📚 Formulae for Sequences | 数列公式
In IGCSE Mathematics, sequences are ordered lists of numbers that follow specific rules. Formulae allow you to find any term or the sum of terms without writing out the whole list. This article covers arithmetic, geometric and quadratic sequences, their nth term and sum formulae, as well as recurrence relations, all essential for the Edexcel exam.
在 IGCSE 数学中，数列是按照特定规则排列的有序数字列表。通过公式，你无需列出所有项就能找到任意项或项的总和。本文涵盖等差、等比和二次数列，它们的第 n 项与求和公式，以及递推关系，这些都是 Edexcel 考试的核心内容。
1. Introduction to Sequences and Their Formulae | 数列及其公式简介
A sequence is a set of numbers written in a particular order. Each number is called a term, and its position is given by n (1st, 2nd, 3rd, ...). A formula for the nth term, often written as T(n) or aₙ, lets you calculate any term directly, saving time and reducing errors.
数列是按特定顺序排列的一组数字。每个数字称为一项，其位置用 n 表示（第1项、第2项、第3项……）。第 n 项公式，常写作 T(n) 或 aₙ，可让你直接计算任意项，节省时间并减少错误。
There are different types of sequences: linear (arithmetic), geometric, and quadratic. Each has a characteristic formula for the nth term. You may also encounter sequences defined by a recurrence relation, where each term depends on previous ones.
数列有不同的类型：线性（等差）、等比和二次数列。每种都有独特的第 n 项公式。你还可能遇到由递推关系定义的数列，其每一项依赖于前一项或前几项。
2. The nth Term and Linear Sequences | 第 n 项与线性数列
Linear sequences have a constant difference between consecutive terms. The nth term formula is of the form a n + b, where a is the common difference and b is a constant. For example, the sequence 5, 8, 11, 14, ... has difference 3, so its nth term is 3n + 2.
线性数列的相邻项之间差值恒定。第 n 项公式形式为 a n + b，其中 a 是公差，b 是常数。例如数列 5, 8, 11, 14, … 的公差为 3，因此第 n 项是 3n + 2。
To find the formula, first calculate the common difference d. Then write d n + (first term – d). Always test your formula with n = 1, 2, 3 to ensure it generates the correct terms.
要找出公式，先计算公差 d，然后写出 d n + (首项 – d)。始终用 n = 1, 2, 3 检验，确保公式生成正确的项。
3. Arithmetic Sequences: Formula for the nth Term | 等差数列：第 n 项公式
rm and the next. The first term is denoted a (or u₁). The nth term is given by aₙ = a + (n – 1)d. This works with positive, negative or fractional differences.
等差数列（算术数列）中，任意项与后一项的差值 d 恒定。首项记作 a（或 u₁）。第 n 项由 aₙ = a + (n – 1)d 给出。该公式适用于正、负或分数公差。
aₙ = a + (n – 1)d
For example, in an arithmetic sequence with a = 7, d = –3, the 20th term is 7 + 19 × (–3) = –. If the sequence is 2, 5, 8, 11, ... you can identify a = 2, d = 3 and write aₙ = 2 + 3(n – 1) = 3n – 1.
例如，在一个等差数列中 a = 7, d = –3，第 20 项是 7 + 19 × (–3) = –50。若数列为 2, 5, 8, 11, …，可识别 a = 2, d = 3，写出 aₙ = 2 + 3(n – 1) = 3n – 1。
4. Arithmetic Sequences: Sum of the First n Terms | 等差数列：前 n 项和
e Sₙ = n/2 (a + l). If only a and the common difference d are known, use Sₙ = n/2 [2a + (n – 1)d].
等差数列前 n 项和，记作 Sₙ，可用两个等价公式求得。若已知首项 a 和末项 l，用 Sₙ = n/2 (a + l)。若只知道 a 和公差 d，则用 Sₙ = n/2 [2a + (n – 1)d]。
Sₙ = n/2 [2a + (n – 1)d]
the sum of the first 30 terms of 4, 7, 10, ... a = 4, d = 3, n = 30. S₃₀ = 30/2 [2×4 + (30 – 7] = 1425. Always check whether the sequence is arithmetic before applying these formulae.
例如，求数列 4, 7, 10, … 前 30 项的和：a = 4, d = 3, n = 30。S₃₀ = 30/2 [2×4 + (30 – 1)×3] = 15 [8 + 87] = 1425。使用这些公式前，务必确认数列为等差数列。
5. Geometric Sequences: Formula for the nth Term | 等比数列：第 n 项公式
ⁿ⁻¹. The ratio r can be a whole number, a fraction or a negative number, leading to terms that grow, shrink or alternate in sign.
等比数列每一步乘以恒定的比例 r。若首项为 a，则第 n 项为 aₙ = a rⁿ⁻¹。比例 r 可以是整数、分数或负数，导致项的增长、缩小或正负交替。
aₙ = a rⁿ⁻¹
× 2⁹ = 1536. For a sequence like 27, 9, 3, 1, ... a = 27, r = 1/3, so aₙ = 27 × (1/3)ⁿ⁻¹.
以数列 3, 6, 12, 24, … 为例，a = 3, r = 2。第 10 项为 3 × 2⁹ = 1536。对于数列 27, 9, 3, 1, …，a = 27, r = 1/3，因此 aₙ = 27 × (1/3)ⁿ⁻¹。
6. Geometric Sequences: Sum of the First n Terms | 等比数列：前 n 项和
The sum of the first n terms of a geometric sequence is given by Sₙ = a(rⁿ – 1) / (r – 1) when r > 1, or Sₙ = a(1 – rⁿ) / (1 – r) when |r| ly equivalent; choose the one that avoids negative denominators or simplifies calculation.
等比数列前 n 项和的公式为：当 r > 1 时，Sₙ = a(rⁿ – 1) / (r – 1)；当 |r| < 1 时，Sₙ = a(1 – rⁿ) / (1 – r)。两种形式代数等价，选择可避免负分母或简化计算的那一种。
Sₙ = a(rⁿ – 1) / (r – 1) (r ≠ 1)
ision by zero. For example, sum of 2 + 6 + 18 + ... first 8 terms: a = 2, r = 3, S₈ = 2(3⁸ – 1) / (3 – 1) = 2(6561 – 1)/2 = 6560.
如果 r = 1，所有项相等，Sₙ = n a。务必检查 r 的值以避免除以零。例如，求 2 + 6 + 18 + … 前 8 项和：a = 2, r = 3，S₈ = 2(3⁸ – 1) / (3 – 1) = 2(6561 – 1)/2 = 6560。
7. Quadratic Sequences: Finding the nth Term | 二次数列：求第 n 项
Quadratic sequences are those whose second difference is constant. The nth term is of the form a n² + b n + c. To find a, b and c, first work out the first differences between consecutive terms, then the second differences. The value of 2a equals the second difference.
二次数列是指二次差分恒定的数列。第 n 项的形式为 a n² + b n + c。要找出 a, b 和 c，先计算相邻项的一次差分，再计算二次差分。2a 的值等于二次差分。
om the original sequence to leave a linear sequence. Find its nth term (b n + c) and combine with a n² to get the full quadratic expression.
得到 a 后，写出 n = 1, 2, 3, … 时 a n² 的数列。从原数列中减去这个数列，剩下一个线性数列。求出它的第 n 项 (b n + c)，然后与 a n² 组合得到完整的二次表达式。
8. Quadratic Sequences: Worked Example | 二次数列：例题解析
efore 2a = 4, so a = 2. The sequence of 2n² is 2, 8, 18, 32, 50. Subtract from original: 4–2=2, 9–8=1, 18–18=0, 31–32=–1, 48–50=–2. This linear sequence is –n + 3. Hence the nth term is 2n² – n + 3.
考虑数列 4, 9, 18, 31, 48, …。一次差分：5, 9, 13, 17。二次差分：4, 4, 4。因此 2a = 4，得 a = 2。2n² 的数列为 2, 8, 18, 32, 50。与原数列相减：4–2=2, 9–8=1, 18–18=0, 31–32=–1, 48–50=–2。该线性数列是 –n + 3。所以第 n 项为 2n² – n + 3。
c, substituting n = 1, 2, 3 gives three equations to solve for a, b and c simultaneously.
另一种方法是用前几项建立方程组。对于二次式 a n² + b n + c，代入 n = 1, 2, 3 得到三个方程，联立解出 a, b 和 c。
9. Using Differences to Identify Sequence Type | 用差分识别数列类型
Calculating differences is a powerful diagnostic tool. If the first differences are constant, the sequence is linear (arithmetic). If the second differences are constant, it is quadratic. If the ratio of consecutive terms is constant, it is geometric.
计算差分是一种强有力的
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