C2 二项式展开真题全解 | Binomial Expansion Sequences & Series 高分突破

Edexcel C2 二项式展开 (Binomial Expansion) 是 A-Level 数学序列与级数 (Sequences & Series) 模块的核心考点，几乎每年必出一道 4-6 分的大题。本文基于 Physics & Maths Tutor 整理的历年真题集，系统梳理题型规律与解题模板，帮你轻松拿下这块”送分题”。

Edexcel C2 Binomial Expansion is a core topic in the Sequences & Series module — almost guaranteed to appear every exam session as a 4–6 mark question. This article, based on Physics & Maths Tutor’s curated past paper collection, systematically breaks down question patterns and solution templates to help you secure these marks with confidence.

🧠 核心公式速查 / Core Formula Quick Reference

二项式展开的核心是二项式定理：对于正整数 n，

(1 + ax)ⁿ = 1 + nC₁(ax) + nC₂(ax)² + nC₃(ax)³ + … + (ax)ⁿ

更常见的形式：(a + b)ⁿ = Σ (nCr) · a^(n-r) · b^r，其中 r 从 0 到 n。

The binomial theorem for positive integer n: (1 + ax)ⁿ expands to 1 + ⁿC₁(ax) + ⁿC₂(ax)² + … + (ax)ⁿ. In general form: (a + b)ⁿ = Σ ⁿCᵣ · aⁿ⁻ʳ · bʳ.

📝 五大经典题型 / 5 Classic Question Types

1. 基础展开：求前 n 项 / Basic Expansion: Find First n Terms

这是最基础也最高频的题型。例如真题第 2 题：”Find the first 3 terms of the binomial expansion of (3 − x)⁶.”

解题步骤：① 识别 a=3, b=−x, n=6；② 依次计算 r=0,1,2 三项；③ 化简合并。答案：729 − 1458x + 1215x²。

This is the most common question type. For (3 − x)⁶, step 1: identify a=3, b=−x, n=6; step 2: compute terms for r=0,1,2; step 3: simplify to get 729 − 1458x + 1215x².

2. 含未知常数的展开 / Expansion with Unknown Constants

真题第 1 题：”Find the first 4 terms of (1 + ax)⁷, where a is a constant.” 这是 Edexcel 的经典套路——先用含 a 的表达式展开，再根据系数条件求解 a。

展开结果：1 + 7ax + 21a²x² + 35a³x³。注意计算 ⁿCᵣ 时的阶乘化简技巧：⁷C₂ = 7×6/2 = 21。

For (1 + ax)⁷: expand to get 1 + 7ax + 21a²x² + 35a³x³. The key is efficient combination calculation: ⁷C₂ = 7×6/2 = 21.

3. 系数关系题 / Coefficient Relationship Problems

这是拉开分数差距的题型。如真题第 3 题：”Given that the coefficient of x² is 6 times the coefficient of x, find the value of k.” 对于 (2 + kx)⁷，x 系数 = ⁷C₁·2⁶·k = 448k，x² 系数 = ⁷C₂·2⁵·k² = 672k²。由 672k² = 6×448k 解得 k = 4。

This is the differentiator question. For (2 + kx)⁷: coeff of x = 448k, coeff of x² = 672k². Setting 672k² = 6×448k gives k = 4. Be careful with factorials and powers!

4. 数值估算题 / Numerical Estimation

真题第 6 题：用 (1 + x/2)¹⁰ 的前四项展开估算 (1.005)¹⁰。令 x = 0.01，代入展开式前三项即可得到 5 位小数的近似值。核心技巧：识别 x 的取值使得代入后的项快速衰减，保证截断误差可控。

Question 6: use the first 4 terms of (1 + x/2)¹⁰ to estimate (1.005)¹⁰ to 5 decimal places. Set x = 0.01 and substitute. Key insight: choose x so that terms decay rapidly, keeping truncation error negligible.

5. 逆推系数求 n 或 a / Reverse-Engineering n or a from Coefficients

真题第 5 题：”(1 + ax)¹⁰ 中 x³ 的系数是 x² 系数的两倍，求 a。” 列出方程 ¹⁰C₃·a³ = 2·¹⁰C₂·a² → 120a³ = 2×45a² → a = 3/4。这类题考察学生能否将文字条件翻译成代数方程。

For (1 + ax)¹⁰ where coeff of x³ = 2× coeff of x²: set up ¹⁰C₃·a³ = 2·¹⁰C₂·a² → 120a³ = 90a² → a = 3/4. This tests the ability to translate word conditions into algebraic equations.

🎯 答题模板 / Solution Template

1. 写出通项公式：Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ʳ · bʳ
2. 逐项计算：r=0 → 第一项（常数项），r=1 → x 项，r=2 → x² 项…
3. 化简系数：注意正负号和幂次，尤其是 (a − bx)ⁿ 形式的符号交替
4. 检查系数：代入小值验算（如 x=0.1），确认与原式近似

Solution template: (1) Write the general term Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ʳ · bʳ; (2) Compute term by term; (3) Simplify coefficients carefully — watch for alternating signs in (a − bx)ⁿ; (4) Verify by plugging in a small value like x = 0.1.

📚 学习建议 / Study Tips

• 熟记 ⁿCᵣ 快速算法：⁷C₃ = (7×6×5)/(3×2×1) = 35，手算比查公式表快得多
• 建立题型直觉：看到”coefficient of x² is n times coefficient of x”立刻反应——列出两个系数的表达式，设等式求解
• 限时刷题：二项式展开题每题控制在 3-5 分钟，追求准确率而非过度检查
• 注意审题：题目要求”ascending powers of x”还是”first n terms”，两者有时等价有时不同
• 结合估算题练习：数值估算题常出现在 C2 的综合题中，与梯形法则、迭代法等知识点联动

Study tips: Master fast ⁿCᵣ calculation (⁷C₃ = 7×6×5/3×2×1); develop pattern recognition for coefficient-relationship problems; practice with a 3–5 minute per question time limit; read questions carefully — “ascending powers” vs. “first n terms” may differ; and practice numerical estimation problems that often link to trapezium rule and iteration methods in C2.

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