Binomial Expansion in IB AQA Mathematics | IB AQA 数学：二项式展开考点精讲

📚 Binomial Expansion in IB AQA Mathematics | IB AQA 数学：二项式展开考点精讲

The binomial expansion is a fundamental algebraic tool used to expand expressions of the form (a + b)^n. In IB and AQA mathematics, it appears across pure mathematics, from basic expansion of positive integer powers to advanced approximations using infinite series for rational exponents. Mastering this topic gives you the ability to compute coefficients, find specific terms, and apply expansions to solve real-world problems.

二项式展开是代数中的基本工具，用于展开形如 (a + b)^n 的表达式。在 IB 和 AQA 数学课程中，它贯穿纯数学的多个板块，从正整数幂的基本展开，到利用有理数指数的无穷级数进行高级近似计算。掌握这一专题，你将能够计算系数、求出特定项，并运用展开式解决实际问题。

1. Introduction to Binomial Expansion | 二项式展开简介

Binomial expansion is the process of writing (a + b)^n as a sum of terms involving powers of a and b. For n as a positive integer, the expansion is finite and contains (n + 1) terms. Historically, the expansion originated from the distributive law and was systematised by Pascal and Newton.

二项式展开是将 (a + b)^n 写成包含 a 和 b 的幂次之和的过程。当 n 为正整数时，展开式是有限的，且含有 (n + 1) 项。从历史上看，展开式源于分配律，并由帕斯卡和牛顿加以系统化。

In the syllabus, you will encounter two main scenarios: expansions where n is a positive integer, and expansions where n is a rational number (including negative integers and fractions) using an infinite binomial series. The latter only converges under certain conditions on x.

在考纲中，你会遇到两种主要情形：n 为正整数时的展开，以及 n 为有理数（包括负整数和分数）时使用无穷二项式级数的展开。后者只有在 x 满足特定条件时才收敛。

2. Pascal’s Triangle and Combinations | 杨辉三角与组合数

For small positive integer exponents, Pascal’s triangle gives a quick way to obtain coefficients. Each entry is the sum of the two entries diagonally above it. The triangle is constructed row by row, with the n-th row corresponding to the coefficients of (a + b)^n.

对于较小的正整数指数，杨辉三角提供了一种快速获得系数的方法。每一个数是它上方左右两个数之和。三角形逐行构建，第 n 行对应 (a + b)^n 的各项系数。

However, for larger n, binomial coefficients are better calculated using the combination formula ⁿCᵣ or C(n, r) = n! / [r! (n − r)!]. Here ⁿCᵣ is read as ‘n choose r’ and gives the coefficient of the term with a^{n−r} b^r.

然而，对于较大的 n，二项式系数最好通过组合数公式 ⁿCᵣ 或 C(n, r) = n! / [r! (n − r)!] 来计算。ⁿCᵣ 读作“n 选 r”，表示含有 a^{n−r} b^r 那一项的系数。

Make sure you are comfortable evaluating ⁿCᵣ by hand using factorial simplification, and also using the nCr function on your calculator where permitted. Note that ⁿC₀ = 1, ⁿCₙ = 1, and ⁿCᵣ = ⁿC_{n−r}.

确保你能够通过阶乘化简手工计算 ⁿCᵣ，并且在允许的情况下熟练使用计算器上的 nCr 函数。注意 ⁿC₀ = 1，ⁿCₙ = 1， ⁿCᵣ = ⁿC_{n−r}。

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

The binomial theorem states that for any positive integer n,

(a + b)ⁿ = ∑_{r=0}^{n} ⁿCᵣ a^{n−r} b^{r}

其中求和符号 ∑_{r=0}^{n} 表示从 r=0 到 r=n 的求和。这一定理为快速展开二项式提供了系统的方法。

Written out in full, the expansion is (a + b)ⁿ = aⁿ + ⁿC₁ a^{n−1} b + ⁿC₂ a^{n−2} b² + … + ⁿCₙ bⁿ. Each term has total degree n, with the powers of a decreasing and powers of b increasing.

完整写出来就是 (a + b)ⁿ = aⁿ + ⁿC₁ a^{n−1} b + ⁿC₂ a^{n−2} b² + … + ⁿCₙ bⁿ。每一项的次数之和为 n，a 的幂次递减，b 的幂次递增。

When the binomial is of the form (1 + x)ⁿ, the expansion simplifies to (1 + x)ⁿ = 1 + nx + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + … . This is particularly useful for rational n.

当二项式为 (1 + x)ⁿ 时，展开式简化为 (1 + x)ⁿ = 1 + nx + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + … 。这对于有理数 n 尤为有用。

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

The (r+1)-th term, or general term, in the expansion of (a + b)ⁿ is given by T_{r+1} = ⁿCᵣ a^{n−r} b^r. Here r runs from 0 to n, so the first term corresponds to r=0, the second to r=1, and so on.

在 (a + b)ⁿ 的展开中，第 (r+1) 项（即通项）为 T_{r+1} = ⁿCᵣ a^{n−r} b^r。这里 r 从 0 取到 n，因此 r=0 对应第一项，r=1 对应第二项，以此类推。

To find a specific coefficient, isolate the required value of r such that the powers match. For instance, in (2 + 3x)⁸, to find the coefficient of x³, set the power of x to 3, which gives r = 3. Then substitute r into the general term formula.

要找出特定项的系数，需要确定 r 的值，使得该项中的未知数幂次相匹配。例如，在 (2 + 3x)⁸ 中求 x³ 的系数，可设 x 的指数为 3，得出 r = 3。然后将 r 代入通项公式。

Always remember to multiply any existing coefficients inside the binomial, e.g., the 3 in 3x, raised to the appropriate power. A common mistake is to forget that (bx)^r contributes b^r to the term.

务必牢记要将二项式内部的系数（例如 3x 中的 3）按相应的幂次乘入。一个常见错误是忘记了 (bx)^r 会贡献 b^r 这一因子。

5. Expanding (a + bx)^n and Finding Specific Terms | 展开 (a + bx)^n 与求特定项

When expanding (a + bx)^n, treat a as the first term and bx as the second term. The general term becomes ⁿCᵣ a^{n−r} (bx)^r = ⁿCᵣ a^{n−r} b^r x^r.

当展开 (a + bx)^n 时，将 a 视为第一项，bx 视为第二项。通项变为 ⁿCᵣ a^{n−r} (bx)^r = ⁿCᵣ a^{n−r} b^r x^r。

To find the term independent of x (the constant term), set the exponent of x to 0 and solve for r. If the solution gives an integer r between 0 and n, the constant term exists; otherwise, there is no constant term.

要求出不含 x 的项（常数项），将 x 的指数设为零并解出 r。若解出的 r 是 0 到 n 之间的整数，则存在常数项；否则展开式中没有常数项。

For example, to find the coefficient of x² in (3 − 2x)⁵, set r=2. The term is ⁵C₂ · 3^{5−2} · (−2x)² = 10 · 27 · 4 x² = 1080 x², so the coefficient is 1080.

例如，求 (3 − 2x)⁵ 中 x² 的系数，设 r=2。该项为 ⁵C₂ · 3^{5−2} · (−2x)² = 10 · 27 · 4 x² = 1080 x²，因此系数为 1080。

6. The Coefficient of x^k and Constant Terms | x^k 系数与常数项

For more complex forms like (a/x² + bx³)ⁿ, the general term involves x^{3r − 2(n−r)}. Set this exponent equal to k to find r, then compute the coefficient. This often leads to solving a linear equation in r.

对于 (a/x² + bx³)ⁿ 这样更复杂的形式，通项含有 x^{3r − 2(n−r)} 的幂次。令该指数等于 k 求出 r，再计算系数。这往往需要解关于 r 的一元一次方程。

When finding the constant term, set the total exponent of x to zero and solve for r. If r is not an integer or lies outside 0..n, the constant term is zero. These problems test your algebraic manipulation skills.

在求常数项时，令 x 的总指数等于零并解出 r。若 r 不是整数或超出 0 到 n 的范围，则常数项为零。这类题目考查代数操作能力。

Always verify that the resulting r is a valid index before computing the term. Also, check for multiple possible terms if the equation is quadratic or has more than one solution, though this is rare in basic problems.

在计算该项之前，务必验证求出的 r 是否为有效下标。若方程为二次方程且有多个解，可能有多项符合条件，不过在基础题中比较少见。

7. Binomial Expansion for Rational Powers | 有理数指数的二项式展开

When n is not a positive integer, but a rational number (such as 1/2, −1, or −2/3), the expansion (1 + x)ⁿ can be written as an infinite series valid for |x| < 1:

当 n 不是正整数，而是有理数（如 1/2、−1 或 −2/3）时，(1 + x)ⁿ 可展开为一个无穷级数，适用范围为 |x| < 1：

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

This series is obtained by extending the binomial theorem, replacing ⁿCᵣ with the generalised coefficient n(n−1)…(n−r+1) / r!. The series continues indefinitely; it is not a finite polynomial.

该级数是通过推广二项式定理得到的，将 ⁿCᵣ 替换为广义系数 n(n−1)…(n−r+1) / r!。级数无限延伸，并非有限的多项式。

For example, (1 + x)^{1/2} ≈ 1 + (1/2)x − (1/8)x² + (1/16)x³ − … for |x| < 1. Notice the factorial denominators and alternating signs when n is a fraction or negative.

例如，对于 |x| < 1，有 (1 + x)^{1/2} ≈ 1 + (1/2)x − (1/8)x² + (1/16)x³ − ... 。需要注意分母中的阶乘，以及 n 为分数或负数时符号交替出现的特点。

8. Validity and Range of x | 展开式的有效性及 x 的范围

The infinite binomial expansion of (1 + x)ⁿ converges only when |x| < 1, or equivalently −1 < x < 1. This condition ensures the terms become smaller and the series tends to a finite limit.

(1 + x)ⁿ 的无穷二项展开式仅在 |x| < 1，即 −1 < x < 1 时收敛。该条件确保各项逐渐变小，级数趋向一个有限极限。

If you are expanding (a + bx)ⁿ with rational n, you must first rewrite it as a^n (1 + (b/a)x)ⁿ. The expansion is then valid for |(b/a)x| < 1, i.e., |x| < |a/b|. Always state the range of validity in your answer.

如果你在展开有理数指数 n 的 (a + bx)ⁿ，必须首先将其改写为 a^n (1 + (b/a)x)ⁿ。此时展开式的有效范围为 |(b/a)x| < 1，即 |x| < |a/b|。在答案中务必注明有效范围。

For example, to expand √(4 + x), write it as 2 (1 + x/4)^{1/2}. The expansion is valid for |x/4| < 1, so |x| < 4. Never forget this crucial step; examiners often test it explicitly.

例如，要展开 √(4 + x)，可写为 2 (1 + x/4)^{1/2}。展开式在 |x/4| < 1 即 |x| < 4 时成立。切莫忘记这一关键步骤；考官常会明确考查。

9. Approximations Using Binomial Expansions | 利用二项式展开求近似值

Binomial expansions are powerful for estimating values like √(1.02), (1.01)^{-3}, or (0.98)^4. By choosing a suitable x and using the series up to the x² or x³ term, you can obtain quick approximations.

二项式展开在估算如 √(1.02)、(1.01)^{-3} 或 (0.98)^4 等数值时非常有效。选择合适的 x 并截取至 x² 或 x³ 项，就能快速得到近似值。

For instance, √(1.02) = (1 + 0.02)^{1/2} ≈ 1 + ½(0.02) − ⅛(0.02)² = 1 + 0.01 − 0.00005 = 1.00995. The error is very small, demonstrating the usefulness of the series.

例如，√(1.02) = (1 + 0.02)^{1/2} ≈ 1 + ½(0.02) − ⅛(0.02)² = 1 + 0.01 − 0.00005 = 1.00995。误差极小，体现了该级数的实用性。

When using approximations, always check that the chosen x falls within the validity range. Also, consider how many terms are needed for the required accuracy—typically up to x² or x³ suffices.

在使用近似值时，务必检查所选 x 是否在有效范围内。同时思考为达到所需精度需要保留几项——通常到 x² 或 x³ 项足矣。

10. Common Exam Pitfalls and Tips | 常见考试陷阱与技巧

Pitfall 1: Forgetting to raise the coefficient of x to the same power r. When expanding (2 − 5x)^7, a common error is to write the x³ term as ⁷C₃ 2⁴ (−5x)³, but then miss the (−5)³ factor. Always incorporate the full binomial factor.

陷阱 1：忘记将 x 的系数也求 r 次方。在展开 (2 − 5x)^7 时，常见错误是写出 x³ 项为 ⁷C₃ 2⁴ (−5x)³，却遗漏了 (−5)³ 这一因子。务必包含完整的二项因子。

Pitfall 2: Misidentifying the general term index r. Remember that T_{r+1} corresponds to ⁿCᵣ. When asked for the 5th term, you need r = 4, not r = 5. Double-check the correspondence between term number and r.

陷阱 2：误用通项下标 r。记住 T_{r+1} 对应 ⁿCᵣ。当题目要求第 5 项时，应取 r = 4，而非 r = 5。务必核对项数与 r 的对应关系。

Pitfall 3: Ignoring the validity condition for infinite series. Giving an expansion without stating ‘for |x| < ...' can lose marks. Always conclude with the range of x for which the series is valid.

陷阱 3：忽略无穷级数的收敛条件。展开后没有写明“当 |x| < ... 时成立”可能会失分。务必在最后注明级数有效的 x 范围。

Tips: Practise rewriting expressions into the form (1 + bx)ᵑ before expanding. Use brackets systematically, and always simplify factorials carefully to avoid numerical errors.

技巧：练习将表达式改写成 (1 + bx)ᵑ 的形式后再展开。有系统地使用括号，并仔细化简阶乘，避免计算错误。

11. Worked Example 1 – Positive Integer Power | 实例精讲 1 – 正整数幂

Question: Find the coefficient of x³ in the expansion of (2 + 3x)⁵.

题目：求 (2 + 3x)⁵ 展开式中 x³ 的系数。

Solution: The general term T_{r+1} = ⁵Cᵣ · 2^{5−r} · (3x)^r = ⁵Cᵣ · 2^{5−r} · 3^r · x^r. For x³, set r = 3. Then T₄ = ⁵C₃ · 2² · 3³ · x³ = 10 · 4 · 27 x³ = 1080 x³. Coefficient is 1080.

解答：通项 T_{r+1} = ⁵Cᵣ · 2^{5−r} · (3x)^r = ⁵Cᵣ · 2^{5−r} · 3^r · x^r。令 r = 3 得到 x³ 项。于是 T₄ = ⁵C₃ · 2² · 3³ · x³ = 10 · 4 · 27 x³ = 1080 x³。系数为 1080。

12. Worked Example 2 – Rational Power and Approximation | 实例精讲 2 – 有理数幂与近似

Question: Expand (1 − 2x)^{-1/2} up to the term in x² and state the range of values of x for which the expansion is valid. Hence estimate 1/√(0.98).

题目：将 (1 − 2x)^{-1/2} 展开至 x² 项，并说明展开式有效的 x 取值范围。由此估计 1/√(0.98) 的值。

Solution: Using (1 + u)ⁿ with n = −1/2 and u = −2x. The series gives: 1 + n u + [n(n−1)/2!] u² + … = 1 + (−1/2)(−2x) + [(−1/2)(−3/2)/2] (4x²) + … = 1 + x + (3/8)(4x²) + … = 1 + x + (3/2)x² + … . Validity: |u| < 1 ⇒ |−2x| < 1 ⇒ |x| < 1/2.

解答：将式子看作 (1 + u)ⁿ，其中 n = −1/2，u = −2x。代入级数公式得：1 + n u + [n(n−1)/2!] u² + … = 1 + (−1/2)(−2x) + [(−1/2)(−3/2)/2] (4x²) + … = 1 + x + (3/8)(4x²) + … = 1 + x + (3/2)x² + … 。有效范围：|u| < 1 ⇒ |−2x| < 1 ⇒ |x| < 1/2。

To estimate 1/√(0.98), rewrite √(0.98) = √(1 − 0.02) = (1 − 0.02)^{1/2}. Then 1/√(0.98) = (1 − 0.02)^{-1/2}. Compare with expansion: we need −2x = −0.02 ⇒ 2x = 0.02 ⇒ x = 0.01. Since |0.01| < 0.5, substitution is valid. Thus (1 − 0.02)^{-1/2} ≈ 1 + 0.01 + (3/2)(0.01)² = 1 + 0.01 + 0.00015 = 1.01015. This is a close approximation to the true value.

为估计 1/√(0.98)，将 √(0.98) 写为 √(1 − 0.02) = (1 − 0.02)^{1/2}。那么 1/√(0.98) = (1 − 0.02)^{-1/2}。与展开式对比，我们需令 −2x = −0.02 ⇒ x = 0.01。因为 |0.01| < 0.5，代入有效。因此 (1 − 0.02)^{-1/2} ≈ 1 + 0.01 + (3/2)(0.01)² = 1 + 0.01 + 0.00015 = 1.01015。这与真实值非常接近。

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Binomial Expansion in IB AQA Mathematics | IB AQA 数学：二项式展开 考点精讲