📚 IB Mathematics: Sequences and Series Core Concepts | IB数学：数列与级数核心考点
Sequences and series form one of the most fundamental topics in IB Mathematics. Whether you are studying Analysis and Approaches (AA) or Applications and Interpretation (AI), understanding how terms are generated and how sums are calculated is essential for success.
数列与级数是IB数学中最基础的知识板块之一。无论你学习的是分析与方法（AA）还是应用与解释（AI），理解项的生成方式以及和的计算方法都是取得高分的关键。
A sequence is an ordered list of numbers, while a series is the sum of the terms of a sequence. In IB exams, you will need to recognise arithmetic and geometric patterns, apply their formulas, and interpret real-world contexts.
数列是按顺序排列的一组数，而级数是数列各项之和。在IB考试中，你需要识别等差与等比规律，灵活运用公式，并解释实际应用情境。
1. Arithmetic Sequences | 等差数列
An arithmetic sequence is a sequence in which the difference between consecutive terms is constant. This constant is called the common difference, usually denoted d.
等差数列是指相邻两项之差恒定的数列。这个常数称为公差，通常用 d 表示。
For an arithmetic sequence with first term u₁ and common difference d, the n-th term is given by:
对于首项为 u₁、公差为 d 的等差数列，其第 n 项公式为：
uₙ = u₁ + (n − 1)d
This formula is also useful when you know two non-consecutive terms: uₙ = uₘ + (n − m)d.
当已知两个不相邻的项时，也可使用推广形式：uₙ = uₘ + (n − m)d。
Example: Find the 20th term of the sequence 3, 7, 11, 15, ...
例题：求数列 3, 7, 11, 15, ... 的第20项。
Here u₁ = 3 and d = 4, so u₂₀ = 3 + 19 × 4 = 79.
这里 u₁ = 3，d = 4，所以 u₂₀ = 3 + 19 × 4 = 79。
2. Geometric Sequences | 等比数列
A geometric sequence is a sequence in which each term is obtained by multiplying the previous term by a fixed non-zero constant, called the common ratio r.
等比数列是指每一项都通过前一项乘以一个固定非零常数而得到的数列，这个常数称为公比 r。
The n-th term of a geometric sequence with first term u₁ and common ratio r is:
首项为 u₁、公比为 r 的等比数列第 n 项为：
uₙ = u₁ rⁿ⁻¹
Example: The 4th term of a geometric sequence is 24 and the 7th term is 192. Find the first term and the common ratio.
例题：等比数列的第4项为24，第7项为192。求首项与公比。
ding the second equation by the first gives r³ = 8, so r = 2. Then u₁ × 8 = 24, so u₁ = 3.
由 u₄ = u₁r³ = 24 和 u₇ = u₁r⁶ = 192。将第二式除以第一式得 r³ = 8，所以 r = 2。于是 u₁ × 8 = 24，因此 u₁ = 3。
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 either of the following equivalent formulas:
等差数列前 n 项和可用下面两个等价公式之一计算：
Sₙ = n/2 [2u₁ + (n − 1)d] = n/2 (u₁ + uₙ)
The first formula is useful when you know the first term and the common difference; the second is useful when you know the first and last terms.
当已知首项与公差时使用第一个公式；当已知首项与末项时使用第二个公式。
Example: Find the sum of the first 50 positive multiples of 3.
例题：求前50个3的正倍数之和。
Here u₁ = 3, d = 3, n = 50. So S₅₀ = 50/2 [2 × 3 + 49 × 3] = 25 × 153 = 3825.
这里 u₁ = 3，d = 3，n = 50。所以 S₅₀ = 50/2 [2 × 3 + 49 × 3] = 25 × 153 = 3825。
4. 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:
公比为 r ≠ 1 的等比数列前 n 项和为：
Sₙ = u₁(1 − rⁿ) / (1 − r)
formula as Sₙ = u₁(rⁿ − 1) / (r − 1); both forms are equivalent and give the same result.
有些教材写成 Sₙ = u₁(rⁿ − 1) / (r − 1)，两种形式等价，计算结果相同。
Example: Find the sum of the first 6 terms of the series 2 + 6 + 18 + ...
例题：求级数 2 + 6 + 18 + ... 的前6项和。
Here u₁ = 2, r = 3, n = 6. Therefore S₆ = 2(1 − 3⁶) / (1 − 3) = 2(1 − 729) / (−2) = 728.
这里 u₁ = 2，r = 3，n = 6。因此 S₆ = 2(1 − 3⁶) / (1 − 3) = 2(1 − 729) / (−2) = 728。
5. Infinite Geometric Series | 无穷等比级数
An infinite geometric series converges to a finite value only when |r| < 1. In that case, the sum to infinity is:
无穷等比级数只有在 |r| < 1 时才收敛于有限值。此时无穷项和为：
S∞ = u₁ / (1 − r)
If |r| ≥ 1, the series diverges and has no finite sum.
若 |r| ≥ 1，则级数发散，不存在有限和。
onsider the series 100 + 50 + 25 + 12.5 + ... Here r = 0.5, so S∞ = 100 / (1 − 0.5) = 200.
例题：考虑级数 100 + 50 + 25 + 12.5 + ...，这里 r = 0.5，所以 S∞ = 100 / (1 − 0.5) = 200。
This result can be understood visually: each new term covers half of the remaining distance to 200.
这一结果也可直观理解：每一项都覆盖了到200剩余距离的一半。
6. Sigma Notation | 求和符号
of writing a series. The expression below represents the sum of the terms from n = 1 to k:
求和符号是书写级数的紧凑方式。下面的表达式表示从 n = 1 到 k 各项之和：
Σn=1k uₙ = u₁ + u₂ + ⋯ + uₖPublished by TutorHao | IB Mathematics Revision Series | aleveler.com
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