📚 Pascal's Triangle: Binomial Coefficients and Exam Applications | 帕斯卡三角形：二项式系数与考试应用
Pascal's triangle is a triangular array of numbers in which each number is the sum of the two numbers directly above it. It appears in algebra, combinatorics and probability, and it is named after the French mathematician Blaise Pascal, although it was known centuries earlier in China and Persia. For A-level students it is not merely a pretty pattern: it is the fastest way to read off binomial coefficients, to expand brackets such as (a + b)ⁿ, and to count combinations without listing every case.
帕斯卡三角形是一个数字三角形阵列，其中每个数字等于它上方两数之和。它出现在代数、组合学和概率中，以法国数学家布莱兹·帕斯卡命名，尽管几个世纪前中国和波斯就已知道这一图形。对 A-level 学生来说，它不只是一个漂亮的图案：它是读取二项式系数、展开 (a + b)ⁿ 这类括号、以及在不逐一列举的情况下计数组合的最快方法。
This article rebuilds the triangle from scratch, explains the notation ⁿCᵣ and the binomial theorem, then explores the patterns that examiners love: symmetry, row sums, Pascal's rule, the hockey-stick identity, Fibonacci diagonals, and the probability applications behind the binomial distribution. Every claim is linked back to a formula you can use in an exam.
本文将从零开始重建这个三角形，解释 ⁿCᵣ 记号与二项式定理，然后逐一梳理考官偏爱的规律：对称性、行和、帕斯卡法则、曲棍球棒恒等式、斐波那契对角线，以及二项分布背后的概率应用。每一条结论都会回扣到考试中可以使用的公式。
1. Introduction to Pascal's Triangle | 帕斯卡三角形简介
Pascal's triangle is built starting with a top row containing a single 1. Each subsequent row begins and ends with 1, and every interior entry is obtained by adding the two entries above it in the previous row. The row number n counts the rows starting from 0 at the top. For example, 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; and so on.
帕斯卡三角形从顶行一个 1 开始构建。后续每行以 1 开头和结尾，每个内部项由上一行中它上方的两个数相加得到。行号 n 从顶行的 0 开始计数。例如，第 0 行：1；第 1 行：1 1；第 2 行：1 2 1；第 3 行：1 3 3 1；第 4 行：1 4 6 4 1；以此类推。
The first thing to fix in your mind is the indexing convention. The top single 1 is row 0, not row 1. The first entry in every row is position 0, not position 1. Almost every mistake in this topic comes from shifting these two counters by one.
首先要记牢的是编号约定。顶部单独的那个 1 是第 0 行，而不是第 1 行。每一行的第一个项是位置 0，而不是位置 1。这个主题里几乎所有的错误都来自把这两个计数器移动了一位。
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
Notice that the numbers quickly become larger than the row number suggests: row 5 reaches 10 at its centre, and row 10 reaches 252. That is why the triangle is useful for small n but the formula ⁿCᵣ is essential for large n.
注意，这些数字很快就比行号本身暗示的要大：第 5 行的中间值已达到 10，而第 10 行的中间值达到 252。这就是为什么三角形对小 n 很方便，而 ⁿCᵣ 公式对大 n 必不可少。
2. Constructing the Triangle | 构造三角形
To construct Pascal' wo numbers above it: the number above and to the left plus the number above and to the right. For row 5, add the pairs from row 4 (1 4 6 4 1): 1 + 4 = 5, 4 + 6 = 10, 6 + 4 = 10, 4 + 1 = 5, giving row 5 as 1 5 10 10 5 1.
要构造帕斯卡三角形，先在顶部写一个 1。然后，对于每一新行，在开头和结尾放置 1。中间的数字通过将上方两数相加得到：左上方数字加上右上方数字。对于第 5 行，将第 4 行 (1 4 6 4 1) 的相邻数对相加：1 + 4 = 5，4 + 6 = 10，6 + 4 = 10，4 + 1 = 5，得到第 5 行为 1 5 10 10 5 1。
ries follow the same rule: 0 + 1 = 1 at the start of a row, and 1 + 0 = 1 at the end. This "zero padding" view explains why the edges are never special cases.
可以把三角形外部空白处看作填满了 0。这样连边缘项也遵循同一条规则：行首 0 + 1 = 1，行尾 1 + 0 = 1。这种"补零"视角解释了为什么边缘项从来不是特例。
The triangle is symmetric because each row reads the same forwards and backwards. This symmetry reflects the combinatorial identity ⁿCᵣ = ⁿCₙ₋ᵣ, which we meet again in Section 5.
三角形是对称的，因为每一行正读和反读相同。这种对称性反映了组合恒等式 ⁿCᵣ = ⁿCₙ₋ᵣ，我们会在第 5 节再次遇到它。
3. Binomial Coefficients and nCr Notation | 二项式系数与 nCr 符号
The numbers in Pascal's triangle are called binomial coefficients. They can be denoted as ⁿCᵣ or C(n, r), where n is the row number and r is the position counted from the left starting at 0. For instance, in row 4 the entries are 1, 4, 6, 4, 1, which correspond to ⁴C₀, ⁴C₁, ⁴C₂, ⁴C₃, ⁴C₄.
帕斯卡三角形中的数字称为二项式系数。它们可以表示为 ⁿCᵣ 或 C(n, r)，其中 n 是行号，r 是从左边开始、从 0 计数的位置。例如，第 4 行的条目为 1, 4, 6, 4, 1，分别对应 ⁴C₀、⁴C₁、⁴C₂、⁴C₃、⁴C₄。
In A-level mathematics, the binomial coefficient ⁿCᵣ counts the number of ways of choosing r objects from n distinct objects when order does not matter. Its factorial form is the working formula for any value of n, including those too large for the triangle:
在 A-level 数学中，二项式系数 ⁿCᵣ 表示从 n 个不同对象中选出 r 个对象的方法数，其中顺序不重要。它的阶乘形式是适用于任何 n 值的工作公式，包括大到无法用三角形表示的情形：
ⁿCᵣ = n! ÷ (r! × (n - r)!)
Substituting n = 4 and r = 2 gives 4! ÷ (2! × 2!) = 24 ÷ 4 = 6, which is exactly the middle entry of row 4. Remember that 0! inition, which is what makes ⁿC₀ = ⁿCₙ = 1 and keeps the edges of the triangle consistent.
代入 n = 4、r = 2 得 4! ÷ (2! × 2!) = 24 ÷ 4 = 6，正好是第 4 行中间那一项。记住 0! = 1 是定义，正是它使 ⁿC₀ = ⁿCₙ = 1，并让三角形的边缘保持一致。
| n \ r | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 1 | - | - | - | - | - |
| 1 | 1 | 1 | - | - | - | - |
| 2 | 1 | 2 | 1 | - | - | - |
| 3 | 1 | 3 | 3 | 1 | - | - |
| 4 | 1 | 4 | 6 | 4 | 1 | - |
| 5 | 1 | 5 | 10 | 10 | 5 | 1 |
4. The Binomial Theorem | 二项式定理
The reason the triangle matters so much in algebra is the binomial theorem. When you expand (a + b)ⁿ for a positive integer n, the coefficients are exactly the entries of row n:
三角形在代数中如此重要的原因是二项式定理。当把 (a + b)ⁿ 展开（n 为正整数）时，各项系数正好是第 n 行的条目：
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ
Equivalently, the general term is ⁿCᵣ aⁿ⁻ʳ bʳ. The powers of a fall from n to 0 while the powers of b rise from 0 to n, and in every term the two powers add up to n. That bookkeeping rule is a quick way to check an expansion before you commit to it.
等价地，通项为 ⁿCᵣ aⁿ⁻ʳ bʳ。a 的幂从 n 降到 0，b 的幂从 0 升到 n，且在每一项中两个指数之和都等于 n。这条记账规则是在你定稿之前快速检查展开式的好方法。
For example, expanding (x + 2)⁵ uses row 5, which is 1 5 10 10 5 1:
例如，展开 (x + 2)⁵ 用到第 5 行，即 1 5 10 10 5 1：
(x + 2)⁵ = x⁵ + 5x⁴(2) + 10x³(2²) + 10x²(2³) + 5x(2⁴) + 2⁵
Simplifying gives x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. Notice how the powers of 2 inside the brackets are essential: forgetting to raise the constant term to the correct power is the single most common error in this type of question.
化简得 x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32。注意括号内 2 的幂是必不可少的：忘记把常数项升到正确的幂，是这类题目中最常见的错误。
5. Symmetry and the Identity nCr = nC(n - r) | 对称性与恒等式
Every row of Pascal's triangle is a palindrome: it reads the same left to right and right to left. In notation, this is written as ⁿCᵣ = ⁿCₙ₋ᵣ. Row 6 is 1 6 15 20 15 6 1, and indeed ⁶C₁ = 6 = ⁶C₅ and ⁶C₂ = 15 = ⁶C₄.
帕斯卡三角形的每一行都是回文：从左到右和从右到左读法相同。用记号写作 ⁿCᵣ = ⁿCₙ₋ᵣ。第 6 行是 1 6 15 20 15 6 1，确实 ⁶C₁ = 6 = ⁶C₅，⁶C₂ = 15 = ⁶C₄。
The combinatorial reason is simple. Choosing which r objects to take from n is the same as choosing which n - r objects to leave behind. The number of leftovers determines the number of selections, so the two counts must be equal. Algebraically, the same result follows immediately from the factorial formula, because r! and (n - r)! simply swap places.
组合上的理由很简单。从 n 个对象中选哪 r 个拿走，等同于决定留下哪 n - r 个。留下的数目决定了选取的数目，所以两个计数必然相等。代数上，同一结果可由阶乘公式立刻得出，因为 r! 与 (n - r)! 只是交换了位置。
This identity saves real time in exams. If you are asked for the coefficient of a high power of b in an expansion, you can look up the low power instead: the coefficient of xⁿ⁻² in (1 + x)ⁿ equals the coefficient of x², namely ⁿC₂.
这个恒等式在考试中确实省时间。如果要求展开式中 b 的高次幂的系数，你可以改为查低次幂：在 (1 + x)ⁿ 中 xⁿ⁻² 的系数等于 x² 的系数，即 ⁿC₂。
6. Pascal's Rule: the Addition Identity | 帕斯卡法则：加法恒等式
The construction rule of the triangle can be written as a formal identity, known as Pascal's rule or Pascal's identity:
三角形的构造规则可以写成一个形式化的恒等式，称为帕斯卡法则或帕斯卡恒等式：
ⁿCᵣ = ⁿ⁻¹Cᵣ₋₁ + ⁿ⁻¹Cᵣ
o the left. Take n = 5, r = 2: ⁵C₂ = 10, while ⁴C₁ = 4 and ⁴C₂ = 6, and indeed 4 + 6 = 10.
用文字说：任一项等于它正上方的项与左上方的项之和。取 n = 5、r = 2：⁵C₂ = 10，而 ⁴C₁ = 4、⁴C₂ = 6，确实 4 + 6 = 10。
The combinatorial proof is elegant. Fix one particular object out of the n. Selections of r objects either include it or exclude it. If it is included, you must choose the remaining r - 1 from the other n - 1 objects, giving ⁿ⁻¹Cᵣ₋₁ ways. If it is excluded, you choose all r from the other n - 1 objects, giving ⁿ⁻¹Cᵣ ways. Adding the two disjoint cases gives the identity.
组合证明很优雅。从 n 个对象中固定某一个特定对象。选取 r 个对象的方案要么包含它，要么不包含它。若包含它，则需从其余 n - 1 个对象中再选 r - 1 个，有 ⁿ⁻¹Cᵣ₋₁ 种；若不包含它，则从其余 n - 1 个对象中选全部 r 个，有 ⁿ⁻¹Cᵣ 种。两个互斥情形相加即得该恒等式。
You can also verify Pascal's rule purely with factorials, writing both terms over the common denominator r!(n - r)! and using n! = n × (n - 1)! This algebraic route is a standard exam-style proof question worth learning line by line.
你也可以纯用阶乘验证帕斯卡法则：把两项通分到公分母 r!(n - r)!，并利用 n! = n × (n - 1)!。这条代数路线是标准的考试证明题，值得逐行掌握。
7. Row Sums, Powers of 2 and Other Patterns | 行和、2 的幂与其他规律
row 1 to 2 = 2¹, row 2 to 4 = 2², row 3 to 8 = 2³, row 4 to 16 = 2⁴ and row 5 to 32 = 2⁵.
第 n 行各数之和为 2ⁿ。第 0 行和为 1 = 2⁰，第 1 行为 2 = 2¹，第 2 行为 4 = 2²，第 3 行为 8 = 2³，第 4 行为 16 = 2⁴，第 5 行为 32 = 2⁵。
ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ
ves the alternating sum, which is zero for every n ≥ 1: row 4 gives 1 - 4 + 6 - 4 + 1 = 0.
把 a = b = 1 代入二项式定理即得。同样代入 a = 1、b = -1 得到交错和，对每个 n ≥ 1 它都为 0：第 4 行给出 1 - 4 + 6 - 4 + 1 = 0。
ives 1331 = 11³, row 4 gives 14641 = 11⁴. The pattern breaks at row 5 only because 1 5 10 10 5 1 contains two-digit entries that require carrying. Second, the middle entries 1, 2, 6, 20, 70, ... are the central binomial coefficients ²ⁿCₙ.
还有两个规律值得记住。第一，11 的幂藏在三角形里：第 0 行给出 1 = 11⁰，第 1 行给出 11 = 11¹，第 2 行给出 121 = 11²，第 3 行给出 1331 = 11³，第 4 行给出 14641 = 11⁴。该规律在第 5 行断裂，仅仅因为 1 5 10 10 5 1 含有两位数的项而需要进位。第二，中间项 1, 2, 6, 20, 70, ... 是中心二项式
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