Pascal's Triangle and the Binomial Theorem | 杨辉三角与二项式定理

📚 Pascal’s Triangle and the Binomial Theorem | 杨辉三角与二项式定理

The binomial theorem is one of the most elegant and practical tools in algebra. It allows us to expand expressions of the form \((a+b)^n\) without tedious multiplication. At the heart of this theorem lies a fascinating triangular array of numbers known as Pascal’s Triangle — a structure that appears throughout mathematics, from combinatorics to probability.

二项式定理是代数中最优雅且最实用的工具之一。它使我们无需繁琐的逐项乘法即可展开形如 \((a+b)^n\) 的表达式。这个定理的核心是一个迷人的三角形数字阵列——杨辉三角（又称帕斯卡三角），这一结构贯穿了整个数学，从组合学到概率论无处不在。

1. What Is Pascal’s Triangle? | 什么是杨辉三角？

Pascal’s Triangle is a triangular arrangement of numbers where each number is the sum of the two numbers directly above it. The triangle begins with a single 1 at the top, and each subsequent row is constructed by adding adjacent numbers from the row above, with 1s placed at both ends.

杨辉三角是一种三角形的数字排列，其中每个数字是其正上方两个数字之和。这个三角形从顶部的单个 1 开始，每一行都由上一行相邻数字相加构成，并在两端放置 1。

Here are the first six rows of Pascal’s Triangle:

以下是杨辉三角的前六行：

Row 0:            1
Row 1:          1  1
Row 2:        1  2  1
Row 3:      1  3  3  1
Row 4:    1  4  6  4  1
Row 5:  1  5 10 10  5  1
Row 6: 1  6 15 20 15  6  1

Notice that row \(n\) contains \(n+1\) numbers. The first and last entries of every row are always 1.

注意第 \(n\) 行包含 \(n+1\) 个数字。每一行的首尾项始终为 1。

2. Historical Roots | 历史渊源

Although this triangle is named after the French mathematician Blaise Pascal (1623–1662), who studied it extensively in his 1654 treatise Treatise on the Arithmetical Triangle, the array was known in China centuries earlier. The Chinese mathematician Yang Hui (杨辉) presented the triangle in his 1261 book Xiangjie Jiuzhang Suanfa (详解九章算法), which is why it is commonly called Yang Hui’s Triangle in Chinese mathematics.

尽管这个三角形以法国数学家布莱兹·帕斯卡（1623–1662）的名字命名——他在1654年出版的《算术三角形论》中对其进行了深入研究——但这个数字阵列早在几个世纪前就已在中国为人所知。中国数学家杨辉在1261年所著的《详解九章算法》中便呈现了这个三角形，因此在中国数学中通常称之为杨辉三角。

The triangle also appeared in the works of the 11th-century Persian poet and mathematician Omar Khayyam and in 13th-century Indian mathematics. Its history reflects how mathematical ideas develop across civilisations.

这个三角形还出现在11世纪波斯诗人兼数学家奥马尔·海亚姆的著作中，以及13世纪的印度数学中。它的历史反映出数学思想如何跨越不同文明而发展。

3. Pascal’s Triangle and Binomial Expansion | 杨辉三角与二项式展开

The key connection between Pascal’s Triangle and the binomial theorem is that the numbers in row \(n\) of the triangle are exactly the coefficients in the expansion of \((a+b)^n\).

杨辉三角与二项式定理之间的关键联系在于：三角形中第 \(n\) 行的数字恰好是 \((a+b)^n\) 展开式中的系数。

Let us examine the first few expansions:

让我们观察前几个展开式：

\((a+b)^0 = 1\)
\((a+b)^1 = a + b\)
\((a+b)^2 = a^2 + 2ab + b^2\)
\((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
\((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)
\((a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\)

The coefficients 1, 3, 3, 1 in \((a+b)^3\) are exactly row 3 of Pascal’s Triangle. Similarly, 1, 4, 6, 4, 1 are row 4. This pattern holds for every positive integer \(n\).

\((a+b)^3\) 中展开式的系数 1, 3, 3, 1 正是杨辉三角的第3行。同样地，1, 4, 6, 4, 1 是第4行。这个模式对每个正整数 \(n\) 都成立。

4. The Combinatorial Coefficients | 组合数系数

The numbers in Pascal’s Triangle are known as binomial coefficients, and they can be expressed using combination notation. The entry in row \(n\), position \(r\) (starting from \(r = 0\)) is denoted \(\binom{n}{r}\) and is read as “n choose r”.

杨辉三角中的数字被称为二项式系数，它们可以用组合记号表示。第 \(n\) 行、第 \(r\) 个位置（从 \(r = 0\) 开始）的项记为 \(\binom{n}{r}\)，读作 “n 选 r”。

\(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)

The factorial \(n!\) (read as “n factorial”) is defined as the product of all positive integers from 1 to \(n\), with \(0! = 1\) by convention. For example:

阶乘 \(n!\) 定义为从1到\(n\) 所有正整数的乘积，按惯例 \(0! = 1\)。例如：

\(\binom{5}{2} = \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = 10\)

Using this formula, we can compute any binomial coefficient without constructing the entire triangle. For instance, \(\binom{10}{3} = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\).

利用这个公式，无需构造整个三角形即可计算任意二项式系数。例如，\(\binom{10}{3} = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)。

5. Statement of the Binomial Theorem | 二项式定理的表述

The binomial theorem gives a general formula for expanding \((a+b)^n\) for any positive integer \(n\):

二项式定理给出 \((a+b)^n\) 对任意正整数 \(n\) 展开的通用公式：

\((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\)

Expanded in full, this becomes:

完整展开后即为：

\((a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n\)

Notice the pattern: the power of \(a\) decreases from \(n\) to 0, while the power of \(b\) increases from 0 to \(n\). The sum of the exponents in each term is always \(n\).

注意其中的模式：\(a\) 的幂从 \(n\) 递减到 0，而 \(b\) 的幂从 0 递增到 \(n\)。每一项中两个指数的总和始终为 \(n\)。

6. Working with the General Term | 通项公式的应用

In many exam problems, you are asked to find a specific term in the expansion rather than the entire expansion. The \((r+1)\)-th term in the expansion of \((a+b)^n\) is given by:

在许多考试题目中，要求找到展开式中的特定项而非完整展开式。\((a+b)^n\) 展开式中的第 \((r+1)\) 项由下式给出：

\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)

For example, to find the 5th term of \((2x+3)^{10}\), set \(r = 4\) (since the indexing starts at \(r = 0\)):

例如，要求 \((2x+3)^{10}\) 的第5项，令 \(r = 4\)（因为索引从 \(r = 0\) 开始）：

\(T_5 = \binom{10}{4} (2x)^6 (3)^4 = 210 \times 64x^6 \times 81 = 1{,}088{,}640x^6\)

A common mistake is miscounting the term index. Remember that the first term corresponds to \(r = 0\), so the \(k\)-th term has \(r = k – 1\).

一个常见错误是数错项的序号。记住第一项对应 \(r = 0\)，因此第 \(k\) 项的 \(r = k – 1\)。

7. Key Properties of Binomial Coefficients | 二项式系数的重要性质

Several properties of Pascal’s Triangle and binomial coefficients are essential for solving problems efficiently.

杨辉三角和二项式系数的几个重要性质对于高效解题至关重要。

Property 1: Symmetry — The coefficients in any row of Pascal’s Triangle are symmetric: \(\binom{n}{r} = \binom{n}{n-r}\). For example, \(\binom{6}{2} = \binom{6}{4} = 15\).

性质1：对称性 — 杨辉三角中任意一行的系数是对称的：\(\binom{n}{r} = \binom{n}{n-r}\)。例如，\(\binom{6}{2} = \binom{6}{4} = 15\)。

Property 2: Sum of a Row — The sum of all entries in row \(n\) of Pascal’s Triangle is \(2^n\). In notation:

性质2：行的和 — 杨辉三角第 \(n\) 行所有项之和为 \(2^n\)。用记号表示：

\(\sum_{r=0}^{n} \binom{n}{r} = 2^n\)

This follows by substituting \(a = 1\) and \(b = 1\) in the binomial theorem.

这可以通过在二项式定理中代入 \(a = 1\) 和 \(b = 1\) 得出。

Property 3: Alternating Sum — Substituting \(a = 1\) and \(b = -1\), we obtain \(\sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0\) for \(n \geq 1\). This means the sum of coefficients at even positions equals the sum at odd positions.

性质3：交错和 — 代入 \(a = 1\) 和 \(b = -1\)，得到 \(\sum_{r=0}^{n} (-1)^r \binom{n}{r} = 0\) 对 \(n \geq 1\) 成立。这意味着偶数位置的系数和等于奇数位置的系数和。

Property 4: Pascal’s Identity — Each interior number equals the sum of the two numbers above it:

性质4：帕斯卡恒等式 — 每个内部数字等于其上方两个数字之和：

\(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\)

For instance, \(\binom{4}{1} + \binom{4}{2} = 4 + 6 = 10 = \binom{5}{2}\).

例如，\(\binom{4}{1} + \binom{4}{2} = 4 + 6 = 10 = \binom{5}{2}\)。

8. Expanding with a Negative or Fractional Index | 负指数与分数指数的展开

In A-Level mathematics, you may also encounter expansions where the index \(n\) is not a positive integer. In such cases, the binomial theorem takes the form of an infinite series, valid for \(|x| < 1\):

在 A-Level 数学中，你还会遇到指数 \(n\) 不是正整数的情况。此时，二项式定理以无穷级数的形式呈现，且要求 \(|x| < 1\) 才能成立：

\((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots

The general term is \(\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}x^r\), where the coefficient is no longer a simple combination number but is computed using the falling factorial.

通项为 \(\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}x^r\)，此时系数不再是简单的组合数，而是通过下降阶乘计算。

Example: Expand \(\frac{1}{1-x}\) using the binomial theorem with \(n = -1\).

示例： 利用二项式定理取 \(n = -1\) 展开 \(\frac{1}{1-x}\)。

\((1-x)^{-1} = 1 + (-1)(-x) + \frac{(-1)(-2)}{2}(-x)^2 + \frac{(-1)(-2)(-3)}{6}(-x)^3 + \cdots = 1 + x + x^2 + x^3 + \cdots

This is the familiar geometric series, valid for \(|x| < 1\). Mastering this form is crucial for approximation questions.

这正是我们熟悉的几何级数，条件为 \(|x| < 1\)。掌握这种形式对于近似计算类题目至关重要。

9. Applications in Probability and Combinatorics | 在概率与组合学中的应用

Binomial coefficients appear naturally when counting the number of ways to choose \(r\) objects from a set of \(n\) objects. This is fundamental in probability problems, especially those involving the binomial distribution.

二项式系数自然出现在从 \(n\) 个物件中选出 \(r\) 个物件的方法数的计数问题中。这在概率问题中非常基础，尤其是涉及二项分布的问题。

For example, the probability of getting exactly 3 heads in 5 coin tosses is:

例如，抛5枚硬币恰好出现3次正面的概率为：

\(P(X = 3) = \binom{5}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2 = 10 \times \frac{1}{32} = \frac{10}{32} = 0.3125\)

In general, if an experiment has success probability \(p\), the probability of exactly \(k\) successes in \(n\) independent trials is \(\binom{n}{k}p^k(1-p)^{n-k}\). This is the binomial distribution.

一般地，如果一次试验的成功概率为 \(p\)，那么在 \(n\) 次独立重复试验中恰好出现 \(k\) 次成功的概率为 \(\binom{n}{k}p^k(1-p)^{n-k}\)。这就是二项分布。

10. Approximation Using the Binomial Theorem | 利用二项式定理做近似计算

One of the most common exam applications is using the binomial expansion to compute numerical approximations. For small values of \(x\), higher-order terms in the expansion become very small and can be neglected.

最常见的考试应用之一是使用二项式展开进行数值近似。当 \(x\) 很小时，展开式中高次项变得非常小，可以忽略不计。

Example: Approximate \(\sqrt{1.04}\).

示例： 近似计算 \(\sqrt{1.04}\)。

\(\sqrt{1.04} = (1+0.04)^{1/2} = 1 + \frac{1}{2}(0.04) + \frac{(1/2)(-1/2)}{2}(0.04)^2 + \cdots\)

\(= 1 + 0.02 – 0.0002 + \cdots \approx 1.0198\)

The actual value of \(\sqrt{1.04}\) is approximately 1.019804, so our approximation is extremely close. The key here is to rewrite the expression in the form \((1+x)^n\) with small \(x\).

\(\sqrt{1.04}\) 的实际值约为 1.019804，因此我们的近似非常接近。关键在于将表达式改写为 \((1+x)^n\) 的形式，且使 \(x\) 很小。

11. Exam Tips and Common Pitfalls | 考试技巧与常见误区

To score well on binomial theorem questions in A-Level exams, keep the following points in mind.

要在 A-Level 考试中做好二项式定理相关题目，请记住以下几点。

Check the power: Always ensure that \((a+b)^n\) is written in the correct form before expanding. If the bracket has a coefficient, factor it out or apply the theorem carefully.
检查指数： 展开前务必确保 \((a+b)^n\) 已写成正确形式。如果括号内有系数，先提取出来或仔细应用定理。

Mind the sign: When \(b\) is negative, the terms alternate in sign. For \((1-x)^n\), the signs are \(+ – + – \cdots\).
注意符号： 当 \(b\) 为负数时，各项符号交替变化。对于 \((1-x)^n\)，符号依次为 \(+ – + – \cdots\)。

Term indexing: The 5th term corresponds to \(r = 4\), not \(r = 5\). This is a classic trap.
项的序号： 第5项对应 \(r = 4\)，而非 \(r = 5\)。这是一个经典陷阱。

Simplify coefficients: Don’t forget to simplify arithmetic: \(\binom{6}{3} = 20\), not 120. The \(r!\) in the denominator must not be omitted.
化简系数： 不要忘记化简运算：\(\binom{6}{3} = 20\) 而不是 120。分母中的 \(r!\) 不能省略。

Validity range: For negative or fractional indices, always state the condition \(|x| < 1\) when required.
成立范围： 对负指数或分数指数，需要时务必注明条件 \(|x| < 1\)。

12. Worked Example | 综合例题

Let us work through a full exam-style question: Find the coefficient of \(x^3\) in the expansion of \(\left(2x + \frac{1}{x}\right)^7\).

让我们完整解答一道考试风格的题目：求 \(\left(2x + \frac{1}{x}\right)^7\) 展开式中 \(x^3\) 项的系数。

The general term with \(r\) is:

含 \(r\) 的通项为：

\(T_{r+1} = \binom{7}{r}(2x)^{7-r}\left(\frac{1}{x}\right)^r = \binom{7}{r}2^{7-r}x^{7-r}x^{-r} = \binom{7}{r}2^{7-r}x^{7-2r}

We require \(7 – 2r = 3\), giving \(r = 2\). Therefore:

令 \(7 – 2r = 3\)，得 \(r = 2\)。因此：

\(\binom{7}{2}2^5 = 21 \times 32 = 672\)

So the coefficient of \(x^3\) is 672.

所以 \(x^3\) 项的系数是 672。

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Pascal’s Triangle and the Binomial Theorem | 杨辉三角与二项式定理