IGCSE WJEC Mathematics: Sequences and Series Key Points | IGCSE WJEC 数学：数列与级数考点精讲

📚 IGCSE WJEC Mathematics: Sequences and Series Key Points | IGCSE WJEC 数学：数列与级数考点精讲

Sequences and series are fundamental topics in the IGCSE WJEC Mathematics syllabus. You need to be confident in identifying patterns, finding the nth term, and calculating sums for both arithmetic and geometric progressions. This revision guide covers all key points, with worked examples and exam tips.

数列与级数是 IGCSE WJEC 数学大纲中的基础主题。你需要熟练识别规律、求通项公式，并计算等差数列与等比数列的和。本复习指南涵盖所有核心考点，并提供范例与考试贴士。

1. What are Sequences and Series? | 什么是数列与级数？

A sequence is an ordered list of numbers that follow a specific rule. Each number in the sequence is called a term. For example, 2, 5, 8, 11, … is a sequence where each term increases by 3. The rule can often be described using an nth term formula, which allows you to calculate any term directly.

序列是按特定规则排列的一串有序数字。序列中的每一个数字称为一项。例如，2, 5, 8, 11, … 是一个每项增加3的序列。这种规则通常可以用一个通项公式（第n项公式）来描述，从而可以直接计算任意一项。

When the terms of a sequence are added together, we obtain a series. For instance, 2 + 5 + 8 + 11 + … is a series. In the IGCSE WJEC exam, you will work with arithmetic series and geometric series, as well as their notation using sigma (Σ).

当序列的各项相加时，就得到了级数。例如，2 + 5 + 8 + 11 + … 就是一个级数。在 IGCSE WJEC 考试中，你将处理等差级数和等比级数，并会用西格玛（Σ）记号来表示它们。

Understanding the difference between a sequence (just a list) and a series (a sum) is essential. Many marks are lost when students confuse the formulae for the nth term with those for the sum.

理解序列（仅仅是一列数）和级数（求和）之间的区别至关重要。许多学生因混淆了求第n项的公式与求和公式而失分。

2. Arithmetic Sequences: Finding the nth Term | 等差数列：求第n项

An arithmetic sequence is one where the difference between consecutive terms is constant. This constant difference is called the common difference, usually denoted by d. If the first term is a, then the sequence can be written as a, a+d, a+2d, a+3d, …

等差数列是指相邻两项之差为常数的数列。这个常数差被称为公差，通常用 d 表示。如果首项为 a，那么该数列可以写成 a, a+d, a+2d, a+3d, …

nth term: aₙ = a + (n – 1)d

第 n 项：aₙ = a + (n – 1)d

To find any term, simply substitute the term number n into the formula. For example, given the arithmetic sequence 3, 7, 11, 15, … we can spot that a = 3 and d = 4. The 20th term is a + (20-1)d = 3 + 19 × 4 = 79.

要求出任一项，只需将项数 n 代入公式。例如，已知等差数列 3, 7, 11, 15, …，可以看出 a = 3，d = 4。第20项为 a + (20-1)d = 3 + 19 × 4 = 79。

You may also be asked to find the first term and common difference if you are given two specific terms. For instance, if the 5th term is 22 and the 10th term is 47, set up simultaneous equations: a + 4d = 22 and a + 9d = 47. Solving gives d = 5 and a = 2.

你也可能需要根据已知的两项来求首项和公差。例如，已知第5项为22，第10项为47，可建立方程组：a + 4d = 22 和 a + 9d = 47。解方程得 d = 5，a = 2。

3. Sum of an Arithmetic Series | 等差级数求和

When we add up the first n terms of an arithmetic sequence, we form an arithmetic series. There are two commonly used formulae for the sum Sₙ:

当我们将等差数列的前 n 项相加时，就构成了等差级数。求和 Sₙ 有两个常用公式：

Sₙ = n/2 (a + l) (where l is the last term)

Sₙ = n/2 (a + l)（其中 l 为末项）

Sₙ = n/2 [2a + (n – 1)d]

Use the first formula when you already know the first and last terms; use the second when you know the first term and common difference. For example, to find the sum of the first 30 terms of the series 5 + 9 + 13 + …, identify a = 5, d = 4, n = 30. Then S₃₀ = 30/2 [2×5 + (30-1)×4] = 15 × [10 + 116] = 15 × 126 = 1890.

当已经知道首项和末项时，使用第一个公式；当知道首项和公差时，使用第二个公式。例如，求级数 5 + 9 + 13 + … 前30项的和：确定 a = 5, d = 4, n = 30。则 S₃₀ = 30/2 [2×5 + (30-1)×4] = 15 × [10 + 116] = 15 × 126 = 1890。

Be careful to count the number of terms correctly. If the wording says “from the 4th term to the 20th term inclusive”, the number of terms is 20 – 4 + 1 = 17. You may need to find the sum of a specific section of a series by treating it as a new arithmetic series.

要小心正确地计算项数。如果题目要求“从第4项到第20项（含）的和”，则项数为 20 – 4 + 1 = 17。有时你需要将某一特定部分视为一个新的等差级数来求和。

4. Geometric Sequences: Finding the nth Term | 等比数列：求第n项

A geometric sequence is one where each term is found by multiplying the previous term by a constant called the common ratio, r. For example, 3, 6, 12, 24, … has a = 3 and r = 2. The terms can be written as a, ar, ar², ar³, …

等比数列是指每一项都等于前一项乘以一个常数（称为公比 r）的数列。例如，3, 6, 12, 24, … 的首项 a=3，公比 r=2。各项可以写成 a, ar, ar², ar³, …

nth term: aₙ = arⁿ⁻¹

第 n 项：aₙ = arⁿ⁻¹

To find the 8th term of the sequence 5, 15, 45, …, first spot r = 3. Then a₈ = 5 × 3⁸⁻¹ = 5 × 3⁷ = 5 × 2187 = 10935. Always use brackets or careful arithmetic when the common ratio is a fraction, e.g., r = ½.

要求数列 5, 15, 45, … 的第8项，首先看出 r=3。则 a₈ = 5 × 3⁸⁻¹ = 5 × 3⁷ = 5 × 2187 = 10935。当公比为分数时（如 r=½），要仔细计算或使用括号。

Just like with arithmetic sequences, you can form simultaneous equations if you are given two terms. For example, if the 3rd term is 36 and the 6th term is 972, then ar² = 36 and ar⁵ = 972. Dividing gives r³ = 27, so r = 3, and a = 4.

与等差数列类似，如果已知两项，可以建立方程组。例如，已知第3项为36，第6项为972，则 ar²=36，ar⁵=972。两式相除得 r³=27，故 r=3，a=4。

5. Sum of a Geometric Series (Finite) | 有限等比级数求和

The sum of the first n terms of a geometric series is given by the formula below, which is valid when r ≠ 1. If r = 1, the series is simply n × a.

等比级数前 n 项的和由以下公式给出，当 r ≠ 1 时适用。若 r=1，级数的和就是 n × a。

Sₙ = a(1 – rⁿ) / (1 – r)

You can also use Sₙ = a(rⁿ – 1) / (r – 1), which is algebraically identical. Choose the version that avoids negative numbers in your working. For the series 2 + 6 + 18 + … up to 5 terms, a = 2, r = 3. S₅ = 2(3⁵ – 1) / (3 – 1) = 2(243 – 1) / 2 = 242.

你也可以使用 Sₙ = a(rⁿ – 1) / (r – 1)，两者代数等价。选择能使计算过程中避免负数的版本。对于级数 2 + 6 + 18 + … 前5项的和，a=2，r=3，S₅ = 2(3⁵ – 1) / (3 – 1) = 2(243 – 1) / 2 = 242。

Many WJEC questions ask you to find how many terms are needed for the sum to exceed a certain value. This requires setting up an inequality with Sₙ and solving, possibly using logs if the common ratio is greater than 1. For example, find the smallest n such that Sₙ > 1000 for a = 2, r = 2. You would solve 2(2ⁿ – 1) > 1000, giving 2ⁿ > 501, so n ≥ 9 (since 2⁹ = 512).

许多 WJEC 的题目会问需要多少项才能使级数的和超过某一数值。这需要列出包含 Sₙ 的不等式并求解，如果公比大于1可能要用到对数。例如，已知 a=2, r=2，求使得 Sₙ > 1000 的最小 n。解 2(2ⁿ – 1) > 1000 得 2ⁿ > 501，因此 n ≥ 9（因 2⁹ = 512）。

6. Sum to Infinity of a Geometric Series | 无穷等比级数的和

If the common ratio r satisfies |r| < 1 (i.e., -1 < r < 1), the terms of the geometric sequence get smaller and smaller. The series is said to converge, and it has a finite sum to infinity.

如果公比 r 满足 |r| < 1（即 -1 < r < 1），等比级数的各项会越来越小，此时级数收敛，具有有限的无穷和。

S∞ = a / (1 – r)

For example, the infinite series 8 + 4 + 2 + 1 + ½ + … has a = 8 and r = ½. Its sum to infinity is 8 / (1 – ½) = 8 / 0.5 = 16. This means that if you kept adding terms forever, the total would approach 16.

例如，无穷级数 8 + 4 + 2 + 1 + ½ + … 的首项 a=8，公比 r=½。其无穷和为 8 / (1 – ½) = 8 / 0.5 = 16。这意味着即使无限地加下去，总和也只会趋近于16。

If |r| ≥ 1, the series diverges – the sum gets infinitely large (or does not settle) and no sum to infinity exists. The WJEC exam often includes a linked question: first prove |r| < 1, then use the formula. Also, the sum to infinity can be used to check answers when calculating partial sums.

若 |r| ≥ 1，级数发散——和会无限增大（或摆动不定），不存在无穷和。WJEC 考试经常包含关联题：先证明 |r| < 1，再使用公式。此外，计算部分和时可用无穷和来粗略检查答案。

7. Using Sigma Notation (Σ) | 使用 Σ 记号

Sigma notation (Σ) provides a compact way of writing a series. The expression below the sigma tells you the starting value of the index (often n or r), and the number above gives the finishing value. The general term is written to the right.

西格玛记号（Σ）提供了一种表示级数的简洁方法。Σ 下方的式子表示索引（通常为 n 或 r）的起始值，上方的数字表示终止值。通项写在右侧。

Σ (from n=1 to 10) of (2n + 3) means (2×1+3) + (2×2+3) + … + (2×10+3)

∑_{n=1}^{10} (2n+3) 表示 (2×1+3) + (2×2+3) + … + (2×10+3)

To evaluate such a sum, you can often recognise it as an arithmetic series. The first term is 5, the last term is 23, and there are 10 terms. So S₁₀ = 10/2 (5 + 23) = 140. Alternatively, you can break the sigma into 2 Σ n + Σ 3 and use standard results, but WJEC IGCSE usually expects the direct arithmetic series approach.

要计算这样的和，通常可以识别出它是一个等差级数。首项为5，末项为23，共有10项。因此 S₁₀ = 10/2 (5 + 23) = 140。你也可以将 Σ 拆分为 2Σn + Σ3 并利用标准结果，但 WJEC IGCSE 通常要求运用等差级数直接求解。

Sigma notation can also represent geometric series. For instance, Σ (from k=1 to n) of 3×2ᵏ⁻¹ is geometric with a = 3, r = 2. The index can start at 0 or 1—always check the pattern of the first few terms to identify a and r correctly.

Σ 记号也可以表示等比级数。例如，∑_{k=1}^{n} 3×2ᵏ⁻¹ 是一个等比级数，a=3, r=2。索引可以从0或1开始——务必检查前几项的模式，以正确识别 a 和 r。

8. Problem Solving with Sequences | 数列问题求解

WJEC questions often combine sequences with algebra. You might be asked to find n given a certain term, or to set up an equation linking terms. For example, the 2nd, 4th and 8th terms of an arithmetic sequence form the first three terms of a geometric sequence. This requires expressing both sequences in terms of a and d, then using the geometric condition (common ratio) to find a connection.

WJEC 题目经常将数列与代数结合。你可能需要根据某一项求 n，或者建立项与项之间的方程。例如，一个等差数列的第2、第4和第8项构成一个等比数列的前三项。这需要先用 a 和 d 表示这两个数列，再利用等比条件（公比相等）找出关系。

Real-life contexts include simple interest (arithmetic) and compound growth or decay (geometric). For instance, a person saves £100 in the first month and increases the saving by £20 each month; the total saved after n months forms an arithmetic series. A radioactive material decaying by 5% per year follows a geometric sequence.

实际情境包括单利（等差）和复利增长或衰减（等比）。例如，某人第一个月存100英镑，之后每月增加20英镑，则 n 个月后的总存款构成等差级数。某放射性物质每年衰减5%，则遵循等比数列。

When tackling word problems, define your variables clearly: first term a, common difference d or ratio r, and the number of terms n. Write down the given information as equations and solve step by step. This structured approach reduces errors.

处理文字题时，要清晰地定义变量：首项 a，公差 d 或公比 r，以及项数 n。将所给信息写成方程，然后逐步求解。这种有条理的方法能减少错误。

9. Exam Technique and Common Pitfalls | 考试技巧与常见陷阱

Know the formulae inside out. You must be able to quote the nth term and sum formulae instantly. Write them down at the start of your exam paper as a memory aid, but ensure you use the correct one for each situation.

彻底掌握公式。 你必须能立刻写出通项公式和求和公式。考试开始时可将它们写在草稿纸上辅助记忆，但要确保针对每种情形使用正确的公式。

Check if a series is arithmetic or geometric. Look for a constant difference or constant ratio. Do not assume; calculate d or r first. A sequence like 2, 4, 8, 14 is neither arithmetic nor geometric, so the standard formulae won’t apply – you would need a different approach.

先检查级数是等差还是等比。 观察是否有常数差或常数比。不要主观臆断；先计算 d 或 r。像 2, 4, 8, 14 这样的数列既不是等差也不是等比，因此标准公式不适用——你需要另寻他法。

Common mistakes: For arithmetic series, forgetting to divide by 2 or using n-1 instead of n for the number of terms. For geometric series, misplacing brackets in the formula can ruin the calculation. Always double-check the sign of r, especially when alternating signs appear (e.g., +, -, +, -).

常见错误： 对于等差级数，忘记除以2，或者项数用了 n-1 而不是 n。对于等比级数，公式中括号位置放错会毁了整道计算。尤其当出现正负交替时（如 +, -, +, -），务必

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IGCSE WJEC Mathematics: Sequences and Series Key Points | IGCSE WJEC 数学：数列与级数 考点精讲