Application of Partial Fractions in Binomial Expansion | 部分分式在二项式展开中的应用

📚 Application of Partial Fractions in Binomial Expansion | 部分分式在二项式展开中的应用

When asked to expand a rational function such as (3x + 1)/((x – 1)(x + 2)) in ascending powers of x, direct binomial expansion is impossible because the denominator is a product of linear factors. The key is to decompose the fraction into simpler partial fractions, then expand each part separately using the binomial series. This technique is a core skill in A-level Mathematics and is frequently tested in pure mathematics papers.

当题目要求将有理函数（如 (3x + 1)/((x – 1)(x + 2))）按 x 的升幂展开时，由于分母是线性因式的乘积，无法直接使用二项式展开。关键在于先将该分式分解为更简单的部分分式，再对每一项分别使用二项级数展开。这一技巧是 A-Level 数学的核心技能，在纯数学试卷中经常出现。

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1. Why Partial Fractions? | 为什么使用部分分式？

The binomial expansion of (1 + x)ⁿ is valid only when the expression is a single power of a linear term. If the denominator contains two or more factors, we cannot apply the formula directly. Partial fractions break a complicated fraction into a sum of simpler fractions, each with a denominator that is linear or quadratic. Each simple fraction can then be rewritten in the form A(1 + kx)⁻¹ or A(1 + kx)⁻², which matches the binomial series pattern.

二项式展开 (1 + x)ⁿ 只在表达式是某个线性项的单次幂时有效。若分母含有两个或以上因式，则无法直接套用公式。部分分式将复杂分式拆成若干个较简单的分式之和，每个分式的分母为线性或二次因式。之后，每个简单分式可改写为 A(1 + kx)⁻¹ 或 A(1 + kx)⁻² 的形式，从而符合二项级数的模式。

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2. Review of the Binomial Expansion | 回顾二项式展开

For a rational exponent n, the binomial series states that for |x| < 1,

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

More generally, (1 + ax)ⁿ can be expanded with each term containing (ax)ᵏ. When n is a negative integer or a fraction, the series is infinite and converges only if |ax| < 1.

对于有理指数 n，二项级数指出当 |x| < 1 时，

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

更一般地，(1 + ax)ⁿ 的展开式中每一项都含有 (ax)ᵏ。当 n 为负整数或分数时，级数是无穷的，并且仅在 |ax| < 1 时收敛。

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3. Partial Fraction Decomposition: Basic Rules | 部分分式分解：基本规则

To decompose a proper rational function, first factorise the denominator completely. Each linear factor (ax + b) contributes a term of the form A/(ax + b). If a linear factor is repeated, such as (ax + b)², then include A/(ax + b) + B/(ax + b)². For a quadratic factor (ax² + bx + c), use (Ax + B)/(ax² + bx + c).

要分解一个真分式，首先将分母完全因式分解。每个线性因式 (ax + b) 对应一项 A/(ax + b)。若某个线性因式是重复的，如 (ax + b)²，则需包含 A/(ax + b) + B/(ax + b)²。对于二次因式 (ax² + bx + c)，则使用 (Ax + B)/(ax² + bx + c)。

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4. Finding the Constants | 求待定常数

There are two standard methods for finding constants: substitution and equating coefficients. For example, to decompose (5x + 7)/((x + 1)(x − 2)), set

(5x + 7)/((x + 1)(x − 2)) = A/(x + 1) + B/(x − 2)

Multiply both sides by (x + 1)(x − 2) to obtain 5x + 7 = A(x − 2) + B(x + 1). Substitute x = 2 to get 17 = 3B, so B = 17/3. Substitute x = −1 to get 2 = −3A, so A = −2/3. Hence the partial fractions are −2/(3(x + 1)) + 17/(3(x − 2)).

求待定常数有两种标准方法：代入法和比较系数法。例如，要分解 (5x + 7)/((x + 1)(x − 2))，令

(5x + 7)/((x + 1)(x − 2)) = A/(x + 1) + B/(x − 2)

两边同时乘以 (x + 1)(x − 2)，得 5x + 7 = A(x − 2) + B(x + 1)。代入 x = 2，得 17 = 3B，所以 B = 17/3。代入 x = −1，得 2 = −3A，所以 A = −2/3。因此部分分式为 −2/(3(x + 1)) + 17/(3(x − 2))。

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5. Rewriting Each Fraction for Binomial Expansion | 将每个分式改写以用于二项式展开

The binomial expansion requires expressions of the form (1 + kx)ⁿ. Therefore each partial fraction must be adjusted so its denominator becomes 1 plus a multiple of x. For a term like A/(ax + b), factor out the constant b:

A/(ax + b) = (A/b) × 1/(1 + (a/b)x) = (A/b)(1 + (a/b)x)⁻¹

For a repeated factor, write A/(ax + b)² = (A/b²)(1 + (a/b)x)⁻². Then apply the binomial series with n = −1 or n = −2.

二项式展开要求表达式形如 (1 + kx)ⁿ。因此，每个部分分式都必须调整，使其分母变成 1 加上 x 的倍数。对于形如 A/(ax + b) 的项，提出常数 b：

A/(ax + b) = (A/b) × 1/(1 + (a/b)x) = (A/b)(1 + (a/b)x)⁻¹

对于重复因式，可写成 A/(ax + b)² = (A/b²)(1 + (a/b)x)⁻²。然后对 n = −1 或 n = −2 使用二项级数。

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6. Expanding Each Partial Fraction | 展开每个部分分式

Using the binomial series with n = −1, we have (1 + t)⁻¹ = 1 − t + t² − t³ + … . With n = −2, (1 + t)⁻² = 1 − 2t + 3t² − 4t³ + … . Substituting t = kx gives a power series in x. For example, if a term is 3/(2 + x), then:

3/(2 + x) = (3/2)(1 + x/2)⁻¹ = (3/2)(1 − x/2 + x²/4 − x³/8 + …)

Each term is expanded about x = 0, so the final result is a Maclaurin series.

利用 n = −1 的二项级数，有 (1 + t)⁻¹ = 1 − t + t² − t³ + …。对于 n = −2，有 (1 + t)⁻² = 1 − 2t + 3t² − 4t³ + …。代入 t = kx 就得到关于 x 的幂级数。例如，若某项为 3/(2 + x)，则：

3/(2 + x) = (3/2)(1 + x/2)⁻¹ = (3/2)(1 − x/2 + x²/4 − x³/8 + …)

每一项都在 x = 0 处展开，因此最终结果是麦克劳林级数。

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7. Combining the Series | 合并级数

After expanding each partial fraction, add the resulting series term by term. Collect like powers of x, keeping terms up to the required order. For example, if one fraction contributes −1 + 2x − 4x² and another contributes 3 − x + 5x², the combined series is 2 + x + x² + … . Care is needed with constants and signs.

展开每个部分分式后，将所得级数逐项相加。合并 x 的各次幂，保留到所需阶数。例如，若一个分式贡献 −1 + 2x − 4x²，另一个贡献 3 − x + 5x²，则合并后的级数为 2 + x + x² + …。务必注意常数项和符号。

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8. Convergence Conditions | 收敛条件

Each partial fraction (1 + kx)⁻¹ or (1 + kx)⁻² has its own convergence condition |kx| < 1, i.e. |x| < 1/|k|. The overall expansion is valid only for x values satisfying all individual conditions simultaneously. Therefore the radius of convergence of the final series is the smallest of these bounds. In an exam, always state the range of validity.

每个部分分式 (1 + kx)⁻¹ 或 (1 + kx)⁻² 都有各自的收敛条件 |kx| < 1，即 |x| < 1/|k|。整个展开式仅在同时满足所有单个条件的 x 值下有效。因此，最终级数的收敛半径是这些界中最小的。在考试中，务必写出有效性范围。

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9. Worked Example: A Single Linear Factor Pair | 完整例题：一对线性因式

Expand, in ascending powers of x up to x³, the function f(x) = (7x − 3)/((1 + 2x)(1 − x)), stating the range of values for which the expansion is valid.

将函数 f(x) = (7x − 3)/((1 + 2x)(1 − x)) 按 x 的升幂展开至 x³ 项，并说明展开式有效的 x 取值范围。

Solution / 解答

Decompose: set

(7x − 3)/((1 + 2x)(1 − x)) = A/(1 + 2x) + B/(1 − x)

Multiplying through: 7x − 3 = A(1 − x) + B(1 + 2x). Substitute x = 1: 4 = 3B ⇒ B = 4/3. Substitute x = −1/2: −13/2 = A(3/2) ⇒ A = −13/3. Thus

f(x) = −13/[3(1 + 2x)] + 4/[3(1 − x)]

Now expand each fraction using n = −1:

(1 + 2x)⁻¹ = 1 − 2x + 4x² − 8x³ + … (valid for |x| < 1/2)

(1 − x)⁻¹ = 1 + x + x² + x³ + … (valid for |x| < 1)

Therefore

f(x) = −13/3(1 − 2x + 4x² − 8x³) + 4/3(1 + x + x² + x³) + …

Combining terms gives f(x) = −3 + 10x − 12x² + 28x³ + … . The expansion is valid for |x| < 1/2.

分解：令

(7x − 3)/((1 + 2x)(1 − x)) = A/(1 + 2x) + B/(1 − x)

两边乘以分母：7x − 3 = A(1 − x) + B(1 + 2x)。代入 x = 1：4 = 3B ⇒ B = 4/3。代入 x = −1/2：−13/2 = A(3/2) ⇒ A = −13/3。因此

f(x) = −13/[3(1 + 2x)] + 4/[3(1 − x)]

利用 n = −1 展开每个分式：

(1 + 2x)⁻¹ = 1 − 2x + 4x² − 8x³ + …（|x| < 1/2 时有效）

(1 − x)⁻¹ = 1 + x + x² + x³ + …（|x| < 1 时有效）

因此

f(x) = −13/3(1 − 2x + 4x² − 8x³) + 4/3(1 + x + x² + x³) + …

合并同类项得 f(x) = −3 + 10x − 12x² + 28x³ + …。该展开式在 |x| < 1/2 时有效。

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10. Worked Example: A Repeated Factor | 完整例题：包含重复因式

Expand f(x) = (2 + 9x)/((1 − x)(1 + 2x)²) up to the x² term.

展开 f(x) = (2 + 9x)/((1 − x)(1 + 2x)²) 至 x² 项。

Solution / 解答

Decompose with the correct form:

(2 + 9x)/((1 − x)(1 + 2x)²) = A/(1 − x) + B/(1 + 2x) + C/(1 + 2x)²

Multiply through by the denominator: 2 + 9x = A(1 + 2x)² + B(1 − x)(1 + 2x) + C(1 − x). Substitute x = 1: 11 = 9A ⇒ A = 11/9. Substitute x = −1/2: 2 − 9/2 = C(3/2) ⇒ −5/2 = 3C/2 ⇒ C = −5/3. To find B, compare coefficients of x²: 0 = 4A − 2B. Since A = 11/9, B = 2A = 22/9. Thus

f(x) = 11/[9(1 − x)] + 22/[9(1 + 2x)] − 5/[3(1 + 2x)²]

Now expand each term:

1/(1 − x) = 1 + x + x² + …

1/(1 + 2x) = 1 − 2x + 4x² + …

1/(1 + 2x)² = 1 − 4x + 12x² + …

Multiplying by the constants and combining:

f(x) = 11/9(1 + x + x²) + 22/9(1 − 2x + 4x²) − 5/3(1 − 4x + 12x²) + …

Constant terms: 11/9 + 22/9 − 5/3 = 33/9 − 15/9 = 18/9 = 2. x terms: 11/9 − 44/9 + 20/3 = −33/9 + 60/9 = 27/9 = 3. x² terms: 11/9 + 88/9 − 20 = 99/9 − 20 = 11 − 20 = −9. Therefore

f(x) = 2 + 3x − 9x² + …

按正确形式分解：

(2 + 9x)/((1 − x)(1 + 2x)²) = A/(1 − x) + B/(1 + 2x) + C/(1 + 2x)²

两边乘以分母：2 + 9x = A(1 + 2x)² + B(1 − x)(1 + 2x) + C(1 − x)。代入 x = 1：11 = 9A ⇒ A = 11/9。代入 x = −1/2：2 − 9/2 = C(3/2) ⇒ −5/2 = 3C/2 ⇒ C = −5/3。为求 B，比较 x² 的系数：0 = 4A − 2B。由于 A = 11/9，故 B = 2A = 22/9。因此

f(x) = 11/[9(1 − x)] + 22/[9(1 + 2x)] − 5/[3(1 + 2x)²]

现在展开每一项：

1/(1 − x) = 1 + x + x² + …

1/(1 + 2x) = 1 − 2x + 4x² + …

1/(1 + 2x)² = 1 − 4x + 12x² + …

乘上常数并合并：

f(x) = 11/9(1 + x + x²) + 22/9(1 − 2x + 4x²) − 5/3(1 − 4x + 12x²) + …

常数项：11/9 + 22/9 − 5/3 = 33/9 − 15/9 = 18/9 = 2。x 项：11/9 − 44/9 + 20/3 = −33/9 + 60/9 = 27/9 = 3。x² 项：11/9 + 88/9 − 20 = 99/9 − 20 = 11 − 20 = −9。因此

f(x) = 2 + 3x − 9x² + …

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11. Common Pitfalls and How to Avoid Them | 常见错误与避坑指南

• Forgetting the repeated factor form: for (ax + b)² you must include both A/(ax + b) and B/(ax + b)², not just one term.

  忘记重复因式的形式：对于 (ax + b)²，必须同时包含 A/(ax + b) 和 B/(ax + b)²，而不是只写一项。

• Not factorising the constant before expanding: always rewrite A/(ax + b) as (A/b)(1 + (a/b)x)⁻¹, otherwise the binomial formula is misapplied.

  展开前没有提出常数：务必把 A/(ax + b) 改写为 (A/b)(1 + (a/b)x)⁻¹，否则会错误使用二项式公式。

• Ignoring convergence conditions: each partial fraction may have a different valid range; the final expansion is only valid on the intersection of all ranges.

  忽略收敛条件：每个部分分式的有效范围可能不同；最终展开式只在整个交集上有效。

• Arithmetic errors when combining series: write out each expansion separately before adding, and align like powers of x carefully.

  合并级数时出现计算错误：先分别写出每个展开式，再相加，并仔细对齐 x 的各次幂。

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12. Exam Tips and Summary | 考试技巧与总结

In the exam, always start by stating the partial fraction form. Use substitution for distinct linear factors and compare coefficients when repeated factors or quadratics appear. Rewrite every fraction in the form (1 + kx)ⁿ before expanding. Collect terms up to the required power and clearly state the range of validity. With practice, partial fractions become a reliable bridge between rational functions and power series.

考试时，先写出部分分式的形式。对于不同的线性因式使用代入法；遇到重复因式或二次因式时使用比较系数法。展开前，将每个分式都改写为 (1 + kx)ⁿ 的形式。合并到题目要求的幂次，并明确写出有效性范围。通过练习，部分分式将成为连接有理函数与幂级数的可靠桥梁。

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