Mixed Exercise 8: Binomial Expansion | 综合练习8:二项式展开

📚 Mixed Exercise 8: Binomial Expansion | 综合练习8:二项式展开

Mixed Exercise 8 in the Pearson Edexcel Pure Mathematics Year 1/AS textbook consolidates the binomial expansion for positive integer powers. This revision guide gives you the key formulas, typical question types, worked solutions and common mistakes in one bilingual framework.

混合练习8 是 Pearson Edexcel 纯数学 Year 1/AS 教材中巩固正整数幂二项式展开的练习。本复习指南将关键公式、常见题型、例题解析和易错点整合在一个双语框架中。


1. Factorial Notation and nCr | 阶乘记号与 nCr

For a positive integer n, factorial is defined as n! = n × (n – 1) × … × 2 × 1, and by convention 0! = 1. The binomial coefficient nCr is then given by nCr = n! / [r!(n – r)!], where 0 ≤ r ≤ n.

对于正整数 n,阶乘定义为 n! = n × (n – 1) × … × 2 × 1,并约定 0! = 1。二项式系数 nCr 则由 nCr = n! / [r!(n – r)!] 给出,其中 0 ≤ r ≤ n。

For example, 5! = 5 × 4 × 3 × 2 × 1 = 120, and 5C2 = 5! / (2! × 3!) = 120 / (2 × 6) = 10.

例如,5! = 5 × 4 × 3 × 2 × 1 = 120,而 5C2 = 5! / (2! × 3!) = 120 / (2 × 6) = 10。


2. Pascal’s Triangle and Coefficient Patterns | 帕斯卡三角与系数规律

Pascal’s triangle gives a fast way to find small binomial coefficients. For n = 0, 1, 2, 3, 4, 5, the rows are 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1.

帕斯卡三角可以快速求出较小的二项式系数。n = 0, 1, 2, 3, 4, 5 时,各行依次为:1;1 1;1 2 1;1 3 3 1;1 4 6 4 1;1 5 10 10 5 1。

Each entry equals the sum of the two entries directly above it. The r-th entry in row n is exactly nCr, so row 5 entries are 5C0, 5C1, 5C2, 5C3, 5C4, 5C5.

每个数等于其上方两个数之和。第 n 行第 r 个数正是 nCr,因此第 5 行的数依次为 5C0、5C1、5C2、5C3、5C4、5C5。

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
n = 5 1 5 10 10 5 1

3. The Binomial Theorem for Positive Integer n | 正整数 n 的二项式定理

If n is a positive integer and a, b are constants, then the binomial expansion is exact for all values of a and b:

若 n 为正整数,a、b 为常数,则二项式展开对任意 a、b 都精确成立:

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

The expansion has n + 1 terms. The powers of a decrease from n to 0, while the powers of b increase from 0 to n. The coefficients are the binomial coefficients nC0, nC1, …, nCn.

该展开式共有 n + 1 项。a 的幂从 n 降到 0,b 的幂从 0 升到 n。系数依次为二项式系数 nC0、nC1、…、nCn。


4. Expanding (a + bx)ⁿ in Practice | 实际展开 (a + bx)ⁿ

To expand (a + bx)ⁿ, identify a, b and n first. Then write out the terms using the binomial theorem and simplify the coefficients carefully.

展开 (a + bx)ⁿ 时,首先确定 a、b 和 n。然后根据二项式定理写出各项,并仔细化简系数。

Example: Expand (2 + 3x)⁴ fully.

例题:完整展开 (2 + 3x)⁴。

(2 + 3x)⁴ = 4C0 × 2⁴ + 4C1 × 2³ × (3x) + 4C2 × 2² × (3x)² + 4C3 × 2 × (3x)³ + 4C4 × (3x)⁴

(2 + 3x)⁴ = 4C0 × 2⁴ + 4C1 × 2³ × (3x) + 4C2 × 2² × (3x)² + 4C3 × 2 × (3x)³ + 4C4 × (3x)⁴

This simplifies to 16 + 4 × 8 × 3x + 6 × 4 × 9x² + 4 × 2 × 27x³ + 1 × 81x⁴, so:

化简得 16 + 4 × 8 × 3x + 6 × 4 × 9x² + 4 × 2 × 27x³ + 1 × 81x⁴,因此:

(2 + 3x)⁴ = 16 + 96x + 216x² + 216x³ + 81x⁴


5. Finding a Specific Coefficient | 求指定项的系数

Edexcel exam questions often ask for only one term rather than the full expansion. For (a + bx)ⁿ, the term containing xʳ is:

Edexcel 考试题通常只要求展开式中的某一项。对于 (a + bx)ⁿ,含 xʳ 的项为:

nCr aⁿ⁻ʳ (bx)ʳ = nCr aⁿ⁻ʳ bʳ xʳ

Therefore the coefficient of xʳ is nCr aⁿ⁻ʳ bʳ. Do not include the xʳ part when stating the coefficient.

因此 xʳ 的系数为 nCr aⁿ⁻ʳ bʳ。在写系数时不要包含 xʳ 部分。

Example: Find the coefficient of x³ in (2 + 3x)⁵.

例题:求 (2 + 3x)⁵ 中 x³ 的系数。

Here n = 5, a = 2, b = 3 and r = 3. The coefficient is 5C3 × 2² × 3³ = 10 × 4 × 27 = 1080.

这里 n = 5,a = 2,b = 3,r = 3。系数为 5C3 × 2² × 3³ = 10 × 4 × 27 = 1080。


6. Symmetry of Binomial Coefficients | 二项式系数的对称性

The symmetry property nCr = nC(n – r) is extremely useful in Mixed Exercise 8. It allows you to replace a large r by a smaller one. For example, 12C10 = 12C2 = 66.

对称性 nCr = nC(n – r) 在混合练习8 中非常有用。它可以将较大的 r 换成较小的 r。例如 12C10 = 12C2 = 66。

A typical question is: given nC8 = nC12, find n. Using symmetry, either 8 = n – 12, giving n = 20, or both sides are the same position, which is impossible here. Hence n = 20.

常见题型为:已知 nC8 = nC12,求 n。利用对称性,可得 8 = n – 12,从而 n = 20;另一种情况是两边位置相同,但这里不可能。因此 n = 20。


7. Solving Equations Involving nCr | 解含 nCr 的方程

You may need to solve equations such as nC2 = 36. Write nC2 = n(n – 1) / 2, so n(n – 1) / 2 = 36, giving n(n – 1) = 72. Since n is a positive integer, n = 9 works because 9 × 8 = 72.

你可能需要解诸如 nC2 = 36 的方程。写出 nC2 = n(n – 1) / 2,则 n(n – 1) / 2 = 36,得到 n(n – 1) = 72。因为 n 是正整数,n = 9 满足 9 × 8 = 72。

Another common type is nC3 = 56. Write nC3 = n(n – 1)(n – 2) / 6 = 56, so n(n – 1)(n – 2) = 336. Testing n = 8 gives 8 × 7 × 6 = 336, so n = 8.

另一种常见题型是 nC3 = 56。写出 nC3 = n(n – 1)(n – 2) / 6 = 56,因此 n(n – 1)(n – 2) = 336。检验 n = 8 得 8 × 7 × 6 = 336,所以 n = 8。


8. Approximations Using Binomial Expansion | 利用二项式展开作近似计算

For positive integer n, the binomial expansion is exact. However, when x is small, the first few terms give a very quick approximation. For example, (1.01)⁵ = (1 + 0.01)⁵ = 1 + 5C1(0.

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