Series: Arithmetic, Geometric and Binomial Expansion | 级数：等差、等比与二项式展开

📚 Series: Arithmetic, Geometric and Binomial Expansion | 级数：等差、等比与二项式展开

In Cambridge A Level Mathematics, the topic Series brings together arithmetic progressions, geometric progressions, sigma notation and binomial expansion. It is assessed both as direct formula questions and as part of applied problems involving savings, population growth, depreciation and algebraic coefficients.

在剑桥 A Level 数学中，级数这一主题整合了等差数列、等比数列、Σ 记号和二项式展开。考试中既有直接套用公式的基础题，也有涉及储蓄、人口增长、折旧和代数系数的综合应用题。

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1. Sequences, Series and Key Notation | 数列、级数与核心记号

A sequence is an ordered list of numbers such as 2, 5, 8, 11, … . A series is the sum of the terms of a sequence, for example 2 + 5 + 8 + 11 + … .

数列是一组有序排列的数，例如 2、5、8、11……。级数则是数列各项之和，例如 2 + 5 + 8 + 11 + ……。

The first term is usually written as a or u₁, the nth term as uₙ, and the sum of the first n terms as Sₙ. In an arithmetic progression, d is the common difference; in a geometric progression, r is the common ratio.

首项通常记作 a 或 u₁，第 n 项记作 uₙ，前 n 项和记作 Sₙ。在等差数列中，d 表示公差；在等比数列中，r 表示公比。

Notation / 记号	Meaning / 含义
a, u₁	first term / 首项
uₙ	nth term / 第 n 项
Sₙ	sum of the first n terms / 前 n 项和
d	common difference / 公差
r	common ratio / 公比

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2. Arithmetic Progressions: The nth Term | 等差数列：第 n 项

An arithmetic progression, or AP, is a sequence in which the difference between any term and the previous term is constant. This constant is the common difference d, so uₙ₊₁ − uₙ = d.

等差数列（AP）是相邻两项之差恒定的数列。这个固定的差称为公差 d，因此 uₙ₊₁ − uₙ = d。

uₙ = a + (n − 1)d

To find the nth term, start with the first term a and add the common difference d exactly (n − 1) times.

求第 n 项时，从首项 a 出发，加上 (n − 1) 个公差 d。

Example: For the AP 3, 7, 11, 15, …, find the 20th term. Here a = 3, d = 4, so u₂₀ = 3 + (20 − 1) × 4 = 3 + 76 = 79.

示例：已知等差数列 3、7、11、15……，求第 20 项。这里 a = 3，d = 4，所以 u₂₀ = 3 + (20 − 1) × 4 = 3 + 76 = 79。

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3. Sum of an Arithmetic Series | 等差数列求和

The sum of the first n terms of an arithmetic series can be found using the first term, the common difference and the number of terms.

等差数列前 n 项和可以通过首项、公差和项数求得。

Sₙ = n/2 × [2a + (n − 1)d]

If the last term l = uₙ is known, the shorter form is useful:

如果已知末项 l = uₙ，可以用更简洁的形式：

Sₙ = n/2 × (a + l)

The formula works because the average of the first and last terms, (a + l)/2, is multiplied by the number of terms n.

该公式的原理是首项与末项的平均值 (a + l)/2 乘以项数 n。

Example: Find the sum of the first 50 terms of 5, 8, 11, 14, … . Here a = 5, d = 3, n = 50, so S₅₀ = 50/2 × [2 × 5 + (50 − 1) × 3] = 25 × [10 + 147] = 25 × 157 = 3925.

示例：求 5、8、11、14……前 50 项的和。这里 a = 5，d = 3，n = 50，所以 S₅₀ = 50/2 × [2 × 5 + (50 − 1) × 3] = 25 × [10 + 147] = 25 × 157 = 3925。

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4. Geometric Progressions: The nth Term | 等比数列：第 n 项

A geometric progression, or GP, is a sequence in which each term is obtained by multiplying the previous term by a constant r, called the common ratio. Thus r = uₙ₊₁ / uₙ.

等比数列（GP）是每一项都等于前一项乘以常数 r 的数列，这个常数称为公比。因此 r = uₙ₊₁ / uₙ。

uₙ = arⁿ⁻¹

The nth term is found by taking the first term a and multiplying by the common ratio r exactly (n − 1) times.

第 n 项等于首项 a 乘以公比 r 的 (n − 1) 次方。

Example: For the GP 2, 6, 18, 54, …, find the 10th term. Here a = 2, r = 3, so u₁₀ = 2 × 3⁹ = 2 × 19683 = 39366.

示例：已知等比数列 2、6、18、54……，求第 10 项。这里 a = 2，r = 3，所以 u₁₀ = 2 × 3⁹ = 2 × 19683 = 39366。

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5. Sum of a Geometric Series | 等比数列求和

The sum of the first n terms of a geometric series with common ratio r ≠ 1 is given by the formula:

公比 r ≠ 1 的等比数列前 n 项和公式为：

Sₙ = a(1 − rⁿ) / (1 − r)

An equivalent form is Sₙ = a(rⁿ − 1) / (r − 1). If r = 1, the series is constant and Sₙ = an.

等价形式为 Sₙ = a(rⁿ − 1) / (r − 1)。如果 r = 1，数列每项都相等，则 Sₙ = an。

Example: Find the sum of the first 8 terms of 3, 6, 12, 24, … . Here a = 3, r = 2, so S₈ = 3(1 − 2⁸) / (1 − 2) = 3(1 − 256) / (−1) = 765.

示例：求 3、6、12、24……前 8 项的和。这里 a = 3，r = 2，所以 S₈ = 3(1 − 2⁸) / (1 − 2) = 3(1 − 256) / (−1) = 765。

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6. Sum to Infinity of a Geometric Series | 等比数列无穷和

A geometric series has a finite sum to infinity only when |r| < 1, because rⁿ tends to 0 as n becomes very large. The infinite sum is:

等比数列只有当 |r| < 1 时才存在有限无穷和，因为当 n 趋近无穷大时 rⁿ 趋向于 0。无穷和公式为：

S∞ = a / (1 − r), for |r| < 1

If |r| ≥ 1, the terms do not become small enough, so the series diverges and has no finite sum to infinity.

如果 |r| ≥ 1，各项不会变得足够小，级数发散，因此没有有限无穷和。

Example: Find the sum to infinity of 8 + 4 + 2 + 1 + … . Here a = 8 and r = 1/2, so S∞ = 8 / (1 − 1/2) = 8 / (1/

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