Binomial Expansions | 二项式展开

📚 Binomial Expansions | 二项式展开

Binomial expansions form a cornerstone of algebra in IB Mathematics, allowing us to write expressions like (a + b)ⁿ as a sum of terms involving powers of a and b. Mastering this topic unlocks shortcuts for polynomial manipulation, probability calculations with combinations, and even approximations in calculus. This article covers everything from the basic Pascal's triangle to the general binomial theorem, with clear examples and common pitfalls highlighted.

二项式展开是IB数学代数的基石，它使我们能够将形如 (a + b)ⁿ 的式子写成一系列包含 a 和 b 幂次的项之和。掌握这一主题，不仅能简化多项式运算，还能用于组合概率计算，甚至微积分中的近似估计。本文从最基本的帕斯卡三角形讲起，一直到一般的二项式定理，配合清晰的示例和常见错误提醒。

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

A binomial is an algebraic expression containing exactly two terms, such as (x + y) or (2p − 3q). Expanding a binomial raised to a positive integer power means multiplying it by itself repeatedly. For example, (a + b)² = a² + 2ab + b², and (a + b)³ = a³ + 3a²b + 3ab² + b³. Notice how the coefficients follow a symmetric pattern, and the powers of a decrease while those of b increase.

二项式是指恰好含有两项的代数式，例如 (x + y) 或 (2p − 3q)。将一个二项式的正整数次幂展开，意味着将它自身重复相乘。例如，(a + b)² = a² + 2ab + b²，而 (a + b)³ = a³ + 3a²b + 3ab² + b³。请注意系数的对称模式，以及 a 的指数逐渐减少、b 的指数逐渐增加。

The fully expanded form of (a + b)ⁿ has n + 1 terms, beginning with aⁿ and ending with bⁿ. The coefficients arise from combinatorial choices, which we will formalise later.

(a + b)ⁿ 的完全展开式共有 n + 1 项，首项为 aⁿ，末项为 bⁿ。各项系数来源于组合选择，我们稍后会进行形式化总结。

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2. Pascal's Triangle and Basic Expansions | 帕斯卡三角形与基本展开

Pascal' ly above it. The rows correspond to n = 0, 1, 2, 3, ... giving the coefficients of (a + b)ⁿ. The first few rows are shown below.

帕斯卡三角形为较小的 n 提供了一种快速确定二项式系数的直观方法。每个数字等于它正上方两个数字之和。每一行对应 n = 0, 1, 2, 3, ...，给出 (a + b)ⁿ 的系数。前几行如下所示。

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

The triangle is symmetric, so the k‑th coefficient from the left equals the k‑th from the right. For (x + 2y)⁴, the coefficients 1, 4, 6, 4, 1 multiply the terms x⁴, x³(2y), x²(2y)², x(2y)³, (2y)⁴, giving the expanded form with additional powers of 2.

三角形是对称的，因此从左数第 k 个系数等于从右数第 k 个系数。对于 (x + 2y)⁴，系数 1, 4, 6, 4, 1 分别与项 x⁴、x³(2y)、x²(2y)²、x(2y)³、(2y)⁴ 相乘，从而得到包含 2 的额外幂次的展开式。

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3. Factorial Notation | 阶乘记号

The factorial of a non‑negative integer n, written n!, is the product of all positive integers from 1 to n. We define 0! = 1. For instance, 4! Factorials grow very rapidly and are fundamental in the formula for binomial coefficients.

非负整数 n 的阶乘，记作 n!，是从 1 到 n 的所有正整数的乘积。我们定义 0! = 1。例如，4! = 4 × 3 × 2 × 1 = 24。阶乘增长极快，它是二项式系数公式的基础。

In algebraic manipulation, n! can often be cancelled with smaller factorials, which helps when finding specific coefficients without fully expanding.

在代数运算中，n! 经常可以和更小的阶乘约分，这有助于在不完全展开的情况下求出特定项的系数。

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4. Combinations and the Binomial Coefficient | 组合与二项式系数

The binomial coefficient, often read as 'n choose k', counts the number of ways to choose k items from a set of n without regard to order. It is denoted variously as ⁿC_k, C(n, k), or by the notation with parentheses. Its formula is:

二项式系数，常读作“n 选 k”，表示从 n 个物品中不计顺序选取 k 个的方法数。它有多种记法：ⁿC_k、C(n, k)，或用括号表示。其公式为：

ⁿC_k = n! / (k! (n − k)!)

For the expansion of (a + b)ⁿ, the coefficient of the term a^n−k b^k is exactly ⁿC_k. This links combinatorics and algebra beautifully. For example, ⁵C₂ = 10, and indeed the coefficient of a³b² in (a + b)⁵ is 10.

在 (a + b)ⁿ 的展开式中，项 a^n−k b^k 的系数恰好是 ⁿC_k。这巧妙地将组合数学与代数联系在一起。例如，⁵C₂ = 10，而 (a + b)⁵ 中 a³b² 的系数正是 10。

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5. The Binomial Theorem for Positive Integer Exponents | 正整数指数的二项式定理

The general binomial theorem states that for any positive integer n,

一般的二项式定理指出，对于任意正整数 n，

(a + b)ⁿ = Σ_k=0^{n n}C_k a^n−k b^k

This compact representation allows us to expand any binomial without drawing Pascal's triangle. Consider (2x − 3)⁴. Using the theorem, the expansion is ⁴C₀(2x)⁴ + ⁴C₁(2x)³(−3) + ⁴C₂(2x)²(−3)² + ⁴C₃(2x)(−3)³ + ⁴C₄(−3)⁴. Evaluating the coefficients and powers yields 16x⁴ − 96x³ + 216x² − 216x + 81.

这个紧凑的表达式使我们无需画出帕斯卡三角形便能展开任意二项式。考虑 (2x − 3)⁴。应用定理，展开式为 ⁴C₀(2x)⁴ + ⁴C₁(2x)³(−3) + ⁴C₂(2x)²(−3)² + ⁴C₃(2x)(−3)³ + ⁴C₄(−3)⁴。计算系数与幂次后得到 16x⁴ − 96x³ + 216x² − 216x + 81。

When the binomial contains a negative sign, the signs of the terms alternate, governed by (−1)^k in the coefficient. Always apply parentheses carefully to the whole term including any subtraction.

当二项式中含有负号时，各项的符号交替变化，受 (−1)^k 支配。请务必用括号将整个项（包括减号）仔细括起。

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6. General Term of a Binomial Expansion | 二项式展开的通项

The (r + 1)‑th term in the expansion of (a + b)ⁿ is given by

(a + b)ⁿ 展开式中的第 (r + 1) 项为

T_r+1 = ⁿC_r a^n−r b^r

la is extremely handy when we only need one particular term, such as the term containing x⁵ in (2x + 1)⁸. First, identify the target power of the variable in a^n−r b^r and solve for r.

其中 r = 0, 1, 2, ..., n。当我们只需要某一个特定项时，这个公式极其方便，例如求 (2x + 1)⁸ 中含有 x⁵ 的项。首先，确定 a^n−r b^r 中变量的目标幂次并解出 r。

For instance, in (x² + 3/x)¹⁰, the general term involves x^2(10−r) times x^−r, giving power 20 − 3r. Setting 20 − 3r equal to a required exponent reveals r and hence the term.

例如，在 (x² + 3/x)¹⁰ 中，通项包含 x^2(10−r) 乘 x^−r，得到 x 的指数为 20 − 3r。令 20 − 3r 等于所需指数即可求出 r，进而得到该项。

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7. Finding Coefficients of Specific Terms | 求特定项的系数

IB exam questions frequently ask for 'the coefficient of x^m' in an expansion. The procedure is: write the general term, simplify the power of x, set it equal to m to find r, then substitute back into the coefficient expression. Do not forget to include any constant multipliers from the binomial terms themselves.

IB考试中经常要求“展开式中 x^m 的系数”。步骤如下：写出通项，化简 x 的指数，令其等于 m 以求出 r，再代回系数表达式。切莫忘记计入二项式各项自身携带的常倍数。

Example: Find the coefficient of x⁶ in (1 + 2x³)⁸. The general term is ⁸C_r (1)^8−r (2x³)^r = ⁸C_r 2^r x^3r. Setting 3r = 6 gives r = 2. The term is ⁸C₂ × 2² x⁶ = 28 × 4 x⁶ = 112x⁶, so the coefficient is 112.

示例：求 (1 + 2x³)⁸ 中 x⁶ 的系数。通项为 ⁸C_r (1)^8−r (2x³)^r = ⁸C_r 2^r x^3r。令 3r = 6 得 r = 2。该项即为 ⁸C₂ × 2² x⁶ = 28 × 4 x⁶ = 112x⁶，故系数为 112。

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8. Sum of Coefficients and Symmetry | 系数和与对称性

A neat trick reveals the sum of all coefficients: set a = 1 and b = 1 in (a + b)ⁿ. The expansion becomes (1 + 1)ⁿ = 2ⁿ, which equals the sum of the binomial coefficients. For (2 + 3x)ⁿ, substituting x = 1 gives the sum of coefficients as 5ⁿ.

求所有系数之和有一个巧妙的方法：在 (a + b)ⁿ 中令 a = 1、b = 1，展开式即变为 (1 + 1)ⁿ = 2ⁿ，这就是二项式系数之和。对于 (2 + 3x)ⁿ，代入 x = 1 得到系数和为 5ⁿ。

Symmetry property: ⁿC_k = ⁿC_n−k. This mirrors the left‑right symmetry of Pascal's triangle and can simplify calculations when k is large by using n − k instead.

对称性：ⁿC_k = ⁿC_n−k。这反映了帕斯卡三角形的左右对称性，并且当 k 较大时可通过用 n − k 替代来简化计算。

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9. Binomial Expansions and Approximations | 二项式展开与近似计算

When the absolute value of x is small, the first few terms of the expansion of (1 + x)ⁿ provide useful approximations. For instance, (1 + x)ⁿ ≈ 1 + nx for small x, ignoring x² and higher powers. This is widely applied in physics and engineering to estimate quantities like (1.002)¹⁰ ≈ 1 + 10 × 0.002 = 1.02.

当 |x| 很小时，(1 + x)ⁿ 展开式的前几项能给出有用的近似。例如，对于小量 x，有 (1 + x)ⁿ ≈ 1 + nx，而忽略 x² 及更高次项。这种方法广泛应用于物理和工程中估算诸如 (1.002)¹⁰ ≈ 1 + 10 × 0.002 = 1.02 之类的量。

In IB examinations, you might be asked to use the binomial expansion up to the x² term to approximate a value and then state an upper bound for the error. Recognise that the remainder terms are of order x³.

在IB考试中，你可能需要利用二项式展开到 x² 项来近似某个数值，然后给出误差的上界。要知道余项是 x³ 数量级的。

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10. Extensions: Rational Exponents (Optional) | 扩展：有理指数（选学）

For IB Mathematics: Analysis and Approaches HL, the binomial theorem extends to negative and fractional exponents using an infinite series. The expansion (1 + x)ⁿ = 1 + nx + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + ... is valid for |x| < 1 when n is not a positive integer. Here n can be 1/2 or −1, giving representations of square roots or reciprocals.

在IB数学分析与方法（HL）中，二项式定理可推广到负指数和分数指数，形成无穷级数。展开式 (1 + x)ⁿ = 1 + nx + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + ... 在 n 不为正整数且 |x| < 1 时成立。此时 n 可以是

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