📚 Partial Fractions & Binomial Expansion | 部分分数与二项式展开：AQA A-Level 数学考点精讲
Partial fractions and binomial expansion are two essential algebraic techniques in the AQA A-Level Mathematics specification. They often appear together in exam questions, where partial fractions are used to simplify a rational expression before applying the binomial expansion. Mastering these methods is crucial for achieving top marks on Paper 1 and Paper 2.
部分分数与二项式展开是 AQA A-Level 数学课程中两个至关重要的代数技巧。在考试题目中，它们经常同时出现：先用部分分数化简有理式，再运用二项式展开。熟练掌握这些方法是冲刺 Paper 1 与 Paper 2 高分的关键。
1. What Are Partial Fractions? | 什么是部分分数？
A partial fraction is a way of splitting a single algebraic fraction into a sum of simpler fractions. For example, the single fraction 5/((x+1)(x+4)) can be decomposed as 1/(x+1) - 1/(x+4). This decomposition makes it easier to integrate, differentiate, or expand the original expression.
部分分数是将一个复杂的有理分式拆分成若干个更简单分式之和的方法。例如，单个分式 5/((x+1)(x+4)) 可以分解为 1/(x+1) - 1/(x+4)。这种分解使得对原表达式进行积分、求导或展开变得更加简单。
To write a fraction in partial fractions, the denominator must first be fully factorised. The form of the decomposition depends on the types of factors present in the denominator.
要将一个分式写成部分分数，首先必须对分母进行完全因式分解。分解的具体形式取决于分母中因式的类型。
- Distinct linear factors → A/(ax+b) + B/(cx+d)
- Repeated linear factors → A/(ax+b) + B/(ax+b)²
- Quadratic factors → (Ax+B)/(ax²+bx+c)
- 不同线性因式 → A/(ax+b) + B/(cx+d)
- 重复线性因式 → A/(ax+b) + B/(ax+b)²
- 二次因式 → (Ax+B)/(ax²+bx+c)
2. Denominators with Distinct Linear Factors | 分母为互不相同的线性因式
When the denominator of a rational expression factorises into distinct linear factors, we write the partial fraction as a sum of terms with unknown constants A, B, C, etc. The standard method is to multiply both sides by the denominator and then substitute suitable values of x to solve for the constants.
当有理表达式的分母可以分解为互不相同的线性因式时，我们将其部分分数写成带有未知常数 A、B、C 等的各项之和。标准方法是：将等式两边同时乘以原分母，然后代入选定的 x 值来求解各常数。
Example 1: Express 5/((x+1)(x+4)) in partial fractions.
5/((x+1)(x+4)) = A/(x+1) + B/(x+4)
Multiplying both sides by (x+1)(x+4):
将等式两边乘以 (x+1)(x+4)：
5 = A(x+4) + B(x+1)
-1 to eliminate B: 5 = 3A → A = 5/3. Substitute x = -4 to eliminate A: 5 = -3B → B = -5/3.
现在代入 x = -1 消去 B：5 = 3A → A = 5/3。代入 x = -4 消去 A：5 = -3B → B = -5/3。
x+1)(x+4)) = (5/3)/(x+1) - (5/3)/(x+4). You can also write this as 5/3(1/(x+1) - 1/(x+4)).
因此：5/((x+1)(x+4)) = (5/3)/(x+1) - (5/3)/(x+4)。也可以写作 5/3(1/(x+1) - 1/(x+4))。
Always check your answer by combining the partial fractions back over a common denominator.
始终可以通过将部分分数重新合并回公分母来检查你的答案是否正确。
3. Denominators with Repeated Linear Factors | 分母含重复线性因式
If the denominator contains a repeated linear factor, such as (x+2)², you must include one term for each power of the factor. For example, for (x+2)² you need A/(x+2) + B/(x+2)².
如果分母包含重复线性因式，例如 (x+2)²，则必须为该因式的每一个幂各写一项。例如，对于 (x+2)²，需要 A/(x+2) + B/(x+2)²。
Example 2: Express (3x+2)/(x(x+1)²) in partial fractions.
例 2：将 (3x+2)/(x(x+1)²) 写成部分分数。
(3x+2)/(x(x+1)²) = A/x + B/(x+1) + C/(x+1)²
Multiply both sides by x(x+1)²:
两边同乘 x(x+1)²：
3x+2 = A(x+1)² + Bx(x+1) + Cx
Substitute x = 0: 2 = A → A = 2. Substitute x = -1: -1 = -C → C = 1.
代入 x = 0：2 = A → A = 2。代入 x = -1：-1 = -C → C = 1。
te any convenient value, say x = 1: 5 = 4(2) + 2B + 1 → 5 = 8 + 2B + 1 → 2B = -4 → B = -2.
为求 B，代入一个方便的值，例如 x = 1：5 = 4(2) + 2B + 1 → 5 = 8 + 2B + 1 → 2B = -4 → B = -2。
Therefore: (3x+2)/(x(x+1)²) = 2/x - 2/(x+1) + 1/(x+1)².
因此：(3x+2)/(x(x+1)²) = 2/x - 2/(x+1) + 1/(x+1)²。
Remember: the numerator in the term with the squared factor is still a constant. Only when dealing with quadratic denominators do we need a linear numerator.
请记住：平方因式对应项中的分子仍然是常数。只有在处理二次分母时，才需要线性的分子。
4. Denominators with Quadratic Factors | 分母含二次因式
When the denominator contains a quadratic factor that cannot be factorised into linear factors (e.g., x²+1), the partial fraction includes a term of the form (Ax+B)/(x²+1).
当分母含有无法因式分解为线性因式的二次因式（例如 x²+1）时，部分分数中需要包含 (Ax+B)/(x²+1) 形式的项。
Example 3: Express (x²+2x+1)/(x(x²+1)) in partial fractions.
例 3：将 (x²+2x+1)/(x(x²+1)) 写成部分分数。
(x²+2x+1)/(x(x²+1)) = A/x + (Bx+C)/(x²+1)
Multiply by x(x²+1):
两边同乘 x(x²+1)：
x²+2x+1 = A(x²+1) + x(Bx+C)
Expanding: x²+2x+1 = Ax² + A + Bx² + Cx. Comparing coefficients:
展开：x²+2x+1 = Ax² + A + Bx² + Cx。比较系数：
- x²: A + B = 1
- x: C = 2
- constant: A = 1
- x²：A + B = 1
- x：C = 2
- 常数项：A = 1
Thus A = 1, B = 0, C = 2. So (x²+2x+1)/(x(x²+1)) = 1/x + 2/(x²+1).
因此 A = 1，B = 0，C = 2。所以 (x²+2x+1)/(x(x²+1)) = 1/x + 2/(x²+1)。
mplified the answer. In general, B could be non-zero, so always write the numerator of the quadratic-fraction term as Bx+C to be safe.
注意本例中 B = 0，这使得答案得以简化。一般情况下，B 可以不为零，所以为了保险起见，二次分式项的分子应始终写为 Bx+C。
5. Improper Fractions | 假分式
If the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is called improper. For such fractions, you must first perform polynomial division (or algebraic long division) before applying partial fractions.
如果分子的次数大于或等于分母的次数，该分式称为假分式。对于此类分式，必须先进行多项式除法（或代数长除法），然后再应用部分分数。
Example 4: Express (x³+2x²)/(x+2)(x-1) in partial fractions.
例 4：将 (x³+2x²)/(x+2)(x-1) 写成部分分数。
expand the denominator: (x+2)(x-1) = x² + x - 2. The numerator has degree 3, so we divide:
首先展开分母：(x+2)(x-1) = x² + x - 2。分子是三次，所以进行除法：
(x³+2x²) ÷ (x²+x-2) = x + 1, with remainder 3x + 2
Therefore:
因此：
(x³+2x²)/((x+2)(x-1)) = x + 1 + (3x+2)/((x+2)(x-1))
Now apply partial fractions to (3x+2)/((x+2)(x-1)). Using the method from Section 1:
现在对 (3x+2)/((x+2)(x-1)) 应用部分分数。使用第 1 节的方法：
(3x+2)/((x+2)(x-1)) = A/(x+2) + B/(x-1)
(x-1) + B(x+2). Substitute x = 1: 5 = 3B → B = 5/3. Substitute x = -2: -4 = -3A → A = 4/3.
两边同乘 (x+2)(x-1)：3x+2 = A(x-1) + B(x+2)。代入 x = 1：5 = 3B → B = 5/3。代入 x = -2：-4 = -3A → A = 4/3。
Final answer: (x³+2x²)/((x+2)(x-1)) = x + 1 + 4/(3(x+2)) + 5/(3(x-1)).
最终答案：(x³+2x²)/((x+2)(x-1)) = x + 1 + 4/(3(x+2)) + 5/(3(x-1))。
Check that the polynomial part (x+1) is retained alongside the partial fractions. Always verify that the original numerator degree is restored when combining all terms.
注意多项式部分 (x+1) 与各分式项一同保留。合并所有项后，务必验证是否恢复出原来的分子次数。
6. Binomial Expansion for Positive Integer Powers | 正整数幂的二项式展开
The binomial theorem for positive integer n gives:
正整数 n 的二项式定理给出：
(a+b)ⁿ = aⁿ + ⁿC₁ aⁿ⁻¹b + ⁿC₂ aⁿ⁻²b² + ... + ⁿCᵣ aⁿ⁻ʳbʳ + ... + bⁿ
where ⁿCᵣ = n!/(r!(n-r)! ficient. This expansion has exactly n+1 terms and is valid for all real values of a and b.
其中 ⁿCᵣ = n!/(r!(n-r)!)，通常写作二项式系数。此展开式恰好有 n+1 项，且对所有实数 a 和 b 都成立。
Example 5: Expand (2x+3)⁴.
例 5：展开 (2x+3)⁴。
Using the theorem:
使用定理：
(2x+3)⁴ = (2x)⁴ + 4(2x)³(3) + 6(2x)²(3)² + 4(2x)(3)³ + (3)⁴
= 16x⁴ + 96x³ + 216x² + 216x + 81.
= 16x⁴ + 96x³ + 216x² + 216x + 81。
In AQA exams, you are often asked to find a particular term, such as the term independent of x or the coefficient of x². To do this, write the general term and set the power of x equal to the required value.
在 AQA 考试中，经常要求你找出某一特定项，例如常数项或 x² 的系数。做法是写出通项，并将 x 的幂设为所需的值。
7. Binomial Expansion for Negative and Fractional Powers | 负数与分数次幂的二项式展开
When n is negative or fractional, the binomial expansion takes the form of an infinite series. The formula commonly used is:
当 n 为负数或分数时，二项式展开变为无穷级数。常用公式是：
(1+x)ⁿ = 1 + nx + n(n-1)/2! x² + n(n-1)(n-2)/3! x³ + ...
This expansion is valid only when |x| < 1, i.e., the modulus of x is strictly less than 1. This radius of convergence is essential in exam questions — you must always state it.
此展开仅在 |x| < 1 时成立，即 x 的绝对值严格小于 1。这个收敛半径在考试中至关重要——你必须在答案中指出。
Example 6: Expand (1+x)⁻² up to the x³ term.
例 6：将 (1+x)⁻² 展开至 x³ 项。
Using n = -2:
取 n = -2：
(1+x)⁻² = 1 + (-2)x + (-2)(-3)/2 x² + (-2)(-3)(-4)/6 x³ + ...
= 1 - 2x + 3x² - 4x³ + ... valid for |x| < 1.
= 1 - 2x + 3x² - 4x³ + ...，当 |x| < 1 时成立。
Watch the signs carefully: with a negative n, the coefficients alternate in sign. Also remember that the binomial coefficient for negative n still uses the same pattern but never terminates.
请仔细留意符号：当 n 为负数时，各项系数的正负号交替出现。还要记住，负 n 的二项式系数遵循相同模式，但该级数永远不会终止。
8. Using Partial Fractions with Binomial Expansion | 部分分数与二项式展开的综合运用
This is a classic AQA exam question style. The question gives a rational function, asks you to express it as partial fractions, and then asks you to use the binomial expansion to find the expansion of the whole expression.
这是一个经典的 AQA 考试题型。题目给出一个有理函数，要求你先将其写成部分分数，然后运用二项式展开来求整个表达式的展开式。
Example 7 rtial fractions, then find the expansion of f(x) in ascending powers of x up to the x³ term.
例 7：将 f(x) = 5/((x+1)(x+4)) 写成部分分数，然后求 f(x) 关于 x 升幂展开式直到 x³ 项。
From Example 1, we have
由例 1，我们有
f(x) = 5/3 (1/(x+1) - 1/(x+4))
a form of (1 + ...)⁻¹. For 1/(x+1) = (1+x)⁻¹. For 1/(x+4) = 1/(4(1+x/4)) = 1/4 (1+x/4)⁻¹.
将每个倒数改写成 (1 + ...)⁻¹ 的形式。1/(x+1) = (1+x)⁻¹。1/(x+4) = 1/(4(1+x/4)) = 1/4 (1+x/4)⁻¹。
Now expand each binomial:
现在逐一展开每个二项式：
(1+x)⁻¹ = 1 - x + x² - x³ + ...
(1+x/4)⁻¹ = 1 - x/4 + x²/16 - x³/64 + ...
Multiplying by 1/4 and subtracting:
乘以 1/4 并相减：
f(x) = 5/3 [ (1 - x + x² - x³) - 1/4(1 - x/4 + x²/16 - x³/64) ]
= 5/3 [ (1 - 1/4) + (-x + x/16) + (x² - x²/64) + (-x³ + x³/256) + ... ]
= 5/3 [ (1 - 1/4) + (-x + x/16) + (x² - x²/64) + (-x³ + x³/256) + ... ]
= 5/3 [ 3/4 - 15x/16 + 63x²/64 - 255x³/256 + ... ]
= 5/3 [ 3/4 - 15x/16 + 63x²/64 - 255x³/256 + ... ]
= 5/4 - 25x/16 + 105x²/64 - 425x³/256 + ...
= 5/4 - 25x/16 + 105x²/64 - 425x³/256 + ...
Each sub-expansion is valid for |x| < 1, so the combined expansion is valid for |x| < 1.
每个子展开式均在 |x| < 1 时收敛，因此合并后的展开式的收敛区间为 |x| < 1。
The key trick is to always factor out constants so that each binomial takes the form (1 + u)ⁿ with |u| < 1.
关键技巧在于，务必先提出常数，使每个二项式都形如 (1 + u)ⁿ 且满足 |u| < 1。
9. Validity of Expansion | 展开式的有效性
For binomial expansions with negative or fractional n, you must always state the range of values of x for which the expansion is valid. The condition is |x| < 1 when expanding (1+x)ⁿ. If you expand (1 + ax)ⁿ, then the condition becomes |ax| < 1, i.e., |x| < 1/|a|.
对于 n 为负数或分数的二项式展开，必须始终说明该展开式成立时 x 的取值范围。展开 (1+x)ⁿ 的条件是 |x| < 1。如果展开 (1+ax)ⁿ，则条件变为 |ax| < 1，即 |x| < 1/|a|。
idity intervals. For example, for f(x) = 5/((x+1)(x+4)), the expansion of (1+x)⁻¹ requires |x| < 1, and the expansion of (1+x/4)⁻¹ requires |x| < 4. The intersection is |x| < 1.
在部分分数与二项式结合的题目中，整体有效区间是各单个有效区间的交集。例如，对于 f(x) = 5/((x+1)(x+4))，(1+x)⁻¹ 的展开需要 |x| < 1，而 (1+x/4)⁻¹ 的展开需要 |x| < 4。两者的交集是 |x| < 1。
When presenting your final answer, always include the validity statement. Missing this can cost you an easy mark.
在给出最终答案时，务必注明有效区间。遗漏这一点可能会让你丢掉本可以轻松获得的分数。
10. Exam-Style Worked Example | 考试风格例题精解
Let us work through a complete AQA-style question from start to finish.
让我们完整地解决一道 AQA 风格的题目。
Question: Given that f(x) = (11x+8)/((2x+1)(x+3)²):
题目：已知 f(x) = (11x+8)/((2x+1)(x+3)²)：
(a) Express f(x) in partial fractions.
（a）将 f(x) 写成部分分数。
(b) Hence find the binomial expansion of f(x) up to and including the x³ term.
（b）由此求 f(x) 的二项式展开，直至并包含 x³ 项。
(c) State the values of x for which your expansion is valid.
（c）说明该展开式成立时 x 的取值范围。
Solution (a):
解答（a）：
(11x+8)/((2x+1)(x+3)²) = A/(2x+1) + B/(x+3) + C/(x+3)²
Multiply by (2x+1)(x+3)²:
两边同乘 (2x+1)(x+3)²：
11x+8 = A(x+3)² + B(2x+1)(x+3) + C(2x+1)
Substitute x = -3: 11(-3)+8 = C(2(-3)+1) → -33+8 = C(-5) → -25 = -5C → C = 5.
代入 x = -3：11(-3)+8 = C(2(-3)+1) → -33+8 = C(-5) → -25 = -5C → C = 5。
= -1/2: 11(-1/2)+8 = A(5/2)² → -11/2 + 16/2 = 25A/4 → 5/2 = 25A/4 → A = (5/2)(4/25) = 2/5.
代入 x = -1/2：11(-1/2)+8 = A(5/2)² → -11/2 + 16/2 = 25A/4 → 5/2 = 25A/4 → A = (5/2)(4/25) = 2/5。
To find B, compare coefficients of x². Expanding the right side:
为求 B，比较两边 x² 的系数。展开右边：
A(x²+6x+9) + B(2x²+7x+3) + C(2x+1)
Coefficient of x²: 0 = A + 2B → 0 = 2/5 + 2B → B = -1/5.
x² 的系数：0 = A + 2B → 0 = 2/5 + 2B → B = -1/5。
Therefore:
因此：
f(x) = (2/5)/(2x+1) - (1/5)/(x+3) + 5/(x+3)²
Solution (b):
解答（b）：
Rewrite each fraction in (1 + u)⁻¹ form:
将每个分式改写成 (1 + u)⁻¹ 形式：
- (2/5)/(2x+1) = (2/5)(1+2x)⁻¹
- -(1/5)/(x+3) = -(1/5)(3(1+x/3))⁻¹ = -(1/15)(1+x/3)⁻¹
- 5/(x+3)² = 5(3(1+x/3))⁻² = (5/9)(1+x/3)⁻²
- (2/5)/(2x+1) = (2/5)(1+2x)⁻¹
- -(1/5)/(x+3) = -(1/5)(3(1+x/3))⁻¹ = -(1/15)(1+x/3)⁻¹
- 5/(x+3)² = 5(3(1+x/3))⁻² = (5/9)(1+x/3)⁻²
Now expand each to x³:
现在将每一项展开至 x³：
(1+2x)⁻¹ = 1 - 2x + 4x² - 8x³ + ...
(1+x/3)⁻¹ = 1 - x/3 + x²/9 - x³/27 + ...
(1+x/3)⁻² = 1 - 2x/3 + (4/3)x²/9? Let's compute carefully:
Using the general binomial formula with n = -2:
使用 n = -2 的一般二项式公式：
(1+x/3)⁻² = 1 + (-2)(x/3) + (-2)(-3)/2 (x/3)² + (-2)(-3)(-4)/6 (x/3)³ + ...
= 1 - 2x/3 + 3(x²/9) - 4(x³/27) + ...
= 1 - 2x/3 + 3(x²/9) - 4(x³/27) + ...
= 1 - 2x/3 + x²/3 - 4x³/27 + ...
= 1 - 2x/3 + x²/3 - 4x³/27 + ...
Combining:
合并：
- 2x + 4x² - 8x³) - (1/15)(1 - x/3 + x²/9 - x³/27) + (5/9)(1 - 2x/3 + x²/3 - 4x³/27) + ...
Now combine like terms. For the constant term:
现在合并同类项。常数项：
2/5 - 1/15 + 5/9 = 18/45 - 3/45 + 25/45 = 40/45 = 8/9
For the x term:
x 项：
(2/5)(-2) + (-1/15)(-1/3) + (5/9)(-2/3) = -4/5 + 1/45 - 10/27
Convert to a common denominator of 135: -108/135 + 3/135 - 50/135 = -155/135 = -31/27.
转换为公分母 135：-108/135 + 3/135 - 50/135 = -155/135 = -31/27。
For the x² term:
x² 项：
(2/5)(4) + (-1/15)(1/9) + (5/9)(1/3) = 8/5 - 1/135 + 5/27
Common denominator 135: 216/135 - 1/135 + 25/135 = 240/135 = 16/9.
公分母 135：216/135 - 1/135 + 25/135 = 240/135 = 16/9。
For the x³ term:
x³ 项：
(2/5)(-8) + (-1/15)(-1/27) + (5/9)(-4/27) = -16/5 + 1/
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