📚 Pascal's Triangle: Binomial Expansion and Coefficients | 帕斯卡三角形：二项式展开与系数
In Edexcel A Level Mathematics, Pascal's triangle is much more than a neat number pattern. It gives you the coefficients in binomial expansions such as (a + b)ⁿ, and it underpins binomial probability calculations. Mastering the triangle and the related nCr notation will make many Pure and Statistics questions faster and more accurate.
在爱德思 A Level 数学中，帕斯卡三角形远不止是一个整齐的数字图案。它为 (a + b)ⁿ 等二项式展开提供系数，并且是二项概率计算的基础。掌握帕斯卡三角形以及相关的 nCr 记号，可以让你在纯数学和统计题中做得更快、更准确。
This revision guide walks through how the triangle is built, how it connects to the binomial theorem, how to find a single coefficient, and how the same numbers reappear in binomial probability. Each section pairs the English explanation with the Chinese version so you can revise in either language.
本篇复习指南将依次讲解三角形如何构造、如何与二项式定理建立联系、如何求单个系数，以及同样的数字如何在二项概率中再次出现。每一节都配有中英对照说明，方便你用任一语言复习。
1. What is Pascal's Triangle? | 什么是帕斯卡三角形？
Pascal's triangle is an infinite triangular array of numbers. Each row starts and ends with 1, and every internal number is the sum of the two numbers directly above it. For positive integer n, the nth row of the triangle gives the coefficients of (1 + x)ⁿ.
帕斯卡三角形是一个无限的三角形数字阵列。每一行以 1 开始并以 1 结束，每一个内部数字都是其正上方两个数字之和。对于正整数 n，三角形的第 n 行给出 (1 + x)ⁿ 的系数。
Although named after Blaise Pascal, the same pattern was known earlier in China, India, and Persia. In A Level exams you will usually use it to expand binomial expressions and to find coefficients quickly.
尽管该三角形以布莱兹·帕斯卡的名字命名，但中国、印度和波斯早在之前就已知道这一规律。在 A Level 考试中，你通常会用它对二项式进行展开并快速求出系数。
Key features to remember:
需要记住的关键特征：
- The triangle is symmetric left to right, so ᵏCᵣ = ᵏC(k − r). | 三角形左右对称，因此 ᵏCᵣ = ᵏC(k − r)。
- Every row begins and ends with 1, because there is only one way to choose none or all of the items. | 每一行以 1 开始并以 1 结束，因为"一个都不选"或"全部选上"都只有一种方式。
- The row number tells you the power you are expanding. | 行号告诉你要展开的幂次。
2. Constructing the Triangle | 构造帕斯卡三角形
To build the triangle, start with a single 1 at the top. Under it, write two 1s. For every new row, place 1 at both ends, then fill each internal position by adding the two numbers above-left and above-right.
要构造三角形，从顶部的一个 1 开始。在它下面写出两个 1。对于每一新行，两端都放 1，然后将左上方和右上方的两个数字相加，填入每个内部位置。
The first six rows are shown below:
前六行如下所示：
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
The same information is often easier to handle in a table, especially when you need a coefficient far along a row:
用表格整理同样的信息通常更好用，尤其是在需要某一行靠后的系数时：
| Power n | 幂次 n | Coefficients | 系数 | Sum of row | 该行之和 |
| 0 | 1 | 1 = 2⁰ |
| 1 | 1, 1 | 2 = 2¹ |
| 2 | 1, 2, 1 | 4 = 2² |
| 3 | 1, 3, 3, 1 | 8 = 2³ |
| 4 | 1, 4, 6, 4, 1 | 16 = 2⁴ |
| 5 | 1, 5, 10, 10, 5, 1 | 32 = 2⁵ |
Notice that row n always contains n + 1 entries, because there are n + 1 terms in the expansion of a two-term bracket raised to the power n.
注意第 n 行总是包含 n + 1 个数，因为一个两项括号的 n 次幂展开后共有 n + 1 项。
3. Row Numbering and the Link to (1 + x)ⁿ | 行号与 (1 + x)ⁿ 的联系
Counting rows from 0 is essential. If you start counting from 1, every coefficient you write down will be one power out. Row 0 is the single 1 at the apex, so row n corresponds to the expansion of (1 + x)ⁿ.
从 0 开始编号行是至关重要的。如果你从 1 开始数，那么写下的每个系数都会错一个幂次。第 0 行是顶点处的单个 1，因此第 n 行对应 (1 + x)ⁿ 的展开式。
The correspondence works like this:
对应关系如下：
(1 + x)ⁿ = 1 + ⁿC₁ x + ⁿC₂ x² + ⁿC₃ x³ + … + xⁿ
le, row 4 reads 1, 4, 6, 4, 1, so (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴. The coefficient of xʳ in (1 + x)ⁿ is exactly the (r + 1)th entry of row n.
例如，第 4 行是 1, 4, 6, 4, 1，所以 (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴。(1 + x)ⁿ 中 xʳ 的系数正好是第 n 行第 (r + 1) 个数。
Setting x = 1 gives a useful check:
令 x = 1 可以得到一个实用的检验：
ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ
This is why the sum of row 5 equals 32, and why the binomial probabilities for a fixed n always add up to 1.
这就是为什么第 5 行之和等于 32，也是为什么固定 n 时所有二项概率之和总为 1。
4. Factorial Notation and nCr | 阶乘记号与 nCr
For larger powers the triangle becomes impractical, so you switch to the combination formula. The number of ways of choosing r objects from n distinct objects is written ⁿCᵣ, and it is defined using factorials.
当幂次较大时，三角形就不实用了，此时要改用组合公式。从 n 个不同对象中选出 r 个的方法数记作 ⁿCᵣ，它由阶乘定义。
ⁿCᵣ = n! ÷ (r! × (n − r)!)
Recall that n! = n × (n − 1) × (n − 2) × … × 2 × 1, and that 0! = 1 by definition. These two facts are needed constantly in binomial work.
回忆一下 n! = n × (n − 1) × (n − 2) × … × 2 × 1，并且按定义 0! = 1。这两个事实在二项式运算中会被反复用到。
Worked mini-example:
小例题：
⁵C₂ = 5! ÷ (2! × 3!) = 120 ÷ (2 × 6) = 10
This matches the third entry in row 5 of the triangle, which is a good way to self-check on a calculator. On most calculators the function appears as nCr, and you enter n first, then r.
这与三角形第 5 行第三个数一致，是用计算器自我检查的好方法。在大多数计算器上该功能显示为 nCr，先输入 n，再输入 r。
5. The Binomial Expansion Formula | 二项式展开公式
The binomial theorem generalises everything above. For a positive integer n and any terms a and b:
二项式定理把上面所有内容一般化。对于正整数 n 以及任意项 a 和 b：
(a + b)ⁿ = Σ (from r = 0 to n) ⁿCᵣ aⁿ⁻ʳ bʳ
Written out in full, the first few terms are:
完整写出来，前几项是：
(a + b)ⁿ = aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + ⁿC₃ aⁿ⁻³ b³ + … + bⁿ
The pattern to memorise is simple: powers of a count down from n to 0, powers of b count up from 0 to n, and the sum of the two powers is always n. The coefficient of each term is the corresponding entry from row n.
要记住的规律很简单：a 的幂从 n 递减到 0，b 的幂从 0 递增到 n，两者幂次之和恒为 n。每一项的系数就是第 n 行对应的那个数。
- Total number of terms in the expansion of (a + b)ⁿ is n + 1. | (a + b)ⁿ 展开后共有 n + 1 项。
- The first term is always aⁿ and the last term is always bⁿ. | 首项总是 aⁿ，末项总是 bⁿ。
- The expansion is symmetric, so the coefficient of the term in aʳbⁿ⁻ʳ equals the coefficient of aⁿ⁻ʳbʳ. | 展开式是对称的，因此 aʳbⁿ⁻ʳ 项的系数与 aⁿ⁻ʳbʳ 项的系数相等。
6. Worked Example: Expanding (1 + 2x)⁵ | 例题：展开 (1 + 2x)⁵
n and the row you need. Here n = 5, so use the coefficients 1, 5, 10, 10, 5, 1 from row 5.
第一步：确定 n 以及所需的那一行。这里 n = 5，因此使用第 5 行的系数 1, 5, 10, 10, 5, 1。
Step 2: write a = 1 and b = 2x, then substitute into the formula.
第二步：令 a = 1、b = 2x，代入公式。
(1 + 2x)⁵ = 1 + 5(2x) + 10(2x)² + 10(2x)³ + 5(2x)⁴ + (2x)⁵
Step 3: simplify each term carefully, remembering to raise the whole of 2x to the power, not just the x.
第三步：逐项仔细化简，记住要把整个 2x 一起乘方，而不是只把 x 乘方。
(1 + 2x)⁵ = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵
both sides, and the coefficients should show the expected symmetry in the pure-power case.
第四步：快速检验。代入 x = 0 时两边都应得到 1，并且系数在纯幂情形下应表现出预期的对称性。
If the bracket is (3 − 2x)⁴, the same method applies but the minus sign alternates: the second term is negative, the third positive, and so on.
如果括号是 (3 − 2x)⁴，方法完全相同，但负号会交替出现：第二项为负、第三项为正，依此类推。
7. Finding a Single Coefficient or Term | 求单个系数或项
Exam questions rarely ask for the whole expansion; they usually ask for one specific coefficient. Use the general term formula and skip everything else.
考题很少要求写出完整展开式，通常只问某一个具体的系数。此时使用通项公式，其余部分可以完全跳过。
General term: ᵏCᵣ × (first part)ᵏ⁻ʳ × (second part)ʳ
Example: find the coefficient of x³ in the expansion of (2 + 5x)⁷.
例：求 (2 + 5x)⁷ 展开式中 x³ 的系数。
- We need the power of x to be 3, so take r = 3. | 我们需要 x 的幂为 3，因此取 r = 3。
- The term is ⁷C₃ × 2⁴ × (5x)³. | 该项为 ⁷C₃ × 2⁴ × (5x)³。
- ⁷C₃ = 35, so the term is 35 × 16 × 125x³ = 70000x³. | ⁷C₃ = 35，所以该项为 35 × 16 × 125x³ = 70000x³。
The coefficient of x³ is 70000
Read the question wording precisely. "Find the coefficient of x³" asks only for the number, while "find the term in x³" requires the whole term including the x³.
务必仔细读题。"求 x³ 的系数"只要求写出那个数字，而"求含 x³ 的项"则要求写出包含 x³ 的整个项。
8. Negative and Fractional Indices | 负指数与分数指数
Edexcel Pure Year 2 extends the binomial theorem to negative and fractional values of n, but only when the bracket is in the form (1 + x)ⁿ and |x| < 1. The triangle cannot be used here, so the nCr formula in its factorial form is replaced by a descending-product form.
爱德思纯数学第二年会把二项式定理推广到负指数和分数指数，但只适用于形如 (1 + x)ⁿ 且 |x| < 1 的括号。此时不能使用三角形，阶乘形式的 nCr 要改写成递降乘积形式。
(1 + x)ⁿ = 1 + nx + n(n − 1)x² ÷ 2! + n(n − 1)(n − 2)x³ ÷ 3! + …
Example with a fractional index:
分数指数示例：
(1 + x)^½ ≈ 1 + ½x − ⅛x² + … for |x| < 1
Check the second coefficient: n(n − 1) ÷ 2! with n = ½ gives (½ × (−½)) ÷ 2 = −¼ ÷ 2 = −⅛, which matches. If the bracket is (1 + 2x)−1, replace every x with 2x before expanding.
检验第二个系数：n = ½ 时 n(n − 1) ÷ 2! 得 (½ × (−½)) ÷ 2 = −¼ ÷ 2 = −⅛，结果吻合。如果括号是 (1 + 2x)−1，展开前要把每个 x 替换成 2x。
- The expansion is infinite, not finite. | 该展开式是无穷级数，不是有限项。
- The condition |x| < 1 must be stated for full marks. | 必须写出条件 |x| < 1 才能拿满分。
- You are normally asked for the first three or four terms only. | 通常只要求写出前三项或前四项。
9. Binomial Probability and the Triangle | 二项概率与三角形
The same coefficients drive binomial probability. If X ~ B(n, p), then the probability of exactly r successes is:
同一批系数也支撑着二项概率。若 X ~ B(n, p)，则恰好取得 r 次成功的概率为：
P(X = r) = ⁿCᵣ × pʳ × (1 − p)ⁿ⁻ʳ
all r sum to 1. The coefficient ⁿCᵣ is the same number you would read from row n of Pascal's triangle.
这正是 (p + (1 − p))ⁿ 的二项展开，而由于 p + (1 − p) = 1，所有 r 的概率之和为 1。系数 ⁿCᵣ 就是你在帕斯卡三角形第 n 行读到的那个数。
Example: X ~ B(5, 0.3). Find P(X = 2).
例：X ~ B(5, 0.3)，求 P(X = 2)。
- ⁵C₂ = 10, taken from row 5: 1, 5, 10, 10, 5, 1. | ⁵C₂ = 10，取自第 5 行：1, 5, 10, 10, 5, 1。
- p² = 0.3² = 0.09 and (1 − p)³ = 0.7³ = 0.343. | p² = 0.3² = 0.09，且 (1 − p)³ = 0.7³ = 0.343。
- P(X = 2) = 10 × 0.09 × 0.343 = 0.3087. | P(X = 2) = 10 × 0.09 × 0.343 = 0.3087。
1) and P(X = 2), or use the binomial cumulative distribution function on your calculator.
对于 P(X ≤ 2) 这类累积概率问题，你需要把 P(X = 0)、P(X = 1) 与 P(X = 2) 相加，或直接使用计算器上的二项累积分布函数。
10. Patterns and Identities Inside the Triangle | 三角形内部的规律与恒等式
Beyond the coefficients, the triangle hides several patterns that examiners like to test indirectly. Recognising them lets you answer "show that" questions without long algebra.
除了系数之外，三角形还隐藏着若干规律，出题人喜欢间接考查它们。识别这些规律能让你不用冗长代数就能回答"证明"类问题。
- Symmetry: each row reads the same forwards and backwards, because ⁿCᵣ = ⁿC(n − r). | 对称性：每一行正读反读相同，因为 ⁿCᵣ = ⁿC(n − r)。
- Row sum: the entries of row n add to 2ⁿ. | 行和：第 n 行的数之和为 2ⁿ。
- Triangular numbers: 1, 3, 6, 10, 15 appear along the third diagonal. | 三角数：1, 3, 6, 10, 15 出现在第三条对角线上。
- Hockey stick rule: add a run of entries down a diagonal and the total is the single entry just below and beyond the last one. | 曲棍球棒法则：沿对角线连续相加，其和等于最后一个数下方偏外的那个单独的数。
- Fibonacci numbers: adding along the shallow diagonals produces 1, 1, 2, 3, 5, 8, … | 斐波那契数列：沿浅对角线相加可得到 1, 1, 2, 3, 5, 8, …
A useful identity linking rows is Pascal's rule, which is just the construction rule written algebraically:
一个连接各行的重要恒等式是帕斯卡法则，它其实就是把构造规则写成代数形式：
ⁿCᵣ + ⁿC(r + 1) = (n + 1)C(r + 1)
11. Exam Technique and Common Errors | 考试技巧与常见错误
Most lost marks in binomial questions come from a small number of repeated slips rather than from not understanding the method.
二项式题目中失分大多来自少数反复出现的疏漏，而不是因为不理解方法。
- Mis-counting rows: count from row 0, not row 1. | 行号数错：要从第 0 行开始数，而不是第 1 行。
- Forgetting to raise the constant: in (2 + 5x)⁷ the 2 must also be raised to the power 7 − r. | 忘记给常数乘方：在 (2 + 5x)⁷ 中，2 也必须取 7 − r 次幂。
- Sign errors: a negative second term alternates in sign through the expansion. | 符号错误：第二项为负时，展开式中符号会交替变化。
- Only raising part of the term: (2x)³ = 8x³, not 2x³. | 只对部分乘方：(2x)³ = 8x³，不是 2x³。
- Omitting the validity condition when the index is negative or fractional. | 当指数为负或分数时，漏写有效范围条件。
t structure even if the final arithmetic slips, and writing the term in the form ᵏCᵣ × aᵏ⁻ʳ × bʳ makes your reasoning clear to the examiner.
先写出通项再化简。即使最后算术出错，正确的结构仍能拿到方法分，把通项写成 ᵏCᵣ × aᵏ⁻ʳ × bʳ 的形式，能让阅卷人清楚看到你的思路。
12. Quick Reference Summary | 速查总结
Keep the following essentials in mind when revising. They cover the overwhelming majority of binomial questions across Pure and Statistics papers.
复习时请牢记以下要点。它们
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