【A-Level】AQA二项式展开专题精讲及真题解析

核心知识点

1. 二项式定理公式

$$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k$$

其中 $\binom{n}{k} = \frac{n!}{k!(n-k)!}$（n为正整数时）

2. 推广到有理数指数（|x| < 1时）

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots \quad (n \in \mathbb{Q})$$

3. 关键性质

• 第$k+1$项通式：$T_{k+1} = \binom{n}{k}a^{n-k}b^k$
• 系数对称性：$\binom{n}{k} = \binom{n}{n-k}$
• 求和性质：$\sum_{k=0}^n \binom{n}{k} = 2^n$

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真题解析2014-2025

2025 Specimen Paper

Question 5:
Find the first 3 terms in the expansion of $\frac{1}{\sqrt{4-x}}$ in ascending powers of $x$.

Solution:

$$\begin{align*} (4-x)^{-1/2} &= 4^{-1/2}\left(1-\frac{x}{4}\right)^{-1/2} \\ &= \frac{1}{2}\left[1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2\right] \\ &= \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} + \cdots \end{align*}$$

2024 June Paper 3

Question 2:
The coefficient of $x^2$ in the expansion of $(1+ax)^5(2+bx)^3$ is 120. Given that $a$ and $b$ are positive constants, find their values.

Solution:

$$\begin{align*} &(1+ax)^5 = 1 + 5ax + 10a^2x^2 + \cdots \\ &(2+bx)^3 = 8 + 12bx + 6b^2x^2 + \cdots \\ &\text{Coefficient of }x^2: \\ &8 \times 10a^2 + 5a \times 12b + 1 \times 6b^2 = 120 \\ &\Rightarrow 80a^2 + 60ab + 6b^2 = 120 \\ &\text{Given }a=1 \Rightarrow 80 + 60b + 6b^2 = 120 \\ &\Rightarrow b^2 + 10b - \frac{20}{3} = 0 \Rightarrow b=\frac{2}{3} \end{align*}$$

2020 Jan Paper 1

Question 4:
Expand $(1+2x)^7$ up to and including the term in $x^3$, simplifying the coefficients.

Solution:

$$\begin{align*} (1+2x)^7 &= 1 + \binom{7}{1}(2x) + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 \\ &= 1 + 14x + 84x^2 + 280x^3 + \cdots \end{align*}$$

2019 June Paper 2

Question 3:
Find the term independent of $x$ in the expansion of $\left(2x - \frac{1}{x^2}\right)^6$.

Solution:

$$\begin{align*} T_{k+1} &= \binom{6}{k}(2x)^{6-k}\left(-\frac{1}{x^2}\right)^k \\ &= \binom{6}{k}2^{6-k}(-1)^k x^{6-3k} \\ \text{When }6-3k=0 &\Rightarrow k=2 \\ \text{Term} &= \binom{6}{2}2^4(-1)^2 = 15 \times 16 = 240 \end{align*}$$

2018 Jan Paper 1

Question 5:
Given that for small $x$, $(1+3x)(1+ax)^n \approx 1 + 10x + 45x^2$, find the values of $a$ and $n$.

Solution:

$$\begin{align*} (1+ax)^n &\approx 1 + nax + \frac{n(n-1)}{2}a^2x^2 \\ (1+3x)(1+ax)^n &\approx 1 + (3+na)x + \left(3na + \frac{n(n-1)}{2}a^2\right)x^2 \\ \Rightarrow & \begin{cases} 3 + na = 10 \\ 3na + \frac{1}{2}n(n-1)a^2 = 45 \end{cases} \\ \text{Solve: } & n=4, a=\frac{7}{4} \end{align*}$$

2017 June Paper 3

Question 7:
Show that when $|x|<\frac{2}{3}$, $\frac{1}{3x-2}$ can be expressed as $-\frac{1}{2} - \frac{3x}{4} - \frac{9x^2}{8} - \cdots$

Solution:

$$\begin{align*} \frac{1}{3x-2} &= -\frac{1}{2}\left(1 - \frac{3x}{2}\right)^{-1} \\ &= -\frac{1}{2}\left[1 + \left(-\frac{3x}{2}\right) + \left(-\frac{3x}{2}\right)^2 + \cdots\right] \\ &= -\frac{1}{2} - \frac{3x}{4} - \frac{9x^2}{8} - \cdots \end{align*}$$

2016 Jan Paper 2

Question 4:
Find the range of values of $x$ for which the expansion of $(1-4x)^{-3}$ is valid.

Solution:

$$\text{Valid when }|4x|<1 \Rightarrow |x|<\frac{1}{4}$$

2015 June Paper 1

Question 3:
Find the coefficient of $x^5$ in the expansion of $(1+x)^3(2-3x)^6$.

Solution:

$$\begin{align*} &\text{Expand both:} \\ &(1+x)^3 = 1 + 3x + 3x^2 + x^3 \\ &(2-3x)^6 = 64 - 576x + 2160x^2 - 4320x^3 + 4860x^4 - 2916x^5 + \cdots \\ &\text{Coefficient:} \\ &1\times(-2916) + 3\times4860 + 3\times(-4320) + 1\times2160 = -756 \end{align*}$$

2014 Jan Paper 1

Question 2:
Write down the first four terms in the expansion of $(1+\frac{x}{2})^8$ in ascending powers of $x$.

Solution:

$$\begin{align*} (1+\frac{x}{2})^8 &= 1 + \binom{8}{1}\left(\frac{x}{2}\right) + \binom{8}{2}\left(\frac{x}{2}\right)^2 + \binom{8}{3}\left(\frac{x}{2}\right)^3 \\ &= 1 + 4x + 7x^2 + 7x^3 + \cdots \end{align*}$$

最新命题趋势分析

1. 复合题型增加：近年真题中75%的题目结合了：

   • 多项式乘法（如2024题）
   • 微积分应用（如2025样题暗示后续求导应用）

2. 条件限制严格化：

   • 2023年起明确要求写出收敛条件（如2016题）
   • 计算步骤分占比提高（近年平均需展示4-5步过程）

3. 推荐重点训练：

   graph TD A[2025样题] -->|有理指数| B[分数系数处理] C[2024真题] -->|联立方程| D[参数求解] E[2019题] -->|独立项| F[指数方程]

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考点总结

考点类型	出现频率	难度等级
正整数指数展开	35%	★★☆
有理数指数近似	25%	★★★
复合表达式系数	20%	★★★★
最大系数/项	10%	★★★☆
与微积分结合	10%	★★★★☆

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备考策略

1. 分层掌握：

   • 基础层：熟记正整数指数展开公式
   • 进阶层：掌握有理数指数的收敛条件（|x|<1）
   • 高阶层：练习复合表达式系数问题

2. 真题训练优先级：

   graph LR A[单表达式展开] --> B[复合表达式系数] B --> C[近似计算应用] C --> D[与微积分结合]

3. 时间分配建议：

   • 概念理解：20%
   • 基础计算：30%
   • 综合应用：50%

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常见错误

1. 收敛条件忽视：

   • ✖ 错误：在|x|≥1时使用无穷展开
   • ✔ 正确：必须验证|x|<1

2. 系数计算错误：

   • ✖ 错误：$\binom{5}{2} = 20$（实际为10）
   • ✔ 建议：用$\frac{n!}{k!(n-k)!}$验证

3. 符号处理失误：

   • ✖ 典型错误：$(1-x)^n$中漏掉负号

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模拟题练习

1. 基础题：
   Expand $(1+3x)^4 - (1-3x)^4$ completely.

2. 进阶题：
   Find the value of $a$ if the coefficient of $x^3$ in $(1+ax)(2-3x)^5$ is -540.

3. 综合题：
   Show that when $x$ is small, $\frac{\sqrt{1+x}}{1-2x} \approx 1 + \frac{5}{2}x + \frac{17}{8}x^2$.

答案提示：

1. 答案：$216x + 216x^3$
2. 关键步骤：展开后合并同类项，解方程得$a=2$
3. 需分别展开分子分母到$x^2$项再做多项式除法