📚 Formulae for Sequences | 数列公式
Sequences are one of the most fundamental topics in IGCSE Mathematics. In this revision article, we will explore how to find the formula (nth term) for different types of sequences, including linear, quadratic, cubic, and geometric sequences. Understanding these formulae allows you to predict any term without listing all preceding terms.
数列是 IGCSE 数学中最基础的话题之一。在本复习文章中，我们将探讨如何为不同类型的数列（包括线性、二次、三次和等比数列）求出公式（第 n 项）。理解这些公式可以帮助你无需列出所有前面的项就能预测任意一项。
1. What is a Sequence? | 什么是数列？
A sequence is an ordered list of numbers that follow a particular rule or pattern. Each number in the sequence is called a term. For example, the sequence 3, 6, 9, 12, … follows the rule “add 3 each time”. The dots (…) indicate that the sequence continues indefinitely.
数列是按照特定规则排列的有序数字列表。数列中的每个数字称为一项。例如，数列 3, 6, 9, 12, … 遵循”每次加 3″的规则。省略号（…）表示数列无限继续下去。
Sequences can be classified into several types based on how the terms change. The most common types you need for the IGCSE Edexcel exam are linear (arithmetic) sequences, quadratic sequences, cubic sequences, and geometric sequences.
根据项的变化方式，数列可分为几种类型。在 Edexcel IGCSE 考试中，你需要掌握最常见的类型：线性（等差）数列、二次数列、三次数列和等比数列。
2. Notation: The nth Term | 符号：第 n 项
The nth term of a sequence is a formula that gives the value of any term when n is substituted with its position number. We usually write the nth term as uₙ or tₙ. The first term is u₁ (when n = 1), the second term is u₂ (when n = 2), and so on.
数列的第 n 项是一个公式，当用位置编号代入 n 时，可以得到任意项的值。我们通常把第 n 项写作 uₙ 或 tₙ。第一项是 u₁（当 n = 1 时），第二项是 u₂（当 n = 2 时），以此类推。
For example, if uₙ = 2n + 1, then:
例如，如果 uₙ = 2n + 1，那么：
| n | 1 | 2 | 3 | 4 |
| uₙ | 3 | 5 | 7 | 9 |
Notice that evaluating the formula for n = 1, 2, 3, 4 produces the sequence 3, 5, 7, 9, …
注意，将 n = 1, 2, 3, 4 代入公式求值，就得到了数列 3, 5, 7, 9, …
3. Linear (Arithmetic) Sequences | 线性（等差）数列
A linear sequence, also known as an arithmetic sequence, is one in which the difference between consecutive terms is constant. This constant difference is called the common difference and is usually denoted by d.
线性数列（也称为等差数列）是指相邻两项之差为常数的数列。这个常数差称为公差，通常用 d 表示。
For example, the sequence 7, 11, 15, 19, … has a common difference of 4 because 11 − 7 = 4, 15 − 11 = 4, and so on.
例如，数列 7, 11, 15, 19, … 的公差为 4，因为 11 − 7 = 4，15 − 11 = 4，依此类推。
The nth term of a linear sequence can always be written in the form:
线性数列的第 n 项公式总可以写成以下形式：
uₙ = an + b
where a is the common difference and b is a constant. The coefficient of n (which is a) always equals the common difference.
其中 a 是公差，b 是一个常数。n 的系数（即 a）始终等于公差。
4. Finding the nth Term of a Linear Sequence | 求线性数列的通项公式
To find the nth term of a linear sequence, follow these three steps:
求线性数列第 n 项公式，请按照以下三个步骤：
- Step 1: Find the common difference by subtracting consecutive terms (this is the value of a).
- 步骤一：用相邻项相减求出公差（这就是 a 的值）。
- Step 2: Write the formula as uₙ = an + b, substituting the value of a you found.
- 步骤二：写出公式 uₙ = an + b，代入你求得的 a 值。
- Step 3: Substitute n = 1 and the first term u₁ into the formula to solve for b.
- 步骤三：将 n = 1 和第一项 u₁ 代入公式，求出 b。
Example: Find the nth term of the sequence 3, 7, 11, 15, …
示例：求数列 3, 7, 11, 15, … 的第 n 项公式。
The common difference is 7 − 3 = 4, so a = 4. Thus uₙ = 4n + b. Using the first term: when n = 1, u₁ = 3, so 3 = 4(1) + b, which gives b = −1. Therefore uₙ = 4n − 1.
公差为 7 − 3 = 4，所以 a = 4。因此 uₙ = 4n + b。利用第一项：当 n = 1 时，u₁ = 3，所以 3 = 4(1) + b，得到 b = −1。因此 uₙ = 4n − 1。
Check: n = 4 gives u₄ = 4(4) − 1 = 15, which matches the fourth term. ✓
验证：n = 4 时，u₄ = 4(4) − 1 = 15，与第四项一致。✓
An alternative form uses the first term u₁ and common difference d directly:
另一种形式直接使用首项 u₁ 和公差 d：
uₙ = u₁ + (n − 1)d
For the same sequence: uₙ = 3 + (n − 1) × 4 = 3 + 4n − 4 = 4n − 1. Both forms give the same result.
对同一数列：uₙ = 3 + (n − 1) × 4 = 3 + 4n − 4 = 4n − 1。两种形式结果相同。
5. Quadratic Sequences | 二次数列
A quadratic sequence is a sequence whose nth term formula contains an n² term. The key characteristic of a quadratic sequence is that the second differences are constant.
二次数列是指第 n 项公式中含有 n² 项的数列。二次数列的关键特征是其二阶差分为常数。
To identify a quadratic sequence, calculate the differences between consecutive terms (first differences), then calculate the differences between those differences (second differences). If the second differences are constant and non-zero, the sequence is quadratic.
要识别二次数列，先计算相邻项之差（一阶差分），再计算这些差分的差（二阶差分）。如果二阶差分恒定且不为零，则该数列是二次数列。
Example: The sequence 4, 9, 18, 31, 48, …
示例：数列 4, 9, 18, 31, 48, …
| uₙ | 4 | 9 | 18 | 31 | 48 |
| First differences | 5 | 9 | 13 | 17 | |
| Second differences | 4 | 4 | 4 |
The second differences are constant at 4, confirming this is a quadratic sequence.
二阶差分恒为 4，确认这是一个二次数列。
6. Finding the nth Term of a Quadratic Sequence | 求二次数列的通项公式
The general form of a quadratic nth term is:
二次数列第 n 项公式的一般形式为：
uₙ = an² + bn + c
To find a, b and c, use the following method:
求 a、b 和 c，使用以下方法：
- Step 1: Find the second difference and divide by 2. This gives the value of a.
- 步骤一：求二阶差分并除以 2，得到 a 的值。
- Step 2: Use the formula 3a + b = first difference (the difference between u₁ and u₂) to find b.
- 步骤二：利用公式 3a + b = 一阶差分（u₁ 与 u₂ 之差）求出 b。
- Step 3: Use u₁ = a + b + c to find c.
- 步骤三：利用 u₁ = a + b + c 求出 c。
Example: Find the nth term of 4, 9, 18, 31, …
示例：求数列 4, 9, 18, 31, … 的第 n 项公式。
Second difference = 4, so a = 4 ÷ 2 = 2. The first difference (between 4 and 9) is 5, so 3a + b = 5, which gives 3(2) + b = 5, so b = −1. Then u₁ = a + b + c gives 4 = 2 + (−1) + c, so c = 3.
二阶差分 = 4，所以 a = 4 ÷ 2 = 2。一阶差分（4 与 9 之间）为 5，所以 3a + b = 5，即 3(2) + b = 5，得 b = −1。然后 u₁ = a + b + c 给出 4 = 2 + (−1) + c，所以 c = 3。
Therefore, uₙ = 2n² − n + 3. Check for n = 3: 2(9) − 3 + 3 = 18 ✓
因此，uₙ = 2n² − n + 3。验证 n = 3：2(9) − 3 + 3 = 18 ✓
Quick check for a: When you have the second difference S, you always have a = S ÷ 2. For example, if the second difference is 6, then a = 3.
快速检验 a：当你得到二阶差分 S 时，总有 a = S ÷ 2。例如，如果二阶差分是 6，那么 a = 3。
7. Cubic Sequences | 三次数列
A cubic sequence has an nth term of the form uₙ = an³ + bn² + cn + d. The distinguishing feature of a cubic sequence is that the third differences are constant.
三次数列的第 n 项公式形式为 uₙ = an³ + bn² + cn + d。三次数列的显著特征是三阶差分恒定。
To find the coefficient a, divide the third difference by 6 (since 3! = 6). The remaining coefficients can be found by setting up simultaneous equations using the first few terms.
要求系数 a，将三阶差分除以 6（因为 3! = 6）。其余系数可通过利用前几项建立联立方程来求解。
Example: Consider the sequence 4, 16, 42, 88, 160, …
示例：考虑数列 4, 16, 42, 88, 160, …
| uₙ | 4 | 16 | 42 | 88 | 160 |
| 1st diff | 12 | 26 | 46 | 72 | |
| 2nd diff | 14 | 20 | 26 | ||
| 3rd diff | 6 | 6 |
The third difference is 6, so a = 6 ÷ 6 = 1. Using simultaneous equations with the first three terms, we can solve for b, c and d, giving uₙ = n³ + n² + 2. Check for n = 2: 8 + 4 + 2 = 14? This doesn’t match. Let us recalculate carefully.
三阶差分为 6，所以 a = 6 ÷ 6 = 1。利用前三项建立联立方程，可以解出 b、c 和 d，得到 uₙ = n³ + n² + 2。验证 n = 2：8 + 4 + 2 = 14？这与原数列不符。让我们仔细重新计算。
Setting up equations: u₁ = a + b + c + d = 4; u₂ = 8a + 4b + 2c + d = 16; u₃ = 27a + 9b + 3c + d = 42. With a = 1, we subtract the first equation from the second: 7 + 3b + c = 12, so 3b + c = 5. Subtracting the second from the third: 19 + 5b + c = 26, so 5b + c = 7. Solving gives b = 1 and c = 2. Then d = 4 − 1 − 1 − 2 = 0. Thus uₙ = n³ + n² + 2n. Check: n = 2 gives 8 + 4 + 4 = 16 ✓
建立方程：u₁ = a + b + c + d = 4；u₂ = 8a + 4b + 2c + d = 16；u₃ = 27a + 9b + 3c + d = 42。由于 a = 1，用第二个方程减去第一个：7 + 3b + c = 12，所以 3b + c = 5。用第三个减去第二个：19 + 5b + c = 26，所以 5b + c = 7。解得 b = 1，c = 2。那么 d = 4 − 1 − 1 − 2 = 0。因此 uₙ = n³ + n² + 2n。验证：n = 2 得 8 + 4 + 4 = 16 ✓
8. Geometric Sequences | 等比数列
A geometric sequence is one where each term is obtained by multiplying the previous term by a constant called the common ratio, denoted by r. For IGCSE Edexcel, you should be able to identify geometric sequences and find the nth term formula.
等比数列是指每一项通过将前一项乘以一个称为公比的常数（记作 r）而得到的数列。对于 Edexcel IGCSE，你应该能够识别等比数列并求其第 n 项公式。
For example, the sequence 2, 6, 18, 54, … is geometric because each term is multiplied by 3. The common ratio is 3.
例如，数列 2, 6, 18, 54, … 是等比数列，因为每一项都乘以 3。公比为 3。
The nth term of a geometric sequence is given by the formula:
等比数列的第 n 项公式为：
uₙ = u₁ · rⁿ⁻¹
where u₁ is the first term, r is the common ratio, and n is the position number.
其中 u₁ 是首项，r 是公比，n 是位置编号。
Example: Find the 10th term of the geometric sequence 5, 10, 20, 40, …
示例：求等比数列 5, 10, 20, 40, … 的第 10 项。
Here u₁ = 5 and r = 2 (since 10 ÷ 5 = 2). So u₁₀ = 5 × 2⁹ = 5 × 512 = 2560.
这里 u₁ = 5，r = 2（因为 10 ÷ 5 = 2）。所以 u₁₀ = 5 × 2⁹ = 5 × 512 = 2560。
To find the common ratio, divide any term by the term before it: r = u₂ ÷ u₁ = u₃ ÷ u₂, and so on.
求公比时，用任意项除以前一项：r = u₂ ÷ u₁ = u₃ ÷ u₂，以此类推。
9. Special Sequences | 特殊数列
Certain sequences appear so frequently in mathematics that you should recognise them immediately and remember their nth term formulae.
有些数列在数学中经常出现，你应该能够立即识别它们并记住它们的第 n 项公式。
| Sequence 数列 | Terms 各项 | nth term 第 n 项 |
| Square numbers 平方数 | 1, 4, 9, 16, 25, … | n² |
| Cube numbers 立方数 | 1, 8, 27, 64, 125, … | n³ |
| Triangular numbers 三角数 | 1, 3, 6, 10, 15, … | n(n + 1) ÷ 2 |
| Powers of 2 2的幂 | 2, 4, 8, 16, 32, … | 2ⁿ |
The Fibonacci sequence 1, 1, 2, 3, 5, 8, … is also important. In this sequence, each term is the sum of the two preceding terms: uₙ = uₙ₋₁ + uₙ₋₂. You are not expected to derive an explicit formula for the Fibonacci sequence at IGCSE level, but you should be able to generate its terms.
斐波那契数列 1, 1, 2, 3, 5, 8, … 也很重要。在这个数列中，每一项是前两项之和：uₙ = uₙ₋₁ + uₙ₋₂。在 IGCSE 阶段，你不需要推导斐波那契数列的显式公式，但你应该能够生成它的各项。
10. Using the Formula: Working Backwards | 利用公式：逆向求项
Often, you will be given the nth term formula and asked to determine whether a particular number appears in the sequence. This involves setting the formula equal to the given number and solving for n.
通常，你会被给第 n 项公式，并要求判断某个特定数字是否出现在数列中。这需要将公式设为等于给定数字并解出 n。
Example: Is 102 a term of the sequence with nth term uₙ = 5n + 2?
示例：102 是否是以 uₙ = 5n + 2 为第 n 项公式的数列中的一项？
Set 5n + 2 = 102. Then 5n = 100, so n = 20. Since n = 20 is a positive integer, 102 is indeed the 20th term of the sequence.
设 5n + 2 = 102。那么 5n = 100，所以 n = 20。由于 n = 20 是正整数，102 确实是该数列的第 20 项。
For quadratic sequences, you may need to solve a quadratic equation. Use factorisation or the quadratic formula, and remember that n must be a positive integer for the number to be a valid term.
对于二次数列，你可能需要解二次方程。使用因式分解或求根公式，并记住 n 必须是正整数，该数字才是有效的项。
Example: Is 55 a term of the sequence uₙ = n² + 2n?
示例：55 是否是数列 uₙ = n² + 2n 的一项？
Set n² + 2n = 55, so n² + 2n − 55 = 0. Factorising: (n + 11)(n − 5) = 0, giving n = 5 or n = −11. Since n must be positive, n = 5. So 55 is the 5th term. ✓
设 n² + 2n = 55，即 n² + 2n − 55 = 0。因式分解：(n + 11)(n − 5) = 0，得 n = 5 或 n = −11。由于 n 必须为正数，取 n = 5。所以 55 是第 5 项。✓
11. Problem Solving with Sequences | 数列综合解题
Exam questions often combine sequences with other topics, such as linear equations or inequalities. You may also be asked to find the nth term from a diagrammatic pattern or a word problem.
考试题目经常将数列与其他主题（如线性方程或不等式）结合。你也可能被要求从图形模式或文字题中求第 n 项公式。
Example (diagram pattern): A pattern of squares is built with matchsticks. Pattern 1 uses 4 matchsticks, Pattern 2 uses 7, Pattern 3 uses 10. Find the nth term formula and the number of matchsticks in Pattern 10.
示例（图形模式）：用火柴棒搭建正方形图案。图案 1 用 4 根火柴棒，图案 2 用 7 根，图案 3 用 10 根。求第 n 项公式以及图案 10 中火柴棒的数量。
The sequence 4, 7, 10, … is linear with common difference 3. So uₙ = 3n + b. Using u₁ = 4: 4 = 3(1) + b, so b = 1. Thus uₙ = 3n + 1. For Pattern 10: u₁₀ = 3(10) + 1 = 31 matchsticks.
数列 4, 7, 10, … 是线性的，公差为 3。所以 uₙ = 3n + b。利用 u₁ = 4：4 = 3(1) + b，所以 b = 1。因此 uₙ = 3n + 1。对于图案 10：u₁₀ = 3(10) + 1 = 31 根火柴棒。
When solving sequence problems, always check your formula against at least two given terms before writing your final answer. This simple verification step can prevent careless errors.
在解数列题时，写出最终答案前，务必用至少两个已知项检验你的公式。这个简单的验证步骤可以防止粗心错误。
12. Summary of Key Formulae | 关键公式总结
The table below summarises all the essential formulae for sequences that you need for your IGCSE Edexcel exam.
下表总结了你参加 Edexcel IGCSE 考试所需的所有数列关键公式。
| Type 类型 | nth term 第 n 项 | Key feature 关键特征 |
| Linear 线性 | uₙ = an + b | First differences constant |
| 一阶差分恒定 | ||
| Quadratic 二次 | uₙ = an² + bn + c | Second differences constant (a = second diff ÷ 2) |
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