📚 Pascal’s Triangle Patterns | 帕斯卡三角形规律
Pascal’s Triangle is one of the most elegant structures in mathematics. Each row is built from the row above it, and yet the triangle encodes a remarkable number of patterns: binomial coefficients, combinations, powers of two, triangular numbers, Fibonacci numbers and many algebraic identities. In the Edexcel A-Level Mathematics syllabus, Pascal’s Triangle appears most directly in the binomial expansion, but understanding its deeper regularities gives you a powerful toolkit for problem solving and proof.
帕斯卡三角形是数学中最优雅的结构之一。每一行都由上一行构造而成，但这三角形却蕴含着大量奇妙的规律：二项式系数、组合数、2 的幂、三角数、斐波那契数以及许多代数恒等式。在 Edexcel A-Level 数学大纲中，帕斯卡三角形最直接出现在二项展开中，理解其深层规律能为你解题与证明提供强大工具。
1. Construction of Pascal’s Triangle | 帕斯卡三角形的构造
Pascal’s Triangle begins with a single 1 at the very top. Each subsequent row is formed by placing 1s at both ends, and every interior number is the sum of the two numbers immediately above it in the previous row. If we number the top row as row 0, the first few rows are shown below.
帕斯卡三角形最顶端是一个 1。接下来每一行的两端都是 1，而中间每个数都等于它上一行正上方的两个数之和。若把最上面一行称为第 0 行，那么最前面的几行如下所示。
| 1 | |||||
| 1 | 1 | ||||
| 1 | 2 | 1 | |||
| 1 | 3 | 3 | 1 | ||
| 1 | 4 | 6 | 4 | 1 | |
| 1 | 5 | 10 | 10 | 5 | 1 |
In general, if the entries of row n are written as p(n, 0), p(n, 1), …, p(n, n), then p(n, 0) = p(n, n) = 1, and for 0 < k < n:
一般地，若把第 n 行的项记为 p(n, 0), p(n, 1), …, p(n, n)，则 p(n, 0) = p(n, n) = 1，且对 0 < k < n：
p(n, k) = p(n − 1, k − 1) + p(n − 1, k)
2. Binomial Expansion Connection | 二项展开的联系
The most important link for A-Level work is the connection between Pascal’s Triangle and the binomial expansion. The coefficients in the expansion of (a + b)n are exactly the numbers in row n of Pascal’s Triangle, starting at row 0.
对于 A-Level 学习而言，最重要的联系就是帕斯卡三角形与二项展开之间的关系。(a + b)n 展开式中的各项系数恰好是帕斯卡三角形第 n 行的数字，其中最上面为第 0 行。
For example, using row 3, which is 1, 3, 3, 1:
例如，利用第 3 行数字 1, 3, 3, 1：
(a + b)3 = a3 + 3a2b + 3ab2 + b3
More generally, the binomial theorem states:
更一般地，二项式定理表明：
(a + b)n = Σr=0n C(n, r) an−r br
where C(n, r) is the entry in row n, position r of Pascal’s Triangle. This is why Pascal’s Triangle is sometimes the fastest way to expand small powers without a calculator.
其中 C(n, r) 就是帕斯卡三角形第 n 行第 r 个位置上的数。因此，在没有计算器时，帕斯卡三角形常常是展开较小幂次最快的方法。
3. Combinations and nCr | 组合数与 nCr
Every number in Pascal’s Triangle is also a combination count. The entry C(n, r) represents the number of ways to choose r objects from a set of n distinct objects, where order does not matter. In Edexcel A-Level notation this is written as nCr or C(n, r).
帕斯卡三角形中的每个数同时也是组合计数。C(n, r) 表示从 n 个不同物体中不按顺序选出 r 个物体的方法数。在 Edexcel A-Level 记号中写作 nCr 或 C(n, r)。
The formula for combinations is:
组合数的公式为：
C(n, r) = n! / (r! (n − r)!)
Notice that row 4 of the triangle, 1, 4, 6, 4, 1, equals C(4, 0), C(4, 1), C(4, 2), C(4, 3), C(4, 4). This interpretation is essential when solving problems about choosing teams, arranging selections or finding binomial coefficients.
注意第 4 行的数字 1, 4, 6, 4, 1 分别等于 C(4, 0), C(4, 1), C(4, 2), C(4, 3), C(4, 4)。这种解释在解决选小组、排列选择或求二项式系数时非常关键。
4. Pascal’s Identity | 帕斯卡恒等式
The rule for building the triangle is really an algebraic identity known as Pascal’s Identity. It states that for any integers n ≥ 1 and 1 ≤ r ≤ n − 1:
构造三角形所用的规则实际上是一个代数恒等式，称为帕斯卡恒等式。它表明对任意整数 n ≥ 1 以及 1 ≤ r ≤ n − 1：
C(n, r) = C(n − 1, r − 1) + C(n − 1, r)
To prove this combinatorially, imagine choosing r people from n people. Fix one particular person, say Alex. If Alex is chosen, then we must choose r − 1 people from the remaining n − 1 people: C(n − 1, r − 1) ways. If Alex is not chosen, then we must choose all r people from the remaining n − 1 people: C(n − 1, r) ways. Adding these two disjoint cases gives the identity.
组合证明如下：设想从 n 个人中选择 r 个人。固定某个人，比如甲。若甲被选中，则需从其余 n − 1 人中再选 r − 1 人，共有 C(n − 1, r − 1) 种方法；若甲没被选中，则必须从其余 n − 1 人中选 r 人，共有 C(n − 1, r) 种方法。把这两种互不相交的情况相加，即得到该恒等式。
5. Sum of Each Row | 每行数字之和
A striking pattern is that the sum of the numbers in row n is exactly 2n. For instance, row 3 is 1 + 3 + 3 + 1 = 8 = 23, and row 4 sums to 1 + 4 + 6 + 4 + 1 = 16 = 24.
一个非常显著的性质是：第 n 行所有数字之和正好是 2n。例如第 3 行 1 + 3 + 3 + 1 = 8 = 23，第 4 行 1 + 4 + 6 + 4 + 1 = 16 = 24。
This pattern follows from the binomial theorem by setting a = b = 1:
这个规律可由二项式定理令 a = b = 1 得到：
(1 + 1)n = 2n = C(n, 0) + C(n, 1) + … + C(n, n)
This identity is often used in probability and counting problems where the total number of subsets of an n-element set is needed.
这个恒等式常用于概率与计数问题，例如计算一个含有 n 个元素的集合一共有多少个子集。
6. Symmetry in Rows | 行的对称性
Every row of Pascal’s Triangle reads the same forwards and backwards. This is because C(n, r) = C(n, n − r). Choosing r objects to keep is the same as choosing n − r objects to discard.
帕斯卡三角形每一行从左往右读与从右往左读完全相同。这是因为 C(n, r) = C(n, n − r)。选出 r 个物体保留，等价于选出 n − r 个物体丢弃。
In binomial expansions, this symmetry means that the coefficients rise to a maximum at the centre and then fall symmetrically. For even n, the largest coefficient is C(n, n/2); for odd n, the two largest coefficients are equal at the centre.
在二项展开中，这种对称性意味着系数先增大到中间最大，然后对称下降。当 n 为偶数时，最大系数是 C(n, n/2)；当 n 为奇数时，中间两个最大系数相等。
7. Diagonal Sequences | 对角线数列
The diagonals of Pascal’s Triangle contain famous number sequences. The first diagonal is always 1. The second diagonal is the natural numbers: 1, 2, 3, 4, … The third diagonal contains the triangular numbers: 1, 3, 6, 10, 15, …
帕斯卡三角形的对角线上蕴含着著名的数列。第一条对角线全是 1。第二条对角线是自然数：1, 2, 3, 4, … 第三条对角线是三角数：1, 3, 6, 10, 15, …
The triangular numbers Tn are given by:
三角数 Tn 的公式为：
Tn = n(n + 1) / 2
Continuing along the next diagonal gives tetrahedral numbers: 1, 4, 10, 20, 35, … These represent the number of objects in a stack of triangular layers. Recognising these sequences can help you spot patterns in exam questions quickly.
再下一条对角线是四面体数：1, 4, 10, 20, 35, … 它们表示由三角形层堆叠而成的物体数量。识别这些数列可以帮助你在考试中快速发现规律。
8. The Hockey Stick Identity | 曲棍球棒恒等式
Another important pattern is called the Hockey Stick Identity. It says that the sum of a diagonal from the edge of the triangle down to a certain point equals the number diagonally below and to the right of the final term.
另一个重要规律叫做曲棍球棒恒等式。它指出：从三角形边缘开始，沿某条对角线往下加到某个数，其总和等于最后一项斜下方右边的那个数。
In formal notation, for k ≤ n:
用正式记号表示，对 k ≤ n：
C(k, k) + C(k + 1, k) + C(k + 2, k) + … + C(n, k) = C(n + 1, k + 1)
For example, starting at the 1 in row 2 and moving down the third diagonal gives 1 + 3 + 6 + 10 = 20, and 20 is exactly C(6, 3), the number below and to the right of 10. The shape formed by the path resembles a hockey stick.
例如，从第 2 行最左侧的 1 开始，沿第三条对角线往下：1 + 3 + 6 + 10 = 20，而 20 恰好是 C(6, 3)，也就是 10 右下方斜向的数字。这条路径形成的形状就像一支曲棍球棒。
9. Fibonacci Numbers | 斐波那契数
Fibonacci numbers also appear in Pascal’s Triangle. If you sum the entries along the shallow diagonals, you obtain the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, …
斐波那契数也出现在帕斯卡三角形中。如果沿“浅对角线”求和，就会得到斐波那契数列：1, 1, 2, 3, 5, 8, 13, …
For example, the first shallow diagonal is just 1. The second is 1. The third shallow diagonal gives 1 + 1 = 2. The fourth gives 1 + 2 = 3. The fifth gives 1 + 3 + 1 = 5. The sixth gives 1 + 4 + 3 = 8.
例如，第一条浅对角线是 1；第二条也是 1；第三条浅对角线给出 1 + 1 = 2；第四条给出 1 + 2 = 3；第五条给出 1 + 3 + 1 = 5；第六条给出 1 + 4 + 3 = 8。
This connection is rarely tested directly in Edexcel, but it is a beautiful illustration of how different areas of pure mathematics are linked, and it can appear in enrichment tasks or interview-style questions.
这一联系虽然很少在 Edexcel 考试中直接考查，但它美丽地展示了纯数学不同分支之间的联系，可能出现在拓展题或面试式问题中。
10. Exam-Style Applications | 考试题型应用
In Edexcel A-Level Maths, Pascal’s Triangle is most often tested through binomial expansions. You may be asked to expand an expression or find a specific coefficient without writing out the whole expansion.
在 Edexcel A-Level 数学中，帕斯卡三角形最常通过二项展开来考查。题目可能要求你展开一个式子，或者在不写出完整展开式的情况下求某一项的系数。
Example: Find the coefficient of x3 in (2 + x)5.
例：求 (2 + x)5 中 x3 的系数。
Row 5 of Pascal’s Triangle is 1, 5, 10, 10, 5, 1. The term involving x3 is C(5, 3) · 22 · x3. Since C(5, 3) = 10 and 22 = 4, the coefficient is 10 × 4 = 40.
帕斯卡三角形第 5 行为 1, 5, 10, 10, 5, 1。含 x3 的项是 C(5, 3) · 22 · x3。因为 C(5, 3) = 10，且 22 = 4，所以系数为 10 × 4 = 40。
You may also be asked to use Pascal’s Identity to simplify expressions such as C(10, 4) + C(10, 5). By the identity, this equals C(11, 5). Knowing the triangle’s structure lets you verify such results by hand.
题目也可能要求你用帕斯卡恒等式化简式子，例如 C(10, 4) + C(10, 5)。根据恒等式，这等于 C(11, 5)。熟悉三角形的结构可以让你手算验证这类结果。
Mastering these patterns will speed up your algebra, deepen your understanding of counting methods, and help you solve a wide range of pure mathematics problems with confidence.
掌握这些规律不仅能加快你的代数计算速度，加深你对计数方法的理解，还能帮助你自信地解决各类纯数学问题。
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