📚 Binomial Expansion Theorem Explained | 二项式展开公式精讲
The binomial theorem is one of the most powerful and frequently tested tools in algebra, combinatorics, and calculus. In this article, we will break down the binomial expansion formula, understand its notation, explore its properties, and apply it to a variety of exam-style problems.
二项式定理是代数、组合数学和微积分中最重要且考试频率最高的工具之一。本文将深入拆解二项式展开公式，理解其符号含义，探索其性质，并通过多种考试风格的题目进行实战演练。
1. The Binomial Theorem Formula | 二项式定理公式
The binomial theorem states that for any positive integer n, the expansion of (a + b)ⁿ can be written as a sum of terms of the form C(n, k) aⁿ⁻ᵏ bᵏ, where k runs from 0 to n. Here, C(n, k) is the binomial coefficient, also written as ⁿCₖ or (ⁿₖ).
二项式定理指出：对任意正整数 n，(a + b)ⁿ 的展开式可以写成 C(n, k) aⁿ⁻ᵏ bᵏ 的形式，其中 k 从 0 取到 n。C(n, k) 称为二项式系数，也记作 ⁿCₖ 或 (ⁿₖ)。
(a + b)ⁿ = Σₖ₌₀ⁿ C(n, k) aⁿ⁻ᵏ bᵏ = C(n,0)aⁿ + C(n,1)aⁿ⁻¹b + C(n,2)aⁿ⁻²b² + … + C(n,n)bⁿ
For example, when n = 3, the expansion is a³ + 3a²b + 3ab² + b³, since C(3,0)=1, C(3,1)=3, C(3,2)=3, and C(3,3)=1.
例如，当 n = 3 时，展开式为 a³ + 3a²b + 3ab² + b³，因为 C(3,0)=1，C(3,1)=3，C(3,2)=3，C(3,3)=1。
2. Factorials and Binomial Coefficients | 阶乘与二项式系数
The binomial coefficient C(n, k) is defined using factorials. The factorial of a non-negative integer n, written n!, is the product of all positive integers from 1 to n. In particular, 0! = 1.
二项式系数 C(n, k) 通过阶乘定义。非负整数 n 的阶乘记作 n!，是从 1 到 n 所有正整数的乘积。特别规定 0! = 1。
C(n, k) = n! / (k! (n − k)!)
This formula gives us a convenient way to compute coefficients without writing out Pascal’s triangle. For instance, C(5, 2) = 5! / (2! × 3!) = 120 / (2 × 6) = 10.
这个公式提供了一种便捷的计算系数的方法，无需逐行写出杨辉三角。例如，C(5, 2) = 5! / (2! × 3!) = 120 / (2 × 6) = 10。
3. Pascal’s Triangle Connection | 杨辉三角的联系
Pascal’s triangle is a triangular array of binomial coefficients. Each row corresponds to a value of n, and each entry is the sum of the two entries directly above it. Row n of Pascal’s triangle gives the coefficients of (a + b)ⁿ in order.
杨辉三角是由二项式系数构成的三角形阵列。每一行对应一个 n 值，每一个数等于其正上方两个数之和。杨辉三角的第 n 行按顺序给出了 (a + b)ⁿ 的展开系数。
| 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 |
| n = 5 | 1 5 10 10 5 1 |
The triangle is symmetric: C(n, k) = C(n, n − k). This follows directly from the factorial formula, and it means that the coefficients read the same from left to right as from right to left.
杨辉三角具有对称性：C(n, k) = C(n, n − k)。这可由阶乘公式直接推出，意味着系数从左到右和从右到左读完全相同。
4. Key Properties of Binomial Coefficients | 二项式系数的关键性质
Several important identities help in simplifying expressions and solving problems. These appear frequently in exams, so we list them here with brief explanations.
几个重要的恒等式有助于化简表达式和解决问题。它们在考试中频繁出现，下面列出并加以简要说明。
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Symmetry identity: C(n, k) = C(n, n − k). This means choosing k items is equivalent to leaving out n − k items.
对称恒等式：C(n, k) = C(n, n − k)。也就是说，选取 k 个元素等价于排除 n − k 个元素。
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Addition identity (Pascal’s rule): C(n, k) + C(n, k+1) = C(n+1, k+1). This is exactly how Pascal’s triangle is built.
加法恒等式（帕斯卡法则）：C(n, k) + C(n, k+1) = C(n+1, k+1)。这正是杨辉三角的构造方式。
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Sum of a row: C(n,0) + C(n,1) + … + C(n,n) = 2ⁿ. This follows by setting a = b = 1 in the binomial theorem.
整行系数和：C(n,0) + C(n,1) + … + C(n,n) = 2ⁿ。令二项式定理中 a = b = 1 即可得到。
5. Finding the General Term | 求通项
For the expansion of (a + b)ⁿ, the term containing bᵏ is called the (k + 1)-th term and is given by Tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ. This general term formula is essential for solving problems that ask for a specific term, such as the middle term or the term independent of x.
在 (a + b)ⁿ 的展开式中，含有 bᵏ 的项称为第 (k + 1) 项，其表达式为 Tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ。通项公式对于求解特定项（如中间项或不含 x 的常数项）至关重要。
Worked example: Find the 5th term in the expansion of (x + 2)¹⁰.
例题：求 (x + 2)¹⁰ 展开式中的第 5 项。
Here n = 10 and the 5th term corresponds to k = 4. So T₅ = C(10, 4) x¹⁰⁻⁴ (2)⁴ = 210 × x⁶ × 16 = 3360 x⁶.
此处 n = 10，第 5 项对应 k = 4。因此 T₅ = C(10, 4) x¹⁰⁻⁴ (2)⁴ = 210 × x⁶ × 16 = 3360 x⁶。
6. Expanding Expressions with Two Terms | 展开两项式表达式
When the expression inside the bracket has more than two terms or includes coefficients, we must carefully substitute. For example, (2x − 3)⁵ can be treated as (a + b)⁵ with a = 2x and b = −3.
当括号内的表达式包含系数或不止两项时，我们需要仔细代入。例如，(2x − 3)⁵ 可视为 a = 2x、b = −3 的 (a + b)⁵。
Worked example: Expand (2x − 3)⁴.
例题：展开 (2x − 3)⁴。
Using the binomial theorem: T₁ = (2x)⁴ = 16x⁴; T₂ = 4(2x)³(−3) = −96x³; T₃ = 6(2x)²(−3)² = 216x²; T₄ = 4(2x)(−3)³ = −216x; T₅ = (−3)⁴ = 81. Therefore the expansion is 16x⁴ − 96x³ + 216x² − 216x + 81.
利用二项式定理：T₁ = (2x)⁴ = 16x⁴；T₂ = 4(2x)³(−3) = −96x³；T₃ = 6(2x)²(−3)² = 216x²；T₄ = 4(2x)(−3)³ = −216x；T₅ = (−3)⁴ = 81。因此展开式为 16x⁴ − 96x³ + 216x² − 216x + 81。
7. Finding the Term Independent of x | 求不含 x 的常数项
A common exam question is to find the constant term in an expansion such as (x + 1/x)⁶. The term independent of x occurs when the powers of x in the numerator and denominator cancel out.
常见考题是求类似 (x + 1/x)⁶ 展开式中的常数项。当分子和分母中 x 的幂相互抵消时，就出现不含 x 的项。
Worked example: Find the constant term in (x + 1/x)⁶.
例题：求 (x + 1/x)⁶ 的常数项。
The general term is Tₖ₊₁ = C(6, k) x⁶⁻ᵏ (1/x)ᵏ = C(6, k) x⁶⁻²ᵏ. For this to be constant, we need 6 − 2k = 0, so k = 3. The term is C(6, 3) = 20.
通项为 Tₖ₊₁ = C(6, k) x⁶⁻ᵏ (1/x)ᵏ = C(6, k) x⁶⁻²ᵏ。要使该项为常数，需 6 − 2k = 0，即 k = 3。该项为 C(6, 3) = 20。
8. Using the Binomial Theorem for Approximations | 利用二项式定理进行近似计算
One of the most practical applications is approximating numbers like (1.01)¹⁰ or (0.99)⁵. We rewrite the base as (1 + x) and take only the first few terms of the expansion, because later terms become very small.
最实际的应用之一是近似计算形如 (1.01)¹⁰ 或 (0.99)⁵ 的数值。我们将底数改写为 (1 + x)，并只取展开式的前几项，因为后面各项变得非常小。
Worked example: Approximate (1.02)⁶ using the first three terms of the binomial expansion.
例题：利用二项展开式的前三项近似计算 (1.02)⁶。
Write 1.02 = 1 + 0.02. Then (1 + 0.02)⁶ ≈ 1 + 6(0.02) + 15(0.02)² = 1 + 0.12 + 0.006 = 1.126. The exact value is about 1.12616, so this approximation is accurate to three decimal places.
将 1.02 写作 1 + 0.02。则 (1 + 0.02)⁶ ≈ 1 + 6(0.02) + 15(0.02)² = 1 + 0.12 + 0.006 = 1.126。精确值约为 1.12616，因此该近似精确到小数点后三位。
9. Binomial Expansion with Negative or Fractional Indices | 负指数或分数指数的二项式展开
For fractional or negative n, the binomial theorem takes an infinite series form, provided |x| < 1:
对于分数或负指数 n，二项式定理给出无穷级数形式，条件为 |x| < 1：
(1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …
This is often tested in A-level and IB exams. For example, (1 + x)⁻¹ = 1 − x + x² − x³ + … for |x| < 1.
这在 A-level 和 IB 考试中经常出现。例如，(1 + x)⁻¹ = 1 − x + x² − x³ + …（其中 |x| < 1）。
Worked example: Find the first four terms of (1 + x)¹ᐟ².
例题：求 (1 + x)¹ᐟ² 的前四项。
Using the formula with n = 1/2: 1 + (1/2)x + (1/2)(−1/2)x²/2 + (1/2)(−1/2)(−3/2)x³/6 = 1 + x/2 − x²/8 + x³/16.
令 n = 1/2，代入公式：1 + (1/2)x + (1/2)(−1/2)x²/2 + (1/2)(−1/2)(−3/2)x³/6 = 1 + x/2 − x²/8 + x³/16。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Students often forget that the coefficient C(n, k) corresponds to k starting from 0, so the “first term” uses k = 0, not k = 1. Another common mistake is ignoring the sign when expanding expressions like (a − b)ⁿ. Always treat b as negative.
学生常常忘记 C(n, k) 的 k 是从 0 开始的，因此“第一项”对应 k = 0，而不是 k = 1。另一个常见错误是展开 (a − b)ⁿ 时忽略符号。始终将 b 视为负数。
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Tip 1: Write down the general term first before solving for a specific term.
技巧一：在求特定项之前，先写出通项公式。
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Tip 2: For constant terms, set the total exponent of x to zero and solve for k.
技巧二：对于常数项
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