Sequences and Series: Key Concepts | 数列与级数考点精讲

📚 Sequences and Series: Key Concepts | 数列与级数考点精讲

Sequences and series form a cornerstone of the IB CCEA Mathematics syllabus, providing a structured way to explore patterns of numbers and their accumulated sums. Mastery of these concepts not only secures solid marks in examinations but also builds analytical skills applicable to real‑world contexts such as finance, population modelling, and computer science.

数列与级数是 IB CCEA 数学大纲的核心支柱，为探索数字规律及其累加和提供了一套严谨的方法。扎实掌握这些概念不仅能确保考试中稳定得分，更能培养可应用于金融、人口建模和计算机科学等实际场景的分析能力。

1. Sequences: Definition and Notation | 数列的定义与表示法

A sequence is an ordered list of numbers following a specific rule. Each number in the sequence is called a term, and we typically denote the nth term by aₙ. The domain of a sequence is the set of positive integers, so terms progress as a₁, a₂, a₃, …, aₙ. Sequences can be finite or infinite, and they are often described by an explicit formula (position‑to‑term) or a recursive relation (term‑to‑term).

数列是一组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项，我们通常用 aₙ 表示第 n 项。数列的定义域是正整数集，因此各项依次为 a₁, a₂, a₃, …, aₙ。数列可以是有限的或无限的，并且常用通项公式（位置与项的对应关系）或递推关系（项与项之间的转换法则）来描述。

• Explicit formula: aₙ = 2n + 1 gives terms 3, 5, 7, 9, …
• Recursive: a₁ = 3, aₙ₊₁ = aₙ + 2 gives the same.
• 通项公式：aₙ = 2n + 1 给出 3, 5, 7, 9, …
• 递推关系：a₁ = 3, aₙ₊₁ = aₙ + 2 给出同样的数列。

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2. Arithmetic Sequences | 等差数列

An arithmetic sequence has a constant difference between successive terms, known as the common difference d. Its defining characteristic is that aₙ₊₁ – aₙ = d for all n. Given the first term a₁, the nth term is aₙ = a₁ + (n – 1)d.

等差数列的相邻项之差恒定，这个差称为公差 d。其特征是对于所有 n 都有 aₙ₊₁ – aₙ = d。已知第一项 a₁，第 n 项的通项公式为 aₙ = a₁ + (n – 1)d。

aₙ = a₁ + (n – 1)d

• Example: 5, 8, 11, 14, … has a₁ = 5, d = 3, so aₙ = 5 + (n – 1)×3 = 3n + 2.
• 示例：5, 8, 11, 14, … 其中 a₁ = 5, d = 3，因此 aₙ = 5 + (n – 1)×3 = 3n + 2。

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3. Arithmetic Series | 等差级数求和

m the beginning and end, resulting in the formula Sₙ = n/2 (a₁ + aₙ) or Sₙ = n/2 [2a₁ + (n – 1)d].

等差级数是等差数列各项的和。前 n 项和记作 Sₙ，可以通过首尾配对的方法求出，得到公式 Sₙ = n/2 (a₁ + aₙ) 或 Sₙ = n/2 [2a₁ + (n – 1)d]。

Sₙ = n/2 (a₁ + aₙ) = n/2 [2a₁ + (n – 1)d]

• For 5 + 8 + 11 + … up to 10 terms, S₁₀ = 10/2 [2×5 + (10-1)×3] = 5 × (10 + 27) = 185.
• 对于数列 5 + 8 + 11 + … 前 10 项，S₁₀ = 10/2 [2×5 + (10-1)×3] = 5 × 37 = 185。

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4. Geometric Sequences | 等比数列

A geometric sequence is one in which each term after the first is obtained by multiplying the previous term by a constant called the common ratio r. Thus, aₙ₊₁ / aₙ = r. With first term a₁, the nth term is aₙ = a₁ × rⁿ⁻¹.

等比数列是从第二项起，每一项与前一项的比值都等于同一个常数（称为公比 r）的数列。因此 aₙ₊₁ / aₙ = r。已知首项 a₁，第 n 项为 aₙ = a₁ × rⁿ⁻¹。

aₙ = a₁ rⁿ⁻¹

• Example: 3, 6, 12, 24, … has a₁ = 3, r = 2, so aₙ = 3 × 2ⁿ⁻¹.
• 示例：3, 6, 12, 24, … 其中 a₁ = 3, r = 2，因此 aₙ = 3 × 2ⁿ⁻¹。

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5. Geometric Series | 等比级数求和

rms of a geometric sequence is called a geometric series. When r ≠ 1, the sum is Sₙ = a₁(1 – rⁿ) / (1 – rmula is essential for computing compound interest growth and repeated percentage changes.

等比数列前 n 项的和称为等比级数。当 r ≠ 1 时，求和公式为 Sₙ = a₁(1 – rⁿ) / (1 – r)。当 r = 1 时，所有项相等，因此 Sₙ = n × a₁。该公式在计算复利增长和重复百分比变化时至关重要。

Sₙ = a₁(1 – rⁿ) / (1 – r), r ≠ 1

• For 3 + 6 + 12 + … up to 5 terms: S₅ = 3(1 – 2⁵) / (1 – 2) = 3(1 – 32) / (–1) = 93.
• 数列 3 + 6 + 12 + … 前 5 项：S₅ = 3(1 – 2⁵) / (1 – 2) = 3(1 – 32) / (–1) = 93。

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6. Infinite Geometric Series | 无穷等比级数

rges to a finite sum only when the absolute value of the common ratio is less than 1, i.e. |r| < 1. In that case, the sum to infinity is S∞ = a₁ / (1 – r). If |r| his concept appears frequently in questions about recurring decimals and fractal patterns.

只有当公比的绝对值小于 1（|r| < 1）时，无穷等比级数才收敛到一个有限和。此时，无穷和公式为 S∞ = a₁ / (1 – r)。如果 |r| ≥ 1，级数发散，没有有限和。这一概念常出现在循环小数和分形图案的题目中。

S∞ = a₁ / (1 – r), |r| < 1

• The series 8 + 4 + 2 + 1 + … has r = ½, so S∞ = 8 / (1 – ½) = 16.
• 级数 8 + 4 + 2 + 1 + … 的公比 r = ½，因此 S∞ = 8 / (1 – ½) = 16。

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7. Sigma Notation | 求和符号 Σ

resent series. The expression Σ (from k = p to q) of f(k) means the sum of f(p) + f(p+1) + … ial series. Being able to read and manipulate sigma notation is vital for problem solving.

求和符号 Σ 提供了一种简洁表示级数的方式。表达式 Σ (k = p 到 q) f(k) 表示 f(p) + f(p+1) + … + f(q) 的和。它广泛用于表达等差与等比级数以及多项式级数。能够读懂和操作 Σ 符号对解题至关重要。

Σ_k=1ⁿ (2k – 1) = 1 + 3 + 5 + … + (2n – 1)

• Σ_k=1⁴ k² = 1² + 2² + 3² + 4² = 30.

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8. Applications: Compound Interest & Finance | 应用：复利与财务

ally grows after n years to P(1 + i)ⁿ. The future value of a series of regular deposits forms a geometric series: A = R[(1 + i)ⁿ – 1] / i, where R is the regular deposit.

等比数列与级数可用于建模储蓄和贷款还款。例如，本金 P 按年利率 i 每年复利一次，n 年后的本利和为 P(1 + i)ⁿ。一系列定期存款的未来值构成一个等比级数：A = R[(1 + i)ⁿ – 1] / i，其中 R 为定期存款额。

• If £1000 is deposited each year at 5% interest, after 10 years the total is 1000[(1.05¹⁰ – 1)/0.05] ≈ £12 577.89.
• 若每年存入 1000 英镑，利率 5%，10 年后的总额约为 1000[(1.05¹⁰ – 1)/0.05] ≈ 12 577.89 英镑。

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9. Convergence and Divergence | 收敛与发散

etric convergence, understanding that a series with terms that do not tend to zero must diverge is a key test. The limit of a sequence aₙ as n → ∞ is also examined: if that limit exists, the sequence converges.

除等比级数外，IB 学生还应能判断给定级数是否收敛。虽然 CCEA 大纲重点考查等比级数的收敛性，但理解“通项不趋于零则级数必发散”是一项关键检验。数列 aₙ 当 n → ∞ 时的极限也是考点：若极限存在，则数列收敛。

• ₙ = 1/n, the sequence converges to 0; the harmonic series Σ 1/n diverges though terms → 0.
• 数列 aₙ = 1/n 收敛于 0；但调和级数 Σ 1/n 发散，尽管其通项趋于 0。

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10. Problem‑Solving Strategies | 解题策略

Many exam questions combine several concepts. Start by identifying whether the given sequence is arithmetic or geometric by checking constant difference or ratio. Write down known values: a₁, d or r, n, Sₙ. Use the relevant formula and solve equations. For word problems, translate the scenario into a sequence – for instance, heights of a bouncing ball form a geometric sequence.

许多考题会综合多个概念。首先通过检查恒定差或恒定比来判断给定数列是等差还是等比。列出已知值：a₁, d 或 r, n, Sₙ。选用相应公式并解方程。对于应用题，将场景转化为数列——例如，弹跳球的高度构成一个等比数列。

• If the 3rd term is 10 and the 6th term is 80, find r: a₃ = a₁r² = 10, a₆ = a₁r⁵ = 80 ⇒ r³ = 8 ⇒ r = 2.
• 如果第 3 项为 10，第 6 项为 80，求公比 r：a₃ = a₁r² = 10, a₆ = a₁r⁵ = 80 ⇒ r³ = 8 ⇒ r = 2。

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11. Common Pitfalls to Avoid | 常见易错点

es, ensure you correctly subtract 1 from n in the power of r when finding aₙ. When summing, check if the series starts at n = 1 or n = 0, as this shifts indices. Finally, for infinite series always verify |r| < 1 before using S∞ = a₁/(1 – r).

学生经常混淆等差和等比级数的公式。记住：等差级数用 d，等比级数用 r。在等比数列中，求 aₙ 时，务必在 r 的指数中正确减去 1。求和时，需检查级数起点是 n = 1 还是 n = 0，因为这会影响序号。最后，使用 S∞ = a₁/(1 – r) 前必须验证 |r| < 1。

Pitfall	Solution
Using S∞ when r > 1	Check |r| < 1 first
Forgetting the (n – 1) in aₙ = a₁rⁿ⁻¹	Count terms: a₂ should be a₁×r, not a₁×r²
Confusing term count in sigma from k = 0	Number of terms = upper – lower + 1

易错点	解决方法
当 r > 1 时使用 S∞	先检查 |r| < 1
忘记 aₙ = a₁rⁿ⁻¹ 中的 (n – 1)	计算项数：a₂ 应为 a₁×r，而非 a₁×r²
混淆 sigma 从 k = 0 开始的项数	项数 = 上限 – 下限 + 1

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12. Summary of Key Formulas | 核心公式一览

for instance Sₙ = n/2 (a₁ + aₙ) from the idea of pairing terms. Below is a concise table.

考试时随身准备好这四个核心公式的参考。通过推导（例如利用首尾配对思想得出 Sₙ = n/2 (a₁ + aₙ)）来加深理解。下表为简明汇总。

Concept	Formula
Arithmetic nth term	aₙ = a₁ + (n – 1)d
Arithmetic sum	Sₙ = n/2 [2a₁ + (n – 1)d]
Geometric nth term	aₙ = a₁ rⁿ⁻¹
Geometric sum	Sₙ = a₁(1 – rⁿ)/(1 – r)
Infinite sum	S∞ = a₁/(1 – r), |r| < 1

概念	公式
等差数列通项	aₙ = a₁ + (n – 1)d
等差数列和	Sₙ = n/2 [2a₁ + (n – 1)d]
等比数列通项	aₙ = a₁ rⁿ⁻¹
等比数列和	Sₙ = a₁(1 – rⁿ)/(1 – r)
无穷和	S∞ = a₁/(1 – r), |r| < 1

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