Expanding (a + bx)^n | 展开 (a + bx)^n

📚 Expanding (a + bx)ⁿ | 展开 (a + bx)ⁿ

Binomial expansion is a cornerstone of A-Level algebra that allows us to write expressions like (a + bx)ⁿ as a sum of terms. Whether n is a positive integer, a fraction, or even negative, mastering the expansion techniques unlocks a wide range of problems from finding specific coefficients to making approximations and handling rational functions. In this article, we will explore the full spectrum of binomial expansions required for Edexcel A-Level Mathematics, providing clear steps, examples, and exam-focused tips.

二项展开是 A-Level 代数的重要基础，它使我们能够将形如 (a + bx)ⁿ 的表达式写成一系列项的和。无论 n 是正整数、分数甚至是负数，掌握展开技巧都能帮助我们解决求特定系数、做近似计算以及处理分式函数等一系列问题。本文全面介绍 Edexcel A-Level 数学所要求的二项展开内容，提供清晰的步骤、示例和应试要点。

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1. Factorials and nCr Notation | 阶乘与组合数记号

In the expansion of (a + bx)ⁿ where n is a positive integer, the coefficients are given by combination numbers, commonly written as ⁿC_r or C(n, r). The combination ⁿC_r is defined for 0 ≤ r ≤ n and equals n! / [r!(n − r)!]. Remember that 0! = 1 and ⁿC₀ = 1.

当 n 为正整数时，(a + bx)ⁿ 的展开系数由组合数给出，通常记作 ⁿC_r 或 C(n, r)。组合数 ⁿC_r 的定义域为 0 ≤ r ≤ n，并且满足 ⁿC_r = n! / [r!(n − r)!]。请记住 0! = 1，并且 ⁿC₀ = 1。

For example, ⁵C₂ = 5! / (2! × 3! imes; 6) = 10. This notation is essential for writing the binomial coefficients compactly.

例如，⁵C₂ = 5! / (2! × 3!) = 120 / (2 × 6) = 10。这种记号对于简洁地表示二项式系数至关重要。

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2. Pascal's Triangle for Positive Integers | 正整数指数的帕斯卡三角形

Pascal' above it. The rows correspond to n = 0, 1, 2, 3, ... and give the coefficients of (1 + x)ⁿ.

帕斯卡三角形为较小的 n 值提供了快速寻找二项式系数的方法。每一个数都是它上方两数之和。行对应 n = 0, 1, 2, 3, ...，并给出 (1 + x)ⁿ 的系数。

The first few rows are:
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

前几行为：
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

For expanding (a + bx)ⁿ, you can still use Pascal's triangle if you first factor out a, but the formula using ⁿC_r is more systematic for larger n.

在展开 (a + bx)ⁿ 时，如果你先把 a 提出来，仍然可以用帕斯卡三角形；但对于较大的 n，使用 ⁿC_r 的公式会更系统。

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

For any positive integer n, the expansion of (a + b)ⁿ is given by:

对任意正整数 n，(a + b)ⁿ 的展开由下式给出：

(a + b)ⁿ = Σ_r=0^{n n}C_r a^n−r b^r

Adapting this to (a + bx)ⁿ, simply treat bx as the second term:

将其应用于 (a + bx)ⁿ，只需把 bx 当作第二项即可：

(a + bx)ⁿ = Σ_r=0^{n n}C_r a^n−r (bx)^r

Each term contains a power of b and x, so the full term becomes ⁿC_r a^n−r b^r x^r.

每一项都含有 b 和 x 的幂次，因此完整的项为 ⁿC_r a^n−r b^r x^r。

For instance, expand (2 + 3x)⁴:

例如，展开 (2 + 3x)⁴：

= ⁴C₀ × 2⁴ × (3x)⁰ + ⁴C₁ × 2³ × (3x)¹ + ⁴C₂ × 2² × (3x)² + ⁴C₃ × 2¹ × (3x)³ + ⁴C₄ × 2⁰ × (3x)⁴
= 16 + 4 × 8 × 3x + 6 × 4 × 9x² + 4 × 2 × 27x³ + 1 × 1 × 81x⁴
= 16 + 96x + 216x² + 216x³ + 81x⁴

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4. The General Term and Coefficient Extraction | 通项与系数提取

The general term in the expansion of (a + bx)ⁿ is often written as T_r+1:

(a + bx)ⁿ 展开中的通项通常记作 T_r+1：

T_r+1 = ⁿC_r a^n−r b^r x^r

This formula is extremely useful when you are asked to find the coefficient of a particular power of x, say x^k. By setting r = k, you can immediately write down that coefficient.

当你需要求某个特定 x 次幂（如 x^k）的系数时，这个公式非常有用。令 r = k，你就能直接写出该系数。

Example: Find the coefficient of x³ in (5 − 2x)⁷.
Here a = 5, b = −2, n = 7. The term with x³ corresponds to r = 3.
Coefficient = ⁷C₃ × 5⁷⁻³ × (−2)³ = 35 × 5⁴ × (−8) = 35 × 625 × (−8) = −175 000.

例题：求 (5 − 2x)⁷ 中 x³ 的系数。
这里 a = 5，b = −2，n = 7。含 x³ 的项对应 r = 3。
系数 = ⁷C₃ × 5⁷⁻³ × (−2)³ = 35 × 5⁴ × (−8) = 35 × 625 × (−8) = −175 000。

Always be careful with signs; a negative b will alternate the sign of terms depending on whether r is even or odd.

务必注意符号；负的 b 会使各项的符号根据 r 是偶数或奇数而交替变化。

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5. Extending to Rational Exponents | 推广到有理数指数

2/3), the binomial expansion becomes an infinite series. The standard form used is (1 + x)ⁿ, and the expansion is valid only for |x| < 1 (or sometimes for specific bounds in more advanced contexts).

当 n 不是正整数时（例如 n = &minus;1、1/2、−2/3），二项展开变为一个无穷级数。标准形式采用 (1 + x)ⁿ，并且展开式仅在 |x| < 1 时有效（在更高级的情境中有时也有特定的范围限制）。

The formula uses a generalised binomial coefficient:

该公式使用了广义二项式系数：

(1 + x)ⁿ = 1 + nx + [n(n − 1)/2!] x² + [n(n − 1)(n − 2)/3!] x³ + …

The series continues infinitely unless n is a positive integer, in which case it terminates. For rational n, you only need the first few terms in most A-Level questions, and the concept of convergence is key.

除非 n 是正整数（此时级数终止），否则级数将无限延续。对于有理数 n，在绝大多数 A-Level 题目中你只需要前几项，而收敛性的概念至关重要。

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6. Converting (a + bx)ⁿ into Standard Form | 将 (a + bx)ⁿ 化为标准形式

To expand (a + bx)ⁿ when n is rational, you must first factor out aⁿ to obtain the required (1 + kx)ⁿ pattern:

当 n 为有理数时，展开 (a + bx)ⁿ 需要先提取因子 aⁿ，得到所需的 (1 + kx)ⁿ 模式：

(a + bx)ⁿ = aⁿ [1 + (b/a)x]ⁿ

Then expand [1 + (b/a)x]ⁿ using the infinite series formula with x replaced by (b/a)x. The overall expansion is the product of aⁿ and the series.

然后将 [1 + (b/a)x]ⁿ 用无穷级数公式展开，其中 x 被替换为 (b/a)x。整个展开就是 aⁿ 与该级数的乘积。

Example: Expand (4 + 3x)⁻¹ up to the term in x².
First write: (4 + 3x)⁻¹ = 4⁻¹ [1 + (3/4)x]⁻¹ = (1/4) [1 + (3/4)x]⁻¹.
Use the expansion (1 + X)⁻¹ ≈ 1 − X + X² for small X.
Substitute X = (3/4)x:
= (1/4)[1 − (3/4)x + (3/4)² x² + …] = (1/4)[1 − (3/4)x + (9/16)x² + …]
= 1/4 − (3/16)x + (9/64)x² + …

例题：将 (4 + 3x)⁻¹ 展开到 x² 项。
首先写为：(4 + 3x)⁻¹ = 4⁻¹ [1 + (3/4)x]⁻¹ = (1/4) [1 + (3/4)x]⁻¹。
对小量 X 使用展开式 (1 + X)⁻¹ ≈ 1 − X + X²。
代入 X = (3/4)x：
= (1/4)[1 − (3/4)x + (3/4)² x² + …] = (1/4)[1 − (3/4)x + (9/16)x² + …]
= 1/4 − (3/16)x + (9/64)x² + …

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7. Validity Conditions for Rational Exponents | 有理数指数的有效性条件

For the expansion of (1 + X)ⁿ to be valid as an infinite series, we require |X| < 1. In the context of (a + bx)ⁿ, this translates to |(b/a)x| < 1, or equivalently |x| < |a/b|, provided a ≠ 0.

对于 (1 + X)ⁿ 作为无穷级数展开的有效性，我们需要 |X| < 1。在 (a + bx)ⁿ 的情境中，这等价于 |(b/a)x| < 1，或者当 a ≠ 0 时，|x| < |a/b|。

Always state the range of x for which your expansion is valid. In many Edexcel questions, you will be asked to find the exact set of x-values that ensure convergence, or to explain why a given value of x can be used in the approximation.

请务必说明你的展开式对哪些 x 值有效。在许多 Edexcel 考题中，你会被要求求出保证收敛的 x 的精确范围，或者解释为什么某个给定的 x 值可以用于近似计算。

For example, in the expansion of (4 + 3x)⁻¹, the validity condition is |(3/4)x| < 1 ⇒ |x| < 4/3.

例如，在 (4 + 3x)⁻¹ 的展开中，有效性条件为 |(3/4)x| < 1 ⇒ |x| < 4/3。

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8. Using Expansions for Approximations | 利用展开式进行近似计算

Binomial expansions are excellent tools for estimating roots, reciprocals, and other irrational values. By substituting a small value of x into a known expansion, you can obtain a polynomial approximation that rapidly converges.

二项展开是估计根式、倒数和其他无理数值的绝佳工具。将较小的 x 代入已知的展开式，你就能得到一个快速收敛的多项式近似。

Example: Use the expansion of √(1 + x) to estimate √4.1.
Write &radic;4.1 = &radic;(4 + 0.1) = &radic;[4(1 + 0.025)] = 2 &radic;(1 + 0.025).
Here n = 1/2: &radic;(1 + X) ≈ 1 + (1/2)X − (1/8)X².
(1/8)(0.000625)] = 2[1 + 0.0125 − 0.000078125] = 2 × 1.012421875 = 2.02484375.

例题：利用 √(1 + x) 的展开式估算 √4.1。
写成 &radic;4.1 = &radic;(4 + 0.1) = &radic;[4(1 + 0.025)] = 2 &radic;(1 + 0.025)。
这里 n = 1/2：√(1 + X) ≈ 1 + (1/2)X − (1/8)X²。
代入 X = 0.025： (1/8)(0.000625)] = 2[1 + 0.0125 − 0.000078125] = 2 × 1.012421875 = 2.02484375。

This approximation is very close to the actual value of about 2.02484567, showing the power of the binomial method.

该近似值与实际值约 2.02484567 非常接近，彰显了二项式方法的强大之处。

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9. Expansion of Rational Functions via Partial Fractions | 通过部分分式展开有理函数

Some expressions are not immediately in the form (a + bx)ⁿ but can be split into partial fractions, each of which can then be expanded binomially. This is a common technique for complicated algebraic fractions, especially when the denominator contains linear factors.

有些表达式并不能直接写成 (a + bx)ⁿ 的形式，但可以拆分为部分分式，之后每一项都可以用二项式展开。对于复杂的分式，尤其是分母包含线性因式时，这是一种常用技巧。

/ [(1 − x)(2 + x)] in ascending powers of x, first decompose it into A/(1 − x) + B/(2 + x). Then write each term as A(1 − x)⁻¹ + (B/2)(1 + x/2)⁻¹ and expand separately, combining terms up to the required power.

例如，要把 f(x) = (3 + 2x) / [(1 − x)(2 + x)] 按 x 的升幂展开，首先将其分解为 A/(1 − x) + B/(2 + x)。接着把每一项写成 A(1 − x)⁻¹ + (B/2)(1 + x/2)⁻¹ 并分别展开，最后合并到所需的幂次即可。

Mastery of partial fractions combined with binomial expansion is often tested in Edexcel Pure Mathematics papers, particularly in the context of series expansions and approximations.

部分分式与二项展开的联合运用经常出现在 Edexcel 纯数试卷中，尤其在级数展开和近似计算的相关题目中。

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10. Finding a Specific Coefficient in Rational Expansions | 求有理指数展开中的特定系数

In questions involving rational n, you may be asked to determine the coefficient of a given power of x, say x³, without expanding every term. Use the generalised term formula for (1 + X)ⁿ: the term in X^r is [n(n−1)...(n−r+1) / r!] X^r. After substitution, multiply by the outer factor aⁿ and any factor from (b/a)^r.

在涉及有理数 n 的题目中，你可能需要求出指定 x 的幂次（如 x³）的系数，而无需展开全部项。使用 (1 + X)ⁿ 的广义通项公式：X^r 的项为 [n(n−1)...(n−r+1) / r!] X^r。代入后，再乘以外层因子 aⁿ 以及来自 (b/a)^r 的因子。

Example: Find the coefficient of x³ in (2 + x)⁻².
Write as 2⁻² (1 + x/2)⁻² = (1/4) (1 + x/2)⁻².
The X³ coefficient in (1 + X)⁻² is [(−2)(−3)(−4)/3!] = &minus;24/6 = &minus;4.
Thus coefficient needed = (1/4) × (−4) × (1/2)³? Wait: the term is (1/4) × (−4) × (x/2)³ = (1/4)(&minus;4)(1/8) x³ = (&minus;1/8) x³.
So coefficient is −1/8.

例题：求 (2 + x)⁻²

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