Sequences and Series | 数列与级数

📚 Sequences and Series | 数列与级数

A sequence is an ordered list of numbers following a specific rule, while a series is the sum of the terms of a sequence. These concepts lie at the heart of A-Level mathematics, forming a bridge between algebra, calculus and real‑world modelling. In Edexcel Pure Mathematics, you are expected to be confident with arithmetic and geometric progressions, sigma notation, sums of powers, binomial expansions and recurrence relations.

数列是按特定规则排列的有序数字列表，而级数是数列各项之和。这两个概念是 A-Level 数学的核心，连接着代数、微积分和现实建模。在 Edexcel 纯数学中，你需要熟练掌握等差数列与等比数列、求和符号 Σ、幂和、二项展开式以及递推关系。

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1. Introduction to Sequences | 数列简介

A sequence is a list of terms written in a definite order. We often denote the nth term by u_n or a_n. Sequences can be finite or infinite. The rule that generates the terms can be given by a position-to-term formula such as u_n = 3n − 1, or by a recurrence relation like u_n+1 = 2u_n + 1 with a starting value.

数列是按特定顺序排列的一系列项，通常用 u_n 或 a_n 表示第 n 项。数列可以是有限的或无限的。产生各项的规则可以通过定位公式给出，例如 u_n = 3n − 1，或者通过递推关系给出，例如 u_n+1 = 2u_n + 1 及初始值。

Increasing, decreasing and periodic sequences are classified by their behaviour. A sequence like 2, 5, 8, 11, … is arithmetic because the difference between consecutive terms is constant.

递增、递减和周期数列是根据其变化特征进行分类的。像 2, 5, 8, 11, … 这样的数列是等差数列，因为连续两项之间的差为常数。

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2. Arithmetic Sequences | 等差数列

An arithmetic sequence has a common difference d between any term and the previous term. If the first term is a, the nth term is given by:

uₙ = a + (n − 1)d

等差数列的特点是任意一项与其前一项的差（公差 d）为常数。若首项为 a，第 n 项的通项公式为：

uₙ = a + (n − 1)d

For example, the sequence with a = 7, d = 4 produces 7, 11, 15, 19, … Here u₅ = 7 + (5−1)×4 = 23. It can be used to find the number of terms when the last term is known.

例如，首项 a = 7，公差 d = 4 的数列生成 7, 11, 15, 19, …，其 u₅ = 7 + (5−1)×4 = 23。当已知末项时，该公式可用于求总项数。

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3. Arithmetic Series | 等差级数求和

The sum of the first n terms of an arithmetic sequence is called an arithmetic series. The sum Sₙ can be found by pairing terms from the beginning and the end:

Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]

等差数列前 n 项之和称为等差级数。求和公式可通过首尾配对得到：

Sₙ = n/2 × (首项 + 末项) = n/2 [2a + (n − 1)d]

Proof: Write Sₙ = a + (a+d) + … + (l−d) + l. Reverse the same sum Sₙ = l + (l−d) + … + (a+d) + a. Adding the two rows gives 2Sₙ = n(a + l), and since l = a + (n−1)d, the second form follows.

证明：写出 Sₙ = a + (a+d) + … + (l−d) + l，再倒序写一次 Sₙ = l + (l−d) + … + (a+d) + a，两式相加得 2Sₙ = n(a + l)，再用 l = a + (n−1)d 即可得到第二种形式。

This identity is extremely useful in solving problems involving total distance, totals in a pattern, or financial calculations with linear growth.

该等式在求解涉及总距离、规律图案总和或线性增长的金融问题时极为有用。

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4. Geometric Sequences | 等比数列

A geometric sequence has a common ratio r between successive terms. If the first term is a, the nth term is:

uₙ = arⁿ⁻¹

等比数列的特点是相邻两项之比（公比 r）为常数。首项为 a 时，第 n 项通项为：

uₙ = arⁿ⁻¹

For instance, the sequence 3, 6, 12, 24, … has a = 3, r = 2, so u₅ = 3 × 2⁵⁻¹ = 48. Geometric sequences can model exponential growth or decay, such as population change, radioactive decay or compound interest.

例如，数列 3, 6, 12, 24, … 中 a = 3, r = 2，故 u₅ = 3 × 2⁵⁻¹ = 48。等比数列常用于模拟指数增长或衰减，如人口变化、放射性衰变或复利。

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5. Geometric Series | 等比级数求和

The sum of the first n terms of a geometric sequence is given by:

Sₙ = a(1 − rⁿ) / (1 − r) for r ≠ 1

等比数列前 n 项求和公式为：

Sₙ = a(1 − rⁿ) / (1 − r) (r ≠ 1)

Derivation: Write Sₙ = a + ar + ar² + … + arⁿ⁻¹. Multiply by r to get rSₙ = ar + ar² + … + arⁿ. Subtract to obtain Sₙ(1 − r) = a(1 − rⁿ).

推导：设 Sₙ = a + ar + ar² + … + arⁿ⁻¹，乘以 r 得 rSₙ = ar + ar² + … + arⁿ，两式相减得 Sₙ(1 − r) = a(1 − rⁿ)。

When |r| < 1, the infinite geometric series converges, and the sum to infinity is:

S∞ = a / (1 − r)

当 |r| < 1 时，无穷等比级数收敛，其无穷和为：

S∞ = a / (1 − r)

This is a crucial idea for recurring decimals and some physical processes that tend to a limiting value.

对于循环小数以及趋于极限值的某些物理过程，这一概念至关重要。

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6. Sigma Notation | 求和符号 Σ

Sigma notation provides a compact way to express a sum. The general form is:

∑ₖ₌ₚⁿ f(k)

求和符号 Σ 提供了表示求和的紧凑形式，一般格式为：

∑ₖ₌ₚⁿ f(k)

Here k is the index of summation, p is the lower limit and n is the upper limit. For example, ∑ₖ₌₁⁵ (2k + 1) means (2×1+1)+(2×2+1)+…+(2×5+1). Properties such as ∑(aₖ + bₖ) = ∑aₖ + ∑bₖ and ∑ c·aₖ = c·∑aₖ simplify manipulation.

其中 k 是求和指标，p 为下限，n 为上限。例如 ∑ₖ₌₁⁵ (2k + 1) 代表 (2×1+1)+(2×2+1)+…+(2×5+1)。利用性质 ∑(aₖ + bₖ) = ∑aₖ + ∑bₖ 以及 ∑ c·aₖ = c·∑aₖ 可以简化运算。

Recognising an arithmetic or geometric series from its sigma form is a common examination skill. For instance, ∑ₖ₌₁ⁿ (3k − 2) is an arithmetic series, while ∑ₖ₌₀ⁿ 5·2ᵏ is geometric.

从 Σ 形式识别等差或等比级数是常见的考试技能。例如 ∑ₖ₌₁ⁿ (3k − 2) 为等差级数，而 ∑ₖ₌₀ⁿ 5·2ᵏ 为等比级数。

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7. Sums of Powers of Natural Numbers | 自然数幂和

Edexcel expects you to memorise and apply three standard sums:

∑ₖ₌₁ⁿ k = n(n + 1)/2

∑ₖ₌₁ⁿ k² = n(n + 1)(2n + 1)/6

∑ₖ₌₁ⁿ k³ = [n(n + 1)/2]²

Edexcel 考纲要求掌握并应用三个标准求和公式：

∑ₖ₌₁ⁿ k = n(n + 1)/2

∑ₖ₌₁ⁿ k² = n(n + 1)(2n + 1)/6

∑ₖ₌₁ⁿ k³ = [n(n + 1)/2]²

These can be combined with the linearity of Σ to evaluate expressions like ∑ (3k² + 2k). For example, ∑ₖ₌₁ⁿ (3k² + 2k) = 3∑k² + 2∑k, which simplifies using the formulas above.

结合 Σ 的线性性质，可以计算诸如 ∑ (3k² + 2k) 的表达式。例如 ∑ₖ₌₁ⁿ (3k² + 2k) = 3∑k² + 2∑k，再代入上述公式即可化简。

Proof by induction is often linked to these results, strengthening your understanding of series manipulation.

这些公式常与数学归纳法证明结合，加深对级数运算的理解。

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8. Binomial Expansion for Positive Integer n | 正整数指数二项展开式

The binomial theorem for positive integer n states:

(a + b)ⁿ = ∑ₖ₌₀ⁿ ⁿCₖ aⁿ⁻ᵏ bᵏ

正整数 n 的二项式定理为：

(a + b)ⁿ = ∑ₖ₌₀ⁿ ⁿCₖ aⁿ⁻ᵏ bᵏ

where ⁿCₖ = n! / [k!(n − k)!] is the binomial coefficient. The expansion has n + 1 terms. Pascal’s triangle can generate coefficients for small n. When a = 1, we get (1 + x)ⁿ = 1 + nx + n(n−1)/2! x² + … + xⁿ.

其中 ⁿCₖ = n! / [k!(n − k)!] 为二项式系数，展开式共有 n + 1 项。对于较小的 n，可利用帕斯卡三角形生成系数。当 a = 1 时，有 (1 + x)ⁿ = 1 + nx + n(n−1)/2! x² + … + xⁿ。

This expansion is exact and valid for all real x. It is used to approximate expressions or to find specific terms without full expansion.

该展开式是精确的，对所有实数 x 成立，常用于逼近表达式或求指定项而无需完全展开。

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9. Binomial Expansion for Rational n | 有理指数二项展开式

When n is not a positive integer (e.g. negative or fractional), the binomial expansion generalises to an infinite series, provided |x| < 1:

(1 + x)ⁿ = 1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + …

当 n 不是正整数（如负数或分数）时，二项展开式推广为无穷级数，且要求 |x| < 1：

(1 + x)ⁿ = 1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + …

The expansion is valid for |x| < 1. For example, (1 + x)⁻¹ = 1 − x + x² − x³ + … with |x| < 1. Similarly, √(1 + x) = (1 + x)^½ = 1 + ½ x − ⅛ x² + …

该展开式在 |x| < 1 时成立。例如 (1 + x)⁻¹ = 1 − x + x² − x³ + …，要求 |x| < 1；同样地，√(1 + x) = (1 + x)^½ = 1 + ½ x − ⅛ x² + …

It is possible to expand expressions like (a + bx)ⁿ by factoring out aⁿ: (a + bx)ⁿ = aⁿ (1 + (b/a)x)ⁿ, then apply the standard series. Questions often ask for the range of validity or for approximations ignoring x³ and higher powers.

对于 (a + bx)ⁿ 这类表达式，可提取因数 aⁿ 化为 aⁿ (1 + (b/a)x)ⁿ，再利用标准展开式。考题常要求给出有效范围，或忽略 x³ 及更高次幂进行近似计算。

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10. Applications and Mixed Problems | 应用与混合题型

Series problems often mix arithmetic and geometric ideas. A typical question might give you the sum of the first few terms and ask for the common ratio or the first term. Another might link a recurrence relation to the sum formula of a series.

级数题目常混合等差与等比概念。典型的考题可能给出前几项之和，要求计算公比或首项；也可能将递推关系与级数求和公式结合起来考查。

Modelling with series includes savings schemes (arithmetic), compound interest (geometric), bouncing ball heights (geometric series summing to infinity), and population forecasts. The key is to identify whether the model has a constant addition or a constant multiplication pattern.

建模应用包括储蓄计划（等差）、复利（等比）、弹跳球高度（等比无尽级数）和人口预测。关键在于识别模型是以恒定的加法还是以恒定的乘法方式变化。

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11. Recurrence Relations | 递推关系

A sequence can be defined by a recurrence relation that gives each term from the previous one, together with an initial term. For example, uₙ₊₁ = 3uₙ − 2, u₁ = 2. This type of sequence may behave in a way that is neither purely arithmetic nor geometric; however, by generating terms we can look for patterns or solve directly.

数列可以由给出后一项与前一项关系的递推公式加上初始项来定义，例如 uₙ₊₁ = 3uₙ − 2，u₁ = 2。这类数列的行为可能既非纯等差也非纯等比，但通过计算前几项可以寻找规律或直接求解。

Sometimes a recurrence relation simplifies to a known type after substitution, or it can be solved to find a formula for uₙ in terms of n. This is explored further in sequences and series linked to iterative methods and proof.

有时递推关系可通过代换化简为已知类型，或直接求解得到关于 n 的通项公式。这在结合迭代法与证明的数列专题中会进一步探讨。

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12. Modelling with Sequences and Series | 数列建模

Real-world phenomena that change in regular steps are ideal for sequence models. An arithmetic model fits linear growth: a person saves £50 more each month. A geometric model fits exponential change: a radioactive substance loses 5% of its mass each year.

现实世界中按固定步骤变化的现象非常适合用数列模型描述。等差模型匹配线性增长：某人每月多存 50 英镑；等比模型匹配指数变化：放射性物质每年损失其质量的 5%。

Set up a sequence, identify a (first term), d or r (common difference or ratio), and then use the term or sum formulas to answer questions about total amounts, thresholds, or long‑term behaviour. Remember to state the limits, for instance that the infinite sum only applies when |r| < 1.

建立数列模型时，确定首项 a、公差 d 或公比 r，然后利用通项或求和公式回答关于总量、阈值或长期行为的问题。务必注明限制条件，例如无穷和仅在 |r| < 1 时适用。

These models train you to translate a written scenario into mathematical language, a skill that is tested regularly in Edexcel applied questions and in the large data set tasks.

这类建模训练你将文字场景翻译成数学语言的能力，这在 Edexcel 的应用题和大数据集任务中频繁考查。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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