📚 Core: Sequences | 核心：序列
A sequence is an ordered list of numbers that follow a particular pattern or rule. Each number in a sequence is called a term. Sequences can be finite or infinite, and they appear in many areas of mathematics, finance, and science. In the IB Mathematics core syllabus, you are expected to recognise arithmetic and geometric sequences, find their general terms, calculate sums of series, and apply these concepts to real-world problems such as compound interest and loan repayments.
序列是一组按照特定规律排列的有序数字，其中每一个数字称为项。序列可以是有限的，也可以是无限的，它在数学、金融和科学等诸多领域都有应用。在IB数学核心课程中，你需要识别等差数列和等比数列，求出通项，计算级数之和，并将这些概念应用到复利、贷款还款等实际问题中。
1. Introduction to Sequences | 序列简介
A sequence is usually written as u₁, u₂, u₃, …, uₙ, where n denotes the position of a term. The nth term is called the general term and is often expressed as a formula in terms of n. Sequences can be defined explicitly by a formula, or recursively by giving the first term and a rule for finding the next term.
序列通常记作 u₁, u₂, u₃, …, uₙ，其中 n 表示项的位置。第 n 项称为通项，通常用一个关于 n 的公式来表示。序列可以通过明确的公式来定义，也可以通过给出首项及一个推导后一项的规则来进行递推定义。
For example, the sequence 2, 5, 8, 11, … ned by adding 3 each time. Another example is the sequence of square numbers: 1, 4, 9, 16, …, whose general term is uₙ = n².
例如序列 2, 5, 8, 11, … 由一个明确的规则构成：每次都加 3。还有一个例子是平方数序列：1, 4, 9, 16, …，其通项为 uₙ = n²。
2. Arithmetic Sequences | 等差数列
An arithmetic sequence is a sequence in which the difference between any term and the previous one is constant. This constant difference is denoted by d and is called the common difference.
等差数列是指从第二项起，每一项与它的前一项的差都等于同一个常数的数列。这个固定的差用 d 表示，称为公差。
by uₙ = u₁ + (n − 1)d. This formula is fundamental and must be memorised for the IB exam.
若首项为 u₁，则第 n 项为 uₙ = u₁ + (n − 1)d。这个公式是基础，IB考试中必须熟记。
Example: Find the 20th term of the arithmetic sequence 7, 12, 17, 22, … Here u₁ = 7 and d = 5, so u₂₀ = 7 + (20 − 1)×5 = 7 + 95 = 102.
例题：求等差数列 7, 12, 17, 22, … 的第 20 项。这里 u₁ = 7，d = 5，所以 u₂₀ = 7 + (20 − 1)×5 = 7 + 95 = 102。
3. Sum of an Arithmetic Series | 等差数列求和
An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first n terms is denoted by Sₙ. There are two common forms for Sₙ:
等差数列各项的和称为等差级数。前 n 项之和记作 Sₙ。Sₙ 有两种常用的表达形式：
Sₙ = n/2 (u₁ + uₙ) and Sₙ = n/2 [2u₁ + (n − 1)d]
Sₙ = n/2 (u₁ + uₙ) 以及 Sₙ = n/2 [2u₁ + (n − 1)d]
Use the first form when you know the first and last terms; use the second when you know the first term and the common difference.
已知首项和末项时使用第一个公式；已知首项和公差时使用第二个公式。
ence with u₁ = 10 and d = 3. S₃₀ = 30/2 [2×10 + (30 − 1)×3] = 15[20 + 87] = 15×107 = 1605.
例如，求等差数列 u₁ = 10，d = 3 的前 30 项和。S₃₀ = 30/2 [2×10 + (30 − 1)×3] = 15[20 + 87] = 15×107 = 1605。
4. Geometric Sequences | 等比数列
A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed, non-zero constant called the common ratio, r. The first term is usually denoted by u₁.
等比数列是指从第二项起，每一项与前一项的比都等于同一个非零常数的数列，这个常数称为公比 r。首项通常记作 u₁。
rmula for the nth term is uₙ = u₁ rⁿ⁻¹. It is essential that the exponent is (n−1), not n.
第 n 项的通项公式为 uₙ = u₁ rⁿ⁻¹，注意指数为 (n−1) 而非 n，这一点非常重要。
For instance, in the sequence 3, 6, 12, 24, …, u₁ = 3, r = 2, thus the 8th term is u₈ = 3×2⁷ = 3×128 = 384.
例如在序列 3, 6, 12, 24, … 中，u₁ = 3，r = 2，因此第 8 项 u₈ = 3×2⁷ = 3×128 = 384。
5. Sum of a Finite Geometric Series | 有限等比数列求和
To find the sum Sₙ of the first n terms of a geometric sequence, use the formula:
要计算等比数列前 n 项的和 Sₙ，可使用公式：
Sₙ = u₁(1 − rⁿ) / (1 − r) for r ≠ 1
Sₙ = u₁(1 − rⁿ) / (1 − r)，其中 r ≠ 1
This formula is derived by multiplying the sum by r and subtracting. Always check that r ≠ 1 before applying it.
该公式的推导方法是将和乘以 r 后再与原式相减。应用前一定注意 r ≠ 1。
Example: Find the sum of the first 12 terms of 2 + 6 + 18 + … , r = 3, n = 12. S₁₂ = 2(1 − 3¹²)/(1 − 3) = 2(1 − 531441)/(−2) = 2(−531440)/(−2) = 531440.
例题：求等比数列 2 + 6 + 18 + … 的前 12 项之和。这里 u₁ = 2，r = 3，n = 12。S₁₂ = 2(1 − 3¹²)/(1 − 3) = 2(1 − 531441)/(−2) = 2(−531440)/(−2) = 531440。
6. Infinite Geometric Series | 无穷等比级数
When the common ratio r satisfies −1 < r < 1 (i.e. |r| etric series converges to a finite sum. The sum to infinity is given by S∞ = u₁ / (1 − r).
当公比 r 满足 −1 < r < 1（即 |r| < 1）时，无穷等比级数会收敛到一个有限的和。无穷项之和的公式为 S∞ = u₁ / (1 − r)。
If |r| 1, the terms do not approach zero, and the series diverges, meaning it has no finite sum.
如果 |r| ≥ 1，则各项不会趋于零，级数发散，意味着它没有有限的和。
For example, the series 12 + 6 + 3 + 1.5 + … has u₁ = 12 and r = 0.5. Since |r| < 1, the sum to infinity is 12/(1 − 0.5) = 24.
例如，级数 12 + 6 + 3 + 1.5 + … 中 u₁ = 12，r = 0.5。因为 |r| < 1，所以无穷项之和为 12/(1 − 0.5) = 24。
7. Sigma Notation (∑) | 求和符号
. The symbol ∑ (capital sigma) is used, with the index and limits written below and above.
求和符号提供了一种简写序列各项和的方式。用希腊大写字母 ∑ 表示，下标和上标分别写出指标变量的起始值与终值。
st 5 terms of an arithmetic sequence can be expressed as ∑ⁿᵢ₌₁ uᵢ or more concretely as ∑⁵k=1 (3k + 2). This means we evaluate 3k+2 for k = 1,2,3,4,5 and add the results.
例如，等差数列前 5 项和可表示为 ∑ⁿᵢ₌₁ uᵢ，更具体地可以写成 ∑⁵k=1 (3k + 2)，意思是将 k = 1,2,3,4,5 分别代入 3k+2 后求总和。
You must be comfortable interpreting and expanding sigma notation, as it is frequently tested in IB exams.
你必须熟练地解读和展开求和符号，因为在IB考试中这种题型很常见。
8. Finding the nth Term from Patterns | 从规律求第n项
Not all sequences are explicitly arithmetic or geometric. Some require you to detect a pattern. For a linear pattern, check the first differences. If the first differences are constant, the sequence is arithmetic.
并非所有序列都是显式的等差或等比数列。有些需要你找出规律。对于线性规律，可以检查一阶差分。如果一阶差分为常数，则序列为等差数列。
odel of the form uₙ = an² + bn + c. You can determine a, b, c by substituting known terms.
如果一阶差分不是常数，但二阶差分为常数，那么该序列遵循二次模型 uₙ = an² + bn + c。可以通过代入已知项来求出 a、b、c 的值。
Example: Given 3, 7, 13, 21, 31, … ferences are 2,2,2, constant. So the nth term is quadratic. Solving gives uₙ = n² + n + 1.
例题：已知序列 3, 7, 13, 21, 31, …，一阶差分为 4,6,8,10，二阶差分为 2,2,2 为常数，因此通项为二次式。计算可得 uₙ = n² + n + 1。
9. Applications: Compound Interest | 应用：复利
ation of geometric sequences. When an amount P is invested at an annual interest rate of r% compounded k times a year, the value after t years is A = P(1 + r/(100k))kt. Each compounding period multiplies the balance by a constant factor, forming a geometric sequence.
复利是等比数列的直接应用。当本金 P 以年利率 r%、每年计息 k 次的复利方式投资时，t 年后的总值为 A = P(1 + r/(100k))kt。每个计息期都会将余额乘以一个固定的因子，从而构成等比数列。
For instance, €5000 invested at 6% per annum compounded monthly gives after 10 years: A = 5000(1 + 6/(100×12))12×10 = 5000(1.005)120. This can be evaluated with a calculator.
例如，5000欧元以年利率6%、每月计息一次的方式投资，10年后的本利和为 A = 5000(1 + 6/(100×12))12×10 = 5000(1.005)120，可用计算器进行求值。
10. Applications: Loan Repayments and Annuities | 应用：贷款还款与年金
repaid by n payments of amount R. The present value equation is L = R × (1 − (1+i)⁻ⁿ) / i.
当贷款以等额分期方式偿还时，每次还款后的未偿还余额可以用等比级数建立模型。设贷款额为 L，每期利率为 i，分 n 期等额偿还，每期还款额 R，则现值方程为 L = R × (1 − (1+i)⁻ⁿ) / i。
This uses the sum of a finite geometric series. You may be asked to find R, n, or the total interest paid.
这里使用了有限等比级数的求和公式。考试中可能会要求计算 R、n 或总支付利息。
Understanding the link between regular payments and geometric series is vital for real-world financial mathematics questions in the IB core.
理解等额还款与等比级数之间的联系，对于解决IB核心课程中的实际金融数学问题至关重要。
11. Recursive Definition of Sequences | 递推定义序列
etic sequence, the recursive rule is uₙ₊₁ = uₙ + d. For a geometric sequence, uₙ₊₁ = r uₙ.
序列也可以通过递推的方式定义，即每一项都用前一项来表示。等差数列的递推公式为 uₙ₊₁ = uₙ + d；等比数列则为 uₙ₊₁ = r uₙ。
d uₙ₊₂ = uₙ₊₁ + uₙ. You should be able to generate terms from a given recurrence relation.
在IB考试中，你可能会遇到诸如斐波那契数列这样的序列，规定 u₁ = 1, u₂ = 1，且 uₙ₊₂ = uₙ₊₁ + uₙ。应能根据给定的递推关系生成后续项。
12. Common Mistakes and Exam Tips | 常见错误与考试技巧
. Also, when using Sₙ = n/2 (u₁ + uₙ), ensure you have the correct last term uₙ, not uₙ₋₁.
一个常见错误是混淆等差数列和等比数列的通项公式。记住：等差数列涉及加固定公差 d，等比数列涉及乘固定公比 r。此外，在使用 Sₙ = n/2 (u₁ + uₙ) 时，要确保使用的是正确的末项 uₙ 而不是 uₙ₋₁。
For infinite geometric series, always check the condition |r| < 1. If you apply the sum to infinity formula when |r| ≥ 1, your answer will be mathematically invalid.
对于无穷等比级数，一定要先检查条件 |r| < 1。如果在 |r| ≥ 1 时使用无穷和公式，所得的答案在数学上是无效的。
In word problems, identify whether a sequence is arithmetic (linear growth) or geometric (exponential growth/decay), then select the appropriate formula. Practice with past paper questions builds confidence.
在应用题中，先要判断是等差数列（线性增长）还是等比数列（指数增长或衰减），然后选取相应的公式。通过练习历年真题可以有效提升信心和熟练度。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图