📚 Binomial Expansions | 二项式展开
Binomial expansions form a cornerstone of algebra, enabling us to rewrite expressions like (a + b)ⁿ or (1 + x)ⁿ as a sum of terms involving powers of a and b. Mastery of this topic is essential for IB Mathematics, as it links combinatorics, algebraic manipulation, and even calculus through infinite series. In this article, we will explore the binomial theorem for positive integer exponents, Pascal’s triangle, the general term formula, and the extension to negative and fractional exponents, all while emphasising practical exam techniques.
二项式展开是代数的基石，它使我们能够将 (a + b)ⁿ 或 (1 + x)ⁿ 这样的表达式写成 a 和 b 的幂次项之和的形式。掌握这个主题对于 IB 数学至关重要，因为它不仅联系了组合学与代数运算，还通过无穷级数与微积分挂钩。本文将深入探讨正整数指数的二项式定理、帕斯卡三角形、通项公式，以及向负指数和分式指数的推广，同时强调实用的考试技巧。
1. What is a Binomial Expansion? | 什么是二项式展开？
increasing or decreasing powers. For example, (x + y)² = x² + 2xy + y² is the simplest non‑ symmetric pattern of coefficients that can be generalised for any positive integer power.
二项式是指包含两项的代数式，例如 (x + y) 或 (2a − 3b)。二项式展开就是将二项式自身重复相乘，并把结果写成一个各项幂次递增或递减的多项式。例如，(x + y)² = x² + 2xy + y² 就是最简单的非平凡情形。这个展开式呈现出对称的系数模式，而这种模式可以推广到任意正整数次幂。
Understanding how to systematically expand (a + b)ⁿ without manually multiplying every factor is the key objective. It saves time and reduces errors, especially for large n. The binomial theorem provides this systematic method using coefficients known as binomial coefficients, which appear in Pascal’s triangle and are expressed via factorials.
系统性地将 (a + b)ⁿ 展开而不必手动逐项相乘，是我们的核心目标。这不仅能节省时间，还能减少错误，尤其当 n 很大时。二项式定理正是通过一种系统的方法，使用名为二项式系数的数来实现展开，这些系数出现在帕斯卡三角形中，并用阶乘表示。
2. Pascal’s Triangle: The Foundation | 帕斯卡三角形：基础
n = 0: a single 1. The next row for n = 1 is 1 1, for n = 2 it is 1 2 1, and so on. The k‑th entry in the n‑ from 0) gives the binomial coefficient for the term aⁿ⁻ᵏ bᵏ in the expansion of (a + b)ⁿ.
帕斯卡三角形是一个数字三角形阵列，其中每个数等于它正上方两数之和。第一行对应 n = 0：一个单独的 1。接下来 n = 1 的行是 1 1，n = 2 是 1 2 1，依此类推。第 n 行中第 k 个数（从 0 开始计数）给出了 (a + b)ⁿ 展开式中项 aⁿ⁻ᵏ bᵏ 的二项式系数。
a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ corresponds to the fourth row of Pascal’s triangle: 1 4 6 4 1. This visual pattern is extremely helpful for small values of n, but for larger exponents it becomes inefficient. Nevertheless, it lays the intuitive groundwork for the binomial theorem.
例如，展开式 (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ 对应着帕斯卡三角形的第四行：1 4 6 4 1。这个直观图形对于较小的 n 值非常有用，然而当指数变大时则效率低下。不过，它仍为二项式定理奠定了直观的基础。
3. Binomial Coefficients and nCr Notation | 二项式系数与 nCr 记法
choose r items from n items without regard to order is denoted ⁿCᵣ or C(n, r) and equals n! / (r!(n − r)!). In the expansion of (a + b)ⁿ, the coefficient of the term aⁿ⁻ʳ bʳ s exactly ⁿCᵣ. This explains why ⁿC₀ = 1 for the first term and ⁿCₙ = 1 for the last term.
二项式系数可以用组合公式计算：从 n 个物品中无序地选出 r 个的方法数记作 ⁿCᵣ 或 C(n, r)，它等于 n! / (r!(n − r)!)。在 (a + b)ⁿ 的展开式中，项 aⁿ⁻ʳ bʳ 的系数正是 ⁿCᵣ。这也解释了为什么首项系数 ⁿC₀ = 1、末项系数 ⁿCₙ = 1。
function, but knowing the factorial formula is essential for algebraic manipulations and when dealing with unknown variables.
理解阶乘记法与帕斯卡三角形之间的关系至关重要。对称性 ⁿCᵣ = ⁿCₙ₋ᵣ 恰好反映了三角形的对称形状。例如，⁵C₂ = 10 = ⁵C₃。计算器通常有 nCr 功能，但掌握阶乘公式对于代数运算和处理未知变量必不可少。
4. The Binomial Theorem for (a+b)ⁿ | (a+b)ⁿ 的二项式定理
nomial theorem for a positive integer n states that (a + b)ⁿ = Σ (from r = 0 to n) ⁿCᵣ aⁿ⁻ʳ bʳ. In expanded form:
(a + b)ⁿ = aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCᵣ aⁿ⁻ʳ bʳ + … + bⁿ
对于正整数 n，二项式定理表述为：(a + b)ⁿ = Σ (r = 0 到 n) ⁿCᵣ aⁿ⁻ʳ bʳ。其展开形式为：
(a + b)ⁿ = aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCᵣ aⁿ⁻ʳ bʳ + … + bⁿ
Each term has a specific structure: the powers of a decrease from n to 0, the powers of b increase from 0 to n, and the sum of the exponents in each term is always n. The coefficients start with 1 and end with 1, following the values given by the corresponding row of Pascal’s triangle or the ⁿCᵣ formula.
每一项都有特定的结构：a 的指数从 n 递减到 0，b 的指数从 0 递增到 n，且每一项的指数之和恒为 n。系数从 1 开始并以 1 结束，遵循帕斯卡三角形对应行的数值或 ⁿCᵣ 公式。
tituting a = 2x and b = −3. Careful handling of signs and powers is vital to avoid errors.
需要特别注意的是，a 和 b 是占位符，它们可以是任意代数式。若 a = 2x、b = −3，则将 a = 2x 和 b = −3 代入即可展开 (2x − 3)⁴。正确处理符号和幂次对避免错误至关重要。
5. Expanding Binomials: Step‑by‑Step Method | 展开二项式：逐步方法
5·16 = 80, and finally 32. The fully expanded form is x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32.
要展开 (x + 2)⁵，先确定 a = x, b = 2, n = 5。系数依次为 ⁵C₀ = 1, ⁵C₁ = 5, ⁵C₂ = 10, ⁵C₃ = 10, ⁵C₄ = 5, ⁵C₅ = 1。然后写出各项：x⁵, 5·x⁴·2, 10·x³·2², 10·x²·2³, 5·x·2⁴, 2⁵。化简系数：5·2 = 10, 10·4 = 40, 10·8 = 80, 5·16 = 80, 最后 32。完整展开式为 x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32。
1)³. Compute: 27y³, 3·9y²·(−1) = −27y², 3·3y·1 = 9y, and −1. Result: 27y³ − 27y² + 9y − 1.
当出现负号或分数时，要始终使用括号。以 (3y − 1)³ 为例，设 a = 3y, b = −1, n = 3。系数为 1, 3, 3, 1。各项为：(3y)³, 3·(3y)²·(−1), 3·(3y)·(−1)², (−1)³。计算得：27y³, 3·9y²·(−1) = −27y², 3·3y·1 = 9y, 与 −1。结果为 27y³ − 27y² + 9y − 1。
6. Finding a Specific Term in an Expansion | 求展开式中的特定项
entire binomial, the general term formula allows you to target a single term. The (r + 1)‑th term in the expansion of (a + b)ⁿ is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ the 5th term (i.e., r = 4) in (x + 3)⁸, compute T₅ = ⁸C₄ x⁸⁻⁴ 3⁴ = 70 · x⁴ · 81 = 5670 x⁴.
不必展开整个二项式，利用通项公式可以锁定某一特定项。(a + b)ⁿ 展开式中的第 (r + 1) 项为 Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ。例如，要求 (x + 3)⁸ 的第 5 项（即 r = 4），计算 T₅ = ⁸C₄ x⁸⁻⁴ 3⁴ = 70 · x⁴ · 81 = 5670 x⁴。
tant term) in an expansion like (2x² − 1/x)⁶, express the general term: Tᵣ₊₁ = ⁶Cᵣ (2x²)⁶⁻ʳ (−1/x)ʳ = ⁶Cᵣ 2⁶⁻ʳ (−1)ʳ x¹²⁻²ʳ⁻ʳ = ⁶Cᵣ 2⁶⁻ʳ (−1)ʳ x¹²⁻³ʳ. Set the exponent of x to zero: 12 − 3r = 0 ⇒ r = 4. Substitute r = 4 to get the constant term: ⁶C₄ 2² (−1)⁴ = 15 · 4 · 1 = 60.
如果要找出 (2x² − 1/x)⁶ 展开式中不含 x 的项（常数项），首先写出通项：Tᵣ₊₁ = ⁶Cᵣ (2x²)⁶⁻ʳ (−1/x)ʳ = ⁶Cᵣ 2⁶⁻ʳ (−1)ʳ x¹²⁻²ʳ⁻ʳ = ⁶Cᵣ 2⁶⁻ʳ (−1)ʳ x¹²⁻³ʳ。令 x 的指数为零：12 − 3r = 0 ⇒ r = 4。代入 r = 4 得到常数项：⁶C₄ 2² (−1)⁴ = 15 · 4 · 1 = 60。
Identifying the correct r value is a key exam skill, often tested in IB papers. Always check whether the question asks for the “term in xᵏ” or the “coefficient of xᵏ”, as the latter requires only the numerical factor.
确定正确的 r 值是一项关键的考试技能，常在 IB 试题中出现。务必看清题目要求的是“xᵏ 的项”还是“xᵏ 的系数”，因为后者只需要提取数值因子。
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