2 an−2b2 + … + nCn bn
4. The General Term (r+1)th Term | 通项（第 r+1 项）
The general term in the expansion of (a + b)n is Tr+1 = nCr an−r br, where r = 0 gives the first term. When the binomial is of the form (a + bx)n, the term containing xr becomes Tr+1 = nCr an−r (bx)r = nCr an−r br xr. This representation is the key to extracting any coefficient you need—simply match the power of x to the required one.
在 (a + b)n 的展开中，通项（第 r+1 项）为 Tr+1 = nCr an−r br，其中 r = 0 给出第一项。若表达式形如 (a + bx)n，则含 xr 的项为 Tr+1 = nCr an−r (bx)r = nCr an−r br xr。这一形式是提取系数的核心工具——只需令 x 的指数等于所需次幂，即可锁定对应的项。
Tr+1 = nCr an−r (bx)r
5. Finding Coefficients of a Specific Power | 求特定次幂的系数
To find the coefficient of xk in (a + bx)n, set r = k in the general term, provided that 0 ≤ k ≤ n and a whole number r results. The coefficient is then nCk an−k bk. Example: in (2 + 3x)5, the coefficient of x3 requires r = 3. Coefficient = 5C3 × 22 × 33 = 10 × 4 × 27 = 1080. Always check whether the question asks for the term or just the coefficient.
要求 (a + bx)n 中 xk 的系数，可直接令通项中的 r = k（前提是 0 ≤ k ≤ n 且 r 为整数）。系数为 nCk an−k bk。例：在 (2 + 3x)5 中，求 x3 的系数，令 r = 3，系数 = 5C3 × 22 × 33 = 10 × 4 × 27 = 1080。解题时务必区分“求某一项”和“求该项的系数”。
6. The Term Independent of x (Constant Term) | 不含 x 的项（常数项）
When the binomial has positive and negative exponents of x, such as (x + 1/x2)9, we often need the constant term (the term independent of x). Set the exponent of x in the general term to zero and solve for r. In this example, the general term is 9Cr x9−r (1/x2)r = 9Cr x9−3r. Solving 9 − 3r = 0 gives r = 3, so the constant term is 9C3 = 84. This method works for any combination of powers.
当二项式中含有 x 的正、负指数时，例如 (x + 1/x2)9，常需要求出常数项（不含 x 的项）。方法是将通项中 x 的指数设为零，解出 r。本例通项为 9Cr x9−r (1/x2)r = 9Cr x9−3r。令 9 − 3r = 0 得 r = 3，常数项即为 9C3 = 84。这一策略适用于任何混合幂的情形。
Set x-exponent = 0 → solve for r → substitute into nCr an−r br
7. Sum of All Binomial Coefficients | 所有二项系数之和
A useful shortcut: to find the sum of all coefficients in the expansion of (a + bx)n, simply substitute x = 1. The expression becomes (a + b)n. For instance, the sum of coefficients in (3 − 2x)4 is (3 − 2)4 = 14 = 1. This trick also helps you check if your expansion is correct. Additionally, by setting x = −1, you can find the alternating sum of coefficients.
一条实用的捷径：要求 (a + bx)n 展开式中所有系数的和，只需令 x = 1，表达式变为 (a + b)n 即可。例如 (3 − 2x)4 的系数和为 (3 − 2)4 = 1。这一技巧也可用于检验展开是否正确。若令 x = −1，还能得到系数交替和（奇数项与偶数项系数之差）。
8. Binomial Expansion with Rational Powers (AS Extension) | 有理数指数的二项展开（AS 延伸）
If your AS syllabus includes the extension to rational n, you will see the infinite series expansion: (1 + x)n = 1 + nx + n(n−1)/2! x2 + n(n−1)(n−2)/3! x3 + … , valid for |x| < 1. For (a + bx)n, rewrite it as an (1 + (b/a)x)n and apply the series, but remember the condition |(b/a)x| < 1. This form is essential for approximating square roots or reciprocals in numerical contexts.
如果你的 AS 大纲包含有理指数的延伸内容，会用到无穷级数展开：(1 + x)n = 1 + nx + n(n−1)/2! x2 + n(n−1)(n−2)/3! x3 + … ，且要求 |x| < 1。对于 (a + bx)n，可先提取 an 写作 an (1 + (b/a)x)n 再展开，但必须验证收敛条件 |(b/a)x| < 1。这类展开常用于近似计算平方根或倒数。
(1 + x)n = 1 + nx + [n(n−1)/2!] x2 + [n(n−1)(n−2)/3!] x3 + … , |x| < 1
9. Common Errors and How to Avoid Them | 常见错误与应对技巧
Error 1: forgetting that r starts at 0. The xk term corresponds to r = k, not r = k−1. Error 2: miscalculating nCr — make sure you use factorial definition correctly, or use the nCr button on your calculator. Error 3: leaving out br when the binomial is (a + bx)n; always raise the entire ‘bx’ to power r. Error 4: sign mistakes with (a − bx)n. Since b is replaced by −b, the term includes (−1)r, leading to alternating signs. Tip: always write the general term first and double-check the exponent simplification.
错误 1：忘记 r 从 0 开始，xk 的项对应 r = k，而非 r = k−1。错误 2：计算 nCr 出错，应正确使用阶乘公式或计算器上的 nCr 功能。错误 3：展开 (a + bx)n 时遗漏 br，必须将整个 ‘bx’ 升到 r 次方。错误 4：处理 (a − bx)n 时符号混乱——由于 b 被替换为 −b，通项会出现 (−1)r，导致符号交替。建议：先写出通项，再仔细化简指数。
10. Worked Exam-Style Questions | 考题实例精解
Question 1: ‘Find the coefficient of x4 in the expansion of (1 + 2x)6.’
General term: Tr+1 = 6Cr 1
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