Edexcel A-Level Maths: Binomial Expansion (Question 116 from PDF) | 爱德思A-Level数学：二项式展开（真题第116题）

📚 Edexcel A-Level Maths: Binomial Expansion (Question 116 from PDF) | 爱德思A-Level数学：二项式展开（真题第116题）

In this article, we explore a classic Edexcel A-Level Mathematics problem on binomial expansion, taken from page 4, question 116 of a compiled revision PDF. We break down the step-by-step method, explain how to handle fractional and negative powers, and show you how to determine the range of validity.

本文深入讲解一道来自爱德思A-Level数学真题汇编PDF第4页第116题的经典二项式展开问题。我们将逐步拆解求解过程，解释如何处理分数指数与负指数，并教你如何确定展开式的有效范围。

1. Understanding the Binomial Theorem for (a + bx)ⁿ | 理解 (a + bx)ⁿ 的二项式定理

The binomial expansion allows us to expand expressions of the form (a + bx)ⁿ, where n is a rational number. When n is not a positive integer, the expansion is an infinite series, and we must ensure |bx/a| < 1 for convergence.

二项式定理可将形如 (a + bx)ⁿ 的式子展开，其中 n 为有理数。当 n 不是正整数时，展开式是一个无穷级数，必须满足条件 |bx/a| < 1 才收敛。

The general form for the expansion of (1 + u)ⁿ is:
(1 + u)ⁿ = 1 + n u + [n(n-1)/2!] u² + [n(n-1)(n-2)/3!] u³ + …
otation. To apply this to (a + bx)ⁿ, we first factor out aⁿ: (a + bx)ⁿ = aⁿ (1 + (b/a)x)ⁿ.

(1 + u)ⁿ 的展开通项为上述形式。要将其应用于 (a + bx)ⁿ，可先提取因子 aⁿ：即 (a + bx)ⁿ = aⁿ (1 + (b/a)x)ⁿ，再利用已知展开式。

2. The Exam Question: Expand (4 + 3x)⁻² up to x³ | 真题再现：展开 (4 + 3x)⁻² 到 x³ 项

Typical Edexcel question 116 asks: "Expand (4 + 3x)⁻² in ascending powers of x, up to and including the term in x³. State the range of values of x for which the expansion is valid." This tests your ability to handle a negative integer exponent and to rewrite the expression in the form (1 + u)ⁿ.

典型的爱德思第116题要求：将 (4 + 3x)⁻² 按 x 升幂展开，至 x³ 项，并说明展开有效的 x 取值范围。这考查了处理负整数指数的能力，以及将表达式化为标准形式 (1 + u)ⁿ 的技巧。

/16) (1 + (3/4)x)⁻². Now the binomial part is in the standard form with u = (3/4)x and n = –2.

首先处理 (4 + 3x)⁻²，提取因子 4⁻²，得到 (1/16)(1 + (3/4)x)⁻²。此时二项式部分化为标准形式，其中 u = (3/4)x，n = –2。

3. Applying the Series Formula with n = –2 | 代入 n = –2 的级数公式

Using the formula (1 + u)⁻² = 1 + (–2)u + [–2(–3)/2!] u² + [–2(–3)(–4)/3!] u³ + … ach coefficient carefully. Note that the signs alternate because of the negative exponent.

利用公式 (1 + u)⁻² 的展开，需仔细计算系数。由于指数为负，各项符号会交替变化。

The term in u: coefficient is –2.
The term in u²: [–2 × –3] / 2 = 6/2 = 3.
The term in u³: [–2 × –3 × –4] / 6 = –24/6 = –4.
Thus (1 + u)⁻² = 1 – 2u + 3u² – 4u³ + …

u 项系数为 –2；u² 项系数为 (–2)(–3)/2! = 3；u³ 项系数为 (–2)(–3)(–4)/3! = –4。因此 (1 + u)⁻² = 1 – 2u + 3u² – 4u³ + …

4. Substituting u = (3/4)x and Simplifying | 代回 u = (3/4)x 并化简

Replace u with (3/4)x in the series:
(1 + (3/4)x)⁻² = 1 – 2(3/4)x + 3(3/4)² x² – 4(3/4)³ x³ + …

将 u 替换为 (3/4)x，得到该级数。

Now compute the numerical parts:
–2 × (3/4) = –3/2.
3 × (9/16) = 27/16.
–4 × (27/64) = –108/64 = –27/16 (after simplifying).
So the bracket expands to 1 – (3/2)x + (27/16)x² – (27/16)x³ + …

计算数值：–2 × (3/4) = –3/2；3 × (3/4)² = 27/16；–4 × (3/4)³ = –108/64 = –27/16。括号内展开为 1 – (3/2)x + (27/16)x² – (27/16)x³ + …

5. Incorporating the Factor 1/16 | 乘上因子 1/16

Remember we had an overall factor of 1/16. Multiply each term of the series by 1/16:
(4 + 3x)⁻² = (1/16)[1 – (3/2)x + (27/16)x² – (27/16)x³ + …]

注意不要遗漏整体因子 1/16，将括号内每一项乘以 1/16。

This gives the final expanded coefficients:
Constant term: 1/16.
Coefficient of x: (1/16) × (–3/2) = –3/32.
Coefficient of x²: (1/16) × (27/16) = 27/256.
Coefficient of x³: (1/16) × (–27/16) = –27/256.
Therefore, (4 + 3x)⁻² = 1/16 – (3/32)x + (27/256)x² – (27/256)x³ + …

得到最终展开系数：常数项 1/16；x 项系数 –3/32；x² 项系数 27/256；x³ 项系数 –27/256。所以结果为上述级数。

6. Determining the Range of Validity | 确定有效范围

mial expansion of (1 + u)ⁿ with n not a positive integer, the expansion is valid only when |u| < 1. Here u = (3/4)x, so we require |(3/4)x| < 1.

当 n 不是正整数时，二项展开仅在 |u| < 1 时有效。此处 u = (3/4)x，需要满足 |(3/4)x| < 1。

Solve the inequality: |x| < 4/3. Hence the expansion is valid for –4/3 < x < 4/3. Some mark schemes also accept |x| < 4/3.

解该不等式得 |x| < 4/3，即 –4/3 < x < 4/3。部分评分标准也接受 |x| < 4/3。

It is essential to state this range clearly, as marks are often allocated for the validity condition. Remember that the original expression (4 + 3x)⁻² is defined for all x except where the base is zero, but the series representation converges only within this radius.

必须明确写出这个范围，因为阅卷时常为有效条件单独给分。注意原表达式在 4 + 3x ≠ 0 时均有定义，但其级数表示仅在该半径内收敛。

7. Checking Special Values and Approximations | 检验特殊值与近似计算

ted series. The original value is (4 + 3×0.2)⁻² = (4.6)⁻² ≈ 0.04726. The series up to x³ gives: 1/16 – (3/32)(0.2) + (27/256)(0.04) – (27/256)(0.008) = 0.0625 – 0.01875 + 0.00421875 – 0.00084375 ≈ 0.047125. The agreement is good, confirming the expansion.

为验证展开式，可在有效区间内取一个小值 x = 0.2，分别代入原式和截断级数。原式值为 (4.6)⁻² ≈ 0.04726；级数至 x³ 的计算结果为 0.047125，两者吻合良好，说明展开正确。

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8. Common Pitfalls and How to Avoid Them | 常见易错点与规避方法

Many students forget to factor out the constant term aⁿ correctly. For (4 + 3x)⁻², they might mistakenly treat the whole expression as (4)⁻²(1 + (3x/4))⁻² but then forget the 4⁻² or misapply the binomial formula without extracting the factor. Always write the standard form explicitly.

很多学生忘记正确提取常数因子 aⁿ。对于 (4 + 3x)⁻²，他们可能错误地直接套用公式而不先提取 4⁻²，或忘记乘回该因子。务必显式写出标准形式。

ishandling the factorial terms when n is negative. The coefficient for u³ is n(n-1)(n-2)/3!. With n = –2, this gives –2 × –3 × –4 ÷ 6 = – tion to the sign of each product. Using a table can help track the signs and coefficients.

另一个错误是处理负指数时的阶乘项。当 n = –2 时，u³ 系数为 –2 × –3 × –4 / 6 = –4，而非 +4。务必留意每一步乘积的符号。可用表格辅助跟踪系数和符号变化。

Term	Coefficient without sign	Sign pattern	Final coefficient
u¹	2	–	–2
u²	(2×3)/2=3	+	+3
u³	(2×3×4)/6=4	–	–4

Such a table clarifies the alternating signs for negative integer powers.

9. Extension: Using the Expansion to Estimate Reciprocal Squares | 拓展：利用展开式估算倒数平方

This expansion can be used to quickly approximate functions like 1/(4+3x)² for small x. This is particularly useful in physics or engineering contexts where a quick estimate is needed without a calculator.

该展开式可用于快速估计类似 1/(4+3x)² 的函数在 x 较小时的取值。在物理或工程领域，常常需要在没有计算器时进行快速估算。

, we write 4.15 = 4 + 3×(0.05). Thus x = 0.05, and we can substitute into the series: 1/16 – (3/32)(0.05) + (27/256)(0.0025) – … This yields a fairly accurate value with minimal computation.

若要估算 1/(4.15)²，可写成 4 + 3×(0.05)，即 x = 0.05，代入级数。只需很少的计算量就可得到较精确的数值。

10. Connecting to the Broader Edexcel A-Level Syllabus | 连接爱德思A-Level大纲整体

Binomial expansion with rational exponents appears in the Edexcel Pure Mathematics 2/3 module. It is frequently examined alongside partial fractions, allowing the expansion of rational functions by first decomposing them. An example would be expanding (2x+1)/[(1+x)(1-2x)] by splitting into partial fractions and then expanding each term.

含有理指数的二项式展开出现在爱德思纯数2/3单元中。常与部分分式结合出题，通过先分解有理函数再展开各项。例如将 (2x+1)/[(1+x)(1-2x)] 拆成部分分式后再展开。

This problem type assesses your algebraic manipulation skills and understanding of convergence. Mastery of the technique shown here will serve you well in questions worth 6 to 8 marks, often part of a larger question on series expansions.

这类题目考查代数操作能力和对收敛的理解。掌握本文展示的技巧，有助于应对占6到8分的题目，常为级数展开大题的一部分。

11. Summary of the Method for Any Similar Question | 通用方法总结

To expand (a + bx)ⁿ where n is not a positive integer:

• Step 1: Rewrite as aⁿ (1 + (b/a)x)ⁿ.
• Step 2: Expand (1 + u)ⁿ using the binomial series up to the desired power.
• Step 3: Multiply the series by aⁿ.
• Step 4: State the validity condition |(b/a)x| < 1, which simplifies to |x| < |a/b|.

展开 (a + bx)ⁿ（n 非正整数）的步骤：

• 步骤1：化为 aⁿ (1 + (b/a)x)ⁿ。
• 步骤2：用二项级数展开 (1 + u)ⁿ 至所需次幂。
• 步骤3：将级数乘上 aⁿ。
• 步骤4：给出有效条件 |(b/a)x| < 1，化简得 |x| < |a/b|。

Practising with different values of a, b, and n – including fractional n like 1/2 or –1/3 – will build fluency. Always double-check the first few coefficients by direct substitution of small x.

通过练习不同的 a、b 和 n 值，包括分数如 1/2 或 –1/3，可提升熟练度。务必用小的 x 值直接代入检验前几项系数。

12. Final Tips for the Exam | 考试提分建议

In the exam, show all steps clearly: the factorisation, the binomial coefficients before simplification, and the final simplified coefficients. Even if you make an arithmetic slip, you can earn method marks. State the range of validity in the form |x| < 4/3 or –4/3 < x < 4/3; both are accepted.

考试中要清晰展示每一步：因式提取、化简前的二项系数、最终简化系数。即使出现计算错误，也能拿到方法分。有效范围可写成 |x| < 4/3 或 –4/3 < x < 4/3，二者均给分。

If the question asks for the expansion up to x³, it is good practice to write the series with a trailing "+ …" to acknowledge it is infinite. Avoid writing an equals sign between the original function and a truncated polynomial without ellipsis, as technically they are not equal except within the radius of convergence and with infinitely many terms.

若题目要求展开到 x³，最好在级数后写上“+ …”以表明其为无穷级数。原函数与截断多项式之间不加省略号地使用等号是不严谨的，技术上仅在收敛半径内且取无穷多项时才相等。

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