📚 IGCSE Edexcel Maths: Binomial Expansion | IGCSE Edexcel 数学：二项式展开考点精讲
Binomial expansion is a cornerstone of IGCSE Edexcel algebra, allowing us to write expressions like (a+b)n as a sum of terms without lengthy multiplication. Mastering this topic gives you a powerful tool for simplifying polynomials, finding specific coefficients, and tackling real-world approximation problems.
二项式展开是 IGCSE Edexcel 代数的重要基础，它能把 (a+b)n 这样的表达式直接写成若干项的和，而无需进行冗长的乘法。掌握这个知识点，你就拥有了简化多项式、求特定系数以及解决实际近似问题的强大工具。
1. What is a Binomial? | 什么是二项式？
A binomial is an algebraic expression containing two terms, such as (x+y) or (2a−3b). The expansion of (a+b)n means writing it as a polynomial with (n+1) terms, each consisting of a coefficient, powers of a, and powers of b.
二项式是包含两个项的代数表达式，比如 (x+y) 或 (2a−3b)。(a+b)n 的展开就是把它写成一个含有 (n+1) 项的多项式，每一项由系数、a 的幂和 b 的幂组成。
For example, (x+2)3 expands to x3 + 6x2 + 12x + 8. Notice that the powers of x decrease from 3 to 0, while the powers of 2 (the constant) increase from 0 to 3.
例如，(x+2)3 展开后得到 x3 + 6x2 + 12x + 8。注意 x 的幂从 3 递减到 0，而常数 2 的幂从 0 递增到 3。
2. Building Coefficients with Pascal's Triangle | 用帕斯卡三角形构建系数
Pascal' ithout using factorials. Each row corresponds to the power n, starting with row 0 for n=0.
帕斯卡三角形可以快速找出二项式展开的系数，而无需使用阶乘。每一行对应幂次 n，从 n=0 的第 0 行开始。
| n=0 | 1 |
| n=1 | 1 1 |
| n=2 | 1 2 1 |
| n=3 | 1 3 3 1 |
| n=4 | 1 4 6 4 1 |
To expand (a+b)4, the coefficients are 1, 4, 6, 4, 1. The terms then follow the pattern a4 b0, a3 b1, a2 b2, a1 b3, a0 b4.
要展开 (a+b)4，系数依次为 1, 4, 6, 4, 1。各项随即按照 a4 b0、a3 b1、a2 b2、a1 b3、a0 b4 的模式排列。
3. Introducing Combinatorial Notation (nCr) | 组合数记法 nCr 入门
While Pascal's triangle is fine for small n, the Edexcel IGCSE syllabus expects you to use nCr notation, also written as C(n,r) or nCr. It represents the number of ways to choose r items from n, and is defined as n! / [r!(n−r)!].
帕斯卡三角形对于较小的 n 来说够用，但 Edexcel IGCSE 大纲要求你使用 nCr 记法，也写作 C(n,r) 或 nCr。它表示从 n 个物品中选出 r 个的组合数，定义为 n! / [r!(n−r)!]。
nCr = n! / [r!(n−r)!]
For instance, 5C2 = 5!/(2!3!) = 10. Notice that nC0 = 1 and nCn = 1. Also, symmetry holds: nCr = nCn−r.
例如，5C2 = 5!/(2!3!) = 10。注意 nC0 = 1 且 nCn = 1。此外，还有对称性：nCr = nCn−r。
4. The Binomial Theorem for Positive Integer n | 正整数指数 n 的二项式定理
The general binomial theorem states that for any positive integer n, the expansion of (a+b)n is a sum from r=0 to r=n of the terms nCr an−r br.
一般二项式定理指出，对于任何正整数 n，(a+b)n 的展开是从 r=0 到 r=n 的各项 nCr an−r br 的总和。
(a + b)n = ∑r=0n nCr an−r br
Here the term r=0 gives the first term an, r=1 gives nC1 an−1 b, and so on. Always remember that the powers of a decrease while those of b increase.
这里 r=0 给出第一项 an，r=1 给出 nC1 an−1 b，以此类推。始终要记住 a 的幂次递减而 b 的幂次递增。
5. Step-by-Step Expansion of (a+b)n | 逐步展开 (a+b)n
Let's expand (2x + 3)4. First identify a = 2x, b = 3, n = 4. The coefficients from n=4 in Pascal's triangle are 1, 4, 6, 4, 1, which match 4C0, 4C1, 4C2, 4C3, 4C4.
我们来展开 (2x + 3)4。先确定 a = 2x, b = 3, n = 4。帕斯卡三角形 n=4 行的系数是 1, 4, 6, 4, 1，它们分别对应 4C0, 4C1, 4C2, 4C3, 4C4。
Now build the terms:
现在构建各项：
r=0: 4C0 (2x)4 (3)0 = 1 × 16x4 × 1 = 16x4
r=0: 4C0 (2x)4 (3)0 = 1 × 16x4 × 1 = 16x4
r=1: 4C1 (2x)3 (3)1 = 4 × 8x3 × 3 = 96x3
r=1: 4C1 (2x)3 (3)1 = 4 × 8x3 × 3 = 96x3
r=2: 4C2 (2x)2 (3)2 = 6 × 4x2 × 9 = 216x2
r=2: 4C2 (2x)2 (3)2 = 6 × 4x2 × 9 = 216x2
r=3: 4C3 (2x)1 (3)3 = 4 × 2x × 27 = 216x
r=3: 4C3 (2x)1 (3)3 = 4 × 2x × 27 = 216x
r=4: 4C4 (2x)0 (3)4 = 1 × 1 × 81 = 81
r=4: 4C4 (2x)0 (3)4 = 1 × 1 × 81 = 81
Hence (2x+3)4 = 16x4 + 96x3 + 216x2 + 216x + 81. Notice how we carefully raised the whole '2x' and '3' to the required powers.
因此，(2x+3)4 = 16x4 + 96x3 + 216x2 + 216x + 81。注意我们是如何仔细地将整个 “2x” 和 “3” 升至所需次幂的。
6. The Special Case of (1+x)n | 特殊情况 (1+x)n 的展开
When the first term is 1, the expansion simplifies enormously because powers of 1 remain 1. The Edexcel IGCSE frequently tests this form, often with a coefficient inside the binomial, such as (1 + 2x)5.
当首项为 1 时，展开会大大简化，因为 1 的任何次幂仍然是 1。Edexcel IGCSE 常考这种形式，而且括号内通常带有系数，比如 (1 + 2x)5。
(1 + x)n = ∑r=0n nCr xr
So (1 + 2x)5 will have terms: 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + ... up to (2x)5. This can be written as 1 + 5(2x) + 10(4x2) + 10(8x3) + 5(16x4) + 1(32x5) = 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5.
因此 (1 + 2x)5 的各项依次为：5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + ... 直至 (2x)5。可写为 1 + 5(2x) + 10(4x2) + 10(8x3) + 5(16x4) + 1(32x5) = 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5。
7. Finding a Specific Term with the General Term Formula | 利用通项公式求指定项
In many exam questions you are asked to find a single term, not the full expansion. The general term Tr+1 is the (r+1)th term and is given by:
许多考题只要求找出某一项，而不是整个展开式。通项 Tr+1 表示第 (r+1) 项，其公式为：
Tr+1 = nCr an−r br
For the expansion of (3 − 2x)6, to find the term in x3 we set a = 3, b = −2x, n = 6. The power of x is r, so we need r = 3. Then T4 = 6C3 (3)6−3 (−2x)3 = 20 × 27 × (−8x3) = −4320 x3.
对于 (3 − 2x)6 的展开，要求含 x3 的项。设 a = 3, b = −2x, n = 6。x 的幂次为 r，所以需要 r = 3。则 T4 = 6C3 (3)6−3 (−2x)3 = 20 × 27 × (−8x3) = −4320 x3。
Always write the general term first, then set the power of x equal to the required index and solve for r. Don't forget to apply the coefficient and sign correctly.
务必先写出通项，然后令 x 的幂等于要求的指数，解出 r。不要忘记正确处理系数和符号。
8. Dealing with Negative Signs and Fractional Coefficients | 处理负号与分数系数
When b contains a negative sign, the term's sign alternates or follows the pattern of (−b)r. Be systematic: substitute the whole 'b' including its negative sign into the general term.
当 b 中含有负号时，该项的符号会交替变化，或遵循 (−b)r 的规律。要系统处理：将包含负号的完整 “b” 代入通项。
For (x − 1/2)4, b = −1/2. The term in x2 occurs when r = 2: T3 = 4C2 x4−2 (−1/2)2 = 6 × x2 × 1/4 = (3/2) x2.
对于 (x − 1/2)4，b = −1/2。含 x2 的项在 r = 2 时得到：T3 = 4C2 x4−2 (−1/2)2 = 6 × x2 × 1/4 = (3/2) x2。
e expansion asked for the constant term, we would need the power of x to be zero, so n−r=0 ⇒ r=4. T5 = 4C4 (−1/2)4 = 1/16.
如果同一展开式要求常数项，我们需要 x 的幂次为零，即 n−r=0 ⇒ r=4。T5 = 4C4 (−1/2)4 = 1/16。
9. Using Expansions for Approximations | 用展开式进行近似计算
IGCSE Edexcel sometimes links binomial expansion to estimation, for example using (1 + x)n with small x. If x is small, higher powers of x become negligible, allowing quick approximations.
IGCSE Edexcel 有时会将二项式展开与估算联系起来，比如利用 (1 + x)n 且 x 很小时进行近似。若 x 很小，x 的高次幂可忽略不计，从而快速得出近似值。
For instance, (1.002)5 can be written as (1 + 0.002)5. Expanding: 1 + 5(0.002) + 10(0.002)2 .01004. The exact value is about 1.01004, showing how accurate the first few terms can be.
例如，(1.002)5 可写成 (1 + 0.002)5。展开得：1 + 5(0.002) + 10(0.002)2 + ... ≈ 1 + 0.01 + 0.00004 = 1.01004。精确值约为 1.01004，可见前几项的近似已经相当准确。
Always state which terms you have omitted and explain why they are small enough to ignore in the context of the required accuracy.
务必要说明你省略了哪些项，并解释在所要求的精度下它们为什么小到可以忽略。
10. Solving Equations Involving Binomial Coefficients | 解含二项式系数的方程
Exam papers may give you a coefficient from an expansion and ask you to find an unknown constant. For example, in the expansion of (2 + kx)5, the coefficient of x3 is 720. Set up an equation using the general term.
试卷可能给出展开式中的某一系数，要你求出未知常数。例如，在 (2 + kx)5 的展开式中，x3 的系数是 720。利用通项建立方程。
The term in x3 occurs at r=3: T4 = 5C3 (2)5−3 (kx)3 = 10 × 4 × k3 x3 = 40 k3 x3. Set 40 k3 = 720 ⇒ k3 = 18 ⇒ k = ∛18. Simplify if necessary.
x3 项在 r=3 时得到：T4 = 5C3 (2)5−3 (kx)3 = 10 × 4 × k3 x3 = 40 k3 x3。令 40 k3 = 720 ⇒ k3 = 18 ⇒ k = ∛18。需要时可化简。
11. Common Mistakes and How to Avoid Them | 常见错误及规避方法
One frequent error is forgetting to raise the coefficient of x to the required power. For instance, in (2x)3, students might write 2x3 instead of 8x3. Always apply the power to both the number and the variable.
一个常见错误是忘记将 x 的系数同样乘方。比如 (2x)3，学生可能写成 2x3 而非 8x3。务必将幂次同时作用于数字和变量。
ion. The first term corresponds to r=0, not r=1. If a question asks for the 5th term, r=4.
另一个易错点是搞错项的位置对应的 r 值。第一项对应 r=0，而非 r=1。如果题目要求第 5 项，则 r=4。
Also, be careful with signs when b is negative. Write the term with parentheses: (−2y)r, and evaluate (negative)r correctly, especially for even/odd r.
此外，当 b 为负时要小心符号。用括号写出该项，如 (−2y)r，并正确计算 (−2y)r，尤其分清 r 的奇偶性。
Finally, always check symmetry: nCr = nCn−r. This can speed up calculations and serve as a quick verification.
最后，要经常使用对称性检查：nCr = nCn−r。这能加速计算，也是个快速验证的方法。
12. Exam-Style Practice and Model Thinking | 考试风格练习与解题思路
Let's consolidate with a typical Edexcel IGCSE question: "Find the term independent of x in the expansion of (2x − 1/x)6." The term independent of x means the power of x is zero.
我们通过一道典型的 Edexcel IGCSE 题目来巩固：求 (2x − 1/x)6 的展开式中与 x 无关的项。与 x 无关的项意味着 x 的幂次为零。
Write a = 2x, b = −1/x = −x−1, n=6. General term: Tr+1 = 6Cr (2x)6−r (−x−1)r = 6Cr 26−r x6−r (−1)r x−r = 6Cr 26−r (−1)r x6−2r.
写出 a = 2x, b = −1/x = −x−1, n=6。通项：Tr+1 = 6Cr (2x)6−r (−x−1)r = 6Cr 26−r x6−r (−1)r x−r = 6Cr 26−r (−1)r x6−2r。
For the term independent of x, set exponent of x to 0: 6−2r = 0 ⇒ r = 3. Then the term is 6C3 23 (−1)3 = 20 × 8 × (−1) = −160.
要求与 x 无关的项，令 x 的指数为 0：6−2r = 0 ⇒ r = 3。该项为 6C3 23 (−1)3 = 20 × 8 × (−1) = −160。
This step-by-step logic, combining algebra with the binomial theorem, will help you tackle any question confidently. Practise with a variety of coefficients, including fractional and negative ones.
这种将代数与二项式定理相结合、步步推导的逻辑，能让你自信地应对任何题目。平时要多练习包含分数系数和负系数的各种题型。
Published by TutorHao | Mathematics
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图