IB Math Topic 1: Number and Algebra | IB数学 Topic 1：数与代数知识点精讲

📚 IB Math Topic 1: Number and Algebra | IB数学 Topic 1：数与代数知识点精讲

Topic 1: Number and Algebra forms the foundation of the IB Diploma Programme Mathematics courses, both Analysis and Approaches (AA) and Applications and Interpretation (AI). It covers sequences and series, exponents and logarithms, the binomial theorem, proof by induction, and complex numbers. A clear understanding of these core topics is essential for success across the syllabus and for tackling real-world modelling problems. This revision guide provides a bilingual point-by-point breakdown of the key concepts, formulas, and methods you will encounter.

Topic 1：数与代数是 IB 文凭数学课程（无论是分析与方法 AA 还是应用与解释 AI）的基础模块。它涵盖数列与级数、指数与对数、二项式定理、数学归纳法以及复数等内容。透彻理解这些核心知识点对于掌握整个大纲和解决实际建模问题至关重要。本复习指南以中英双语逐点拆解核心概念、公式与方法，帮助你系统备考。

1. Arithmetic Sequences | 等差数列

An arithmetic sequence is a list of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by d. The general term formula is u_n = u₁ + (n − 1)d, where u₁ is the first term and n is the term number. You can use this formula to find any term directly without listing all previous terms.

等差数列是指相邻两项的差恒定的数列。这个恒定的差值称为公差，记作 d。通项公式为 u_n = u₁ + (n − 1)d，其中 u₁ 为首项，n 为项数。利用该公式可以直接求出任意一项，而不必逐一推算。

For example, consider the sequence 3, 7, 11, 15, … Here d = 4 and u₁ = 3, so the 20th term is u₂₀ = 3 + 19×4 = 79.

例如，数列 3, 7, 11, 15, … 的公差 d=4，首项 u₁=3，因此第20项 u₂₀ = 3 + 19×4 = 79。

2. Arithmetic Series | 等差级数

The sum of the first n terms of an arithmetic sequence is called an arithmetic series. There are two useful forms for the sum S_n: S_n = n/2 (u₁ + u_n) and S_n = n/2 [2u₁ + (n − 1)d]. These formulas reflect that the sum is the average of the first and last terms multiplied by the number of terms.

等差数列前 n 项的和称为等差级数。求和公式 S_n 有两种常用形式：S_n = n/2 (u₁ + u_n) 以及 S_n = n/2 [2u₁ + (n − 1)d]。这两个公式的本质是首项与末项的平均值乘以项数。

For instance, find the sum of the first 30 terms of 5 + 9 + 13 + … Here u₁ = 5, d = 4, so u₃₀ = 5 + 29×4 = 121, and S₃₀ = 30/2 (5 + 121) = 15 × 126 = 1890.

例如，求数列 5 + 9 + 13 + … 前30项的和。这里 u₁=5, d=4，因此 u₃₀=5+29×4=121，S₃₀ = 30/2 (5+121) = 15×126 = 1890。

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3. Geometric Sequences | 等比数列

In a geometric sequence, each term is obtained by multiplying the previous term by a constant called the common ratio, r. The general term is given by u_n = u₁ · rⁿ⁻¹. It is crucial to check that r ≠ 0, and the behaviour of the sequence depends on whether |r| < 1, |r| > 1, or r = 1.

在等比数列中，每一项都等于前一项乘以一个常数，称为公比 r。通项公式为 u_n = u₁ · rⁿ⁻¹。必须注意 r≠0，数列的性质取决于 |r| < 1、|r| > 1 或 r=1。

Example: 2, 6, 18, 54, … Here u₁ = 2, r = 3, so u₆ = 2 × 3⁵ = 2 × 243 = 486.

例如：数列 2, 6, 18, 54, … 中，u₁=2, r=3，因此 u₆ = 2 × 3⁵ = 2×243 = 486。

4. Geometric Series | 等比级数

The sum of the first n terms of a geometric sequence is S_n = u₁(1 − rⁿ)/(1 − r) for r ≠ 1. An alternative form is S_n = u₁(rⁿ − 1)/(r − 1), which is often used when r > 1. The sum to infinity S_∞ exists only when |r| < 1 and is given by S_∞ = u₁/(1 − r).

等比数列前 n 项的和公式为 S_n = u₁(1 − rⁿ)/(1 − r)，其中 r≠1。当 r>1 时，也常使用 S_n = u₁(rⁿ − 1)/(r − 1)。无穷级数求和 S_∞ 仅在 |r|<1 时存在，公式为 S_∞ = u₁/(1 − r)。

For a ball dropped from 10 m bouncing to 60% of its previous height each time, the total vertical distance travelled until it stops is a geometric series with infinite sum. The descending distances: 10, 6, 3.6, ... are a geometric sequence with u₁ nce = 10/(1−0.6) + same for ascending (starting from 6) = 10/0.4 + 6/0.4 = 25 + 15 = 40 m.

一个从10 m高度落下的球每次反弹至前次高度的60%，直到停止所经过的总垂直距离可表示为无穷等比级数。下降距离：10, 6, 3.6, … 是首项 u₁=10, r=0.6 的等比数列，总下降距离=10/(1−0.6)=25 m；上升距离从6开始，总和=6/(1−0.6)=15 m；总距离=40 m。

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5. Sigma Notation | 求和符号

Sigma notation (Σ) provides a compact way to write the sum of a sequence. The expression Σ_k=mⁿ a_k means the sum of terms a_k xpand the notation, identify the general term, and evaluate sums using known formulas for arithmetic and geometric sequences.

求和符号 Σ 用于紧凑地表示数列求和。Σ_k=mⁿ a_k 表示对 a_k 从 k=m 到 k=n 求和。你需要能够展开符号、识别通项，并运用等差数列与等比数列的求和公式计算。

For linear sequences like 5 + 8 + 11 + ... + 62, identify that the general term is 3k + 2 (with k starting from 1 to 20) and use the arithmetic sum formula.

对于线性数列如 5+8+11+...+62，可识别通项为 3k+2（k从1到20），进而用等差级数求和公式计算。

Σ_k=1ⁿ (3k + 2) = 3 Σ k + Σ 2 = 3·n(n+1)/2 + 2n

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6. Exponents and Logarithms | 指数与对数

Exponents and logarithms are inverse operations. If a^x = b, then x = log_ab, provided a > 0, a ≠ 1, and b > 0. The natural exponential function e^x and the natural logarithm ln x (log_e) play a central role in modelling growth and decay.

指数与对数互为逆运算。若 a^x = b，则 x = log_ab，其中 a>0, a≠1, b>0。自然指数函数 e^x 和自然对数 ln x（即 log_e）在增长与衰减建模中占据核心地位。

The laws of exponents: a^m · aⁿ = a^m+n, a^m/aⁿ = a^m−n, (a^m)ⁿ = a^mn, a⁰ = 1.

指数运算法则：a^m · aⁿ = a^m+n，a^m/aⁿ = a^m−n，(a^m)ⁿ = a^mn，a⁰=1。

The laws of logarithms: log_a(xy) = log_ax + log_ay; log_a(x/y) = log_ax − log_ay; log_a(x^k) = k·log_ax.

对数运算法则：log_a(xy) = log_ax + log_ay；log_a(x/y) = log_ax − log_ay；log_a(x^k) = k·log_ax。

Changing base: log_ab = log_cb / log_ca. This is especially useful for evaluating unfamiliar bases on a calculator.

换底公式：log_ab = log_cb / log_ca。这在计算非常见底数的对数时尤其有用。

Solve 3^2x−1 = 5. Take natural log: (2x−1)ln3 = ln5 → 2x−1 = ln5/ln3 → x = ½(ln5/ln3 + 1).

解方程 3^2x−1 = 5，取自然对数：(2x−1)ln3 = ln5 → 2x−1 = ln5/ln3 → x = ½(ln5/ln3 + 1)。

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7. Binomial Theorem | 二项式定理

The binomial theorem allows expansion of expressions of the form (a + b)ⁿ for positive integer n. The general term is given by T_r+1 = C(n, r) a^n−r b^r, where C(n, r) = n!/(r!(n−r)!) is the binomial coefficient, also written as _nC_r or (ⁿ_r). Pascal's triangle can be used for small n to find coefficients quickly.

二项式定理用于展开形如 (a + b)ⁿ（n为正整数）的表达式。通项为 T_r+1 = C(n, r) a^n−r b^r，其中二项式系数 C(n, r) = n!/(r!(n−r)!)，也写作 _nC_r 或 (ⁿ_r)。对于较小的 n，可用帕斯卡三角形快速找到系数。

For example, (2x − 3)⁴ = C(4,0)(2x)⁴ + C(4,1)(2x)³(−3) + C(4,2)(2x)²(−3)² + C(4,3)(2x)(−3)³ + C(4,4)(−3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81.

例如 (2x − 3)⁴ = C(4,0)(2x)⁴ + C(4,1)(2x)³(−3) + C(4,2)(2x)²(−3)² + C(4,3)(2x)(−3)³ + C(4,4)(−3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81。

The binomial theorem can also be extended to rational and negative exponents using the generalised binomial series, which is infinite and converges when |b/a| < 1.

二项式定理可推广至有理指数和负指数，形成无穷级数，当 |b/a| < 1 时收敛。

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8. Proof by Induction | 数学归纳法

e base case (show the statement holds for n = 1) and the inductive step (assume true for n = k and then prove it for n = k + 1). Concluding correctly is essential to complete the proof.

数学归纳法是一种证明对所有正整数成立的命题的有力方法。它包含两个主要步骤：基础情形（证明 n=1 时命题成立）和归纳步骤（假设 n=k 成立，证明 n=k+1 成立）。正确下结论才算完成证明。

Prove that 1 + 3 + 5 + … + (2n−1) = n². Base case n=1: LHS = 1, RHS = 1² = 1, true. Inductive hypothesis: assume true for n=k: 1+3+…+(2k−1) = k². For n=k+1, LHS = [1+3+…+(2k−1)] + (2(k+1)−1) = k² + 2k+1 = (k+1)². Thus true for n=k+1. By induction, the statement holds for all n ∈ ℕ.

证明 1+3+5+…+(2n−1)=n²。基础情形 n=1：左边=1，右边=1²=1，成立。归纳假设：设 n=k 时成立，即 1+3+…+(2k−1)=k²。则 n=k+1 时，左边 = [1+3+…+(2k−1)] + (2(k+1)−1) = k² + 2k+1 = (k+1)²。故 n=k+1 也成立。由归纳法，命题对所有正整数 n 成立。

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9. Complex Numbers | 复数

Complex numbers extend the real number system with the imaginary unit i, where i² a + bi, with real part a and imaginary part b. Operations follow algebraic rules, remembering that i² = −1. The complex conjugate is z̄ = a − bi, useful for division and finding moduli.

复数通过引入虚数单位 i（满足 i² = −1）扩展了实数系统。复数记作 z = a + bi，其中 a 为实部，b 为虚部。运算遵循代数规则，并要记住 i²=−1。复共轭为 z̄ = a − bi，在除法和求模中十分有用。

The modulus of z is |z| = √a² + b² real axis. Complex numbers can be expressed in polar form: z = r(cosθ + i sinθ), where r = |z|.

复数的模为 |z| = √a² + b²。辐角 θ 是线段 OZ 与正实轴的夹角。复数可用极形式表示：z = r(cosθ + i sinθ)，其中 r = |z|。

Multiplication and division in polar form: if z₁ = r₁cisθ₁ and z₂ = r₂cisθ₂, then z₁z₂ = r₁r₂cis(θ₁+θ₂) and z₁/z₂ = (r₁/r₂)cis(θ₁−θ₂).

极形式的乘除法：若 z₁ = r₁cisθ₁，z₂ = r₂cisθ₂，则 z₁z₂ = r₁r₂cis(θ₁+θ₂)，z₁/z₂ = (r₁/r₂)cis(θ₁−θ₂)。

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10. Applications of Sequences and Series | 数列与级数的应用

Sequences and series model many real-life situations: arithmetic growth (simple interest, linear depreciation), geometric growth (compound interest, population growth, radioactive decay), and infinite processes (bouncing ball, fractals). Interpreting the context to identify u₁, d or r, and n is a key skill.

数列与级数可用于模拟许多现实情境：等差数列描述线性增长（如单利、线性折旧），等比数列描述指数变化（如复利、人口增长、放射性衰变），无穷级数可处理反复过程（如弹跳球、分形）。根据背景识别首项、公差或公比以及项数是关键技能。

When dealing with financial applications, note that payments or deposits often form an annuity, which is a geometric series. The formula for the future value of an ordinary annuity uses the sum of a geometric progression.

在金融应用中，等额支付的年金往往构成等比级数。普通年金的终值公式即源于等比数列求和。

Modelling with exponential functions: N(t) = N₀e^kt for continuous growth, or N(t) = N₀b^t for discrete steps. Solving for time often requires logarithms.

指数函数建模：连续增长使用 N(t) = N₀e^kt，离散增长使用 N(t) = N₀b^t。求解时间常需使用对数。

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