📚 Number and Algebra: From Sequences to Complex Numbers | 数与代数：从数列到复数
The Number and Algebra topic forms the backbone of IB Mathematics, linking pattern recognition, financial modelling, algebraic manipulation, and abstract reasoning. For both SL and HL students, mastering sequences, exponents, binomial expansions and complex numbers enables a deeper understanding of calculus, statistics, and real‑world applications. This article covers all essential SL content while extending into HL concepts such as counting principles, complex numbers and proof by induction.
数与代数是IB数学的主干，将模式识别、金融建模、代数运算和抽象推理连接起来。无论是SL还是HL学生，掌握数列、指数、二项式展开和复数都能为微积分、统计和实际应用打下坚实基础。本文涵盖所有SL核心内容，并延伸到HL的计数原理、复数和数学归纳法证明。
1. Arithmetic Sequences | 等差数列
difference is denoted by d. If the first term is a₁, the nth term is given by uₙ = a₁ + (n – e sequence helps in modelling uniform change, from weekly savings to construction spacing.
等差数列是一列相邻两项之差恒定的数，这个公差记作d。若首项为a₁，则第n项通项公式为uₙ = a₁ + (n – 1)d。认识到数列的线性特征有助于对均匀变化（如每周储蓄或建筑间距）进行建模。
, simply plug n into the formula. For example, the 20th term of the sequence 5, 8, 11, 14, … has a₁ = 5, d = 3, giving u₂₀ = 5 + 19·3 = 62. Always double‑check that the sequence is arithmetic by testing a few differences.
要求出任一项，只需将n代入公式。例如，数列5, 8, 11, 14, …的第20项有a₁ = 5, d = 3，得u₂₀ = 5 + 19·3 = 62。务必先通过检验相邻项差来确认数列是否为等差数列。
2. Arithmetic Series | 等差级数
sum Sₙ can be computed using the pairing method: Sₙ = n/2 (a₁ + aₙ) or Sₙ = n/2 [2a₁ + (n – is handy when the last term is known, the second when only the common difference is given.
等差数列前n项之和称为等差级数。和Sₙ可用配对法计算：Sₙ = n/2 (a₁ + aₙ) 或 Sₙ = n/2 [2a₁ + (n – 1)d]。两种形式都很有用：已知末项时用前者，仅知公差时用后者。
Sₙ = n/2 [2a₁ + (n – 1)d]
For instance, the sum of the first 40 odd numbers (1, 3, 5, … relationship appears frequently in IB exam questions, often combined with sigma notation.
例如，前40个奇数之和（1, 3, 5, …）有a₁ = 1, d = 2, n = 40，因此S₄₀ = 40/2 [2·1 + 39·2] = 20 × 80 = 1600。这一关系常出现在IB试题中，常与求和符号∑结合。
3. Geometric Sequences | 等比数列
ⁿ⁻¹. Geometric growth models explosions or decay, such as population increase or radioactive half‑life. The value of r determines behaviour: |r| > 1 leads to growth, 0 < r < 1 leads to decay, and r < 0 causes alternating signs.
等比数列相邻项之比r恒定。首项为u₁时，第n项通项为uₙ = u₁ rⁿ⁻¹。几何增长可模拟激增或衰减，如人口增长或放射性半衰期。r的取值决定行为：|r| > 1为增长，0 < r < 1为衰减，r < 0则出现正负交替。
₁ r² = 12 and u₅ = u₁ r⁴ = 48. Dividing gives r² = 4, so r = ±2. Then u₁ can be recovered.
求缺失项时，可用公式建立方程。若数列第3项为12、第5项为48，则有u₃ = u₁ r² = 12，u₅ = u₁ r⁴ = 48。两式相除得r² = 4，故r = ±2，进而可求出u₁。
4. Geometric Series and Sum to Infinity | 等比级数与无穷和
The sum of the first n terms of a geometric sequence is Sₙ = u₁ (1 – rⁿ)/(1 – r) for r ≠ 1. When |r| < 1, the term rⁿ approaches 0 as n grows, so the sum to infinity exists: S∞ = u₁/(1 – ula underpins recurring decimals, present value of perpetuities, and fractal measurements.
等比数列前n项之和为 Sₙ = u₁ (1 – rⁿ)/(1 – r) （r ≠ 1）。当 |r| < 1 时，随着n增大，rⁿ趋于0，因此无穷和存在：S∞ = u₁/(1 – r)。该公式是循环小数、永续年金现值以及分形度量等的基础。
For example, the infinite geometric series 8 + 4 + 2 + 1 + … has u₁ = 8, r = ½, so S∞ = 8/(1 – e range of r for convergence or to convert recurring decimals to fractions using this sum.
例如，无穷等比级数8 + 4 + 2 + 1 + …有u₁ = 8, r = ½，故S∞ = 8/(1 – ½) = 16。IB考题常要求找出使级数收敛的r的范围，或利用此和将循环小数化为分数。
5. Financial Applications | 金融数学应用
and compounding periods per year k, the future value after t years is FV = PV × (1 + r/k)kt. This is a geometric sequence in discrete time. Regular savings plans generate geometric series, making the sum formulas essential.
数列与级数直接用于复利、贷款摊还和年金估值建模。设年利率为r，每年复利k次，则t年后的终值 FV = PV × (1 + r/k)kt，这是离散时间下的等比数列。定期储蓄计划则生成等比级数，因此求和公式至关重要。
eposits PMT, the total after n years is the sum of a geometric series: S = PMT × ((1 + i)ⁿ – 1)/i, where i is the interest rate per period.
在SL考试中，要能熟练运用复利公式并通过对数变形求解时间或利率。对于每年存入PMT的储蓄计划，n年后的总额为等比级数之和：S = PMT × ((1 + i)ⁿ – 1)/i，其中i为每期利率。
6. Exponents and Logarithms | 指数与对数
Exponential expressions aˣ 718 creates the natural logarithm ln x = logₑ x, ubiquitous in calculus and growth models.
指数表达式aˣ和对数函数logₐ x互逆。基本法则包括：aᵐ × aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, logₐ (xy) = logₐ x + logₐ y, 以及 logₐ (xⁿ) = n logₐ x。自然底数 e ≈ 2.718 衍生出自然对数 ln x = logₑ x，在微积分和增长模型中广泛使用。
lving exponential equations, take logs on both sides and apply the power rule. For example, to solve 3ˣ = 20, take ln: x ln 3 = ln 20, so x ≈ 2.7268.
换底公式logₐ b = log_c b / log_c a 常被考查，它可将任意对数转换为以10或e为底。求解指数方程时，两边取对数并运用幂法则。例如，解3ˣ = 20，取自然对数：x ln 3 = ln 20，得 x ≈ 2.7268。
7. Binomial Theorem (SL) | 二项式定理（SL）
For positive integer n, the expansion of (a + b)ⁿ is Σₖ₌₀ⁿ C(n,k) aⁿ⁻ᵏ bᵏ, where C(n,k) is the binomial coefficient, often written as ⁿCₖ or n!/(k!(n–k)!). The coefficients follow Pascal's triangle. In SL, n is limited to small positive integers, and the focus is on finding specific terms without full expansion.
对于正整数n，(a + b)ⁿ的展开式为 Σₖ₌₀ⁿ C(n,k) aⁿ⁻ᵏ bᵏ，其中C(n,k)为二项式系数，常写作 ⁿCₖ 或 n!/(k!(n–k)!)。系数遵循杨辉三角（帕斯卡三角）。在SL中，n限于较小的正整数，重点是不完全展开而求指定项。
To find the term containing xʳ in (2x – 3)⁵, set the general term C(5,k)·(2x)⁵⁻ᵏ·(–3)ᵏ and equate the exponent of x to r. Always simplify coefficients carefully and be mindful of signs arising from negative terms.
求(2x – 3)⁵中含xʳ的项，可设通项 C(5,k)·(2x)⁵⁻ᵏ·(–3)ᵏ 并令x的指数等于r。仔细化简系数，并注意负项带来的符号变化。
8. Scientific Notation and Error Estimation | 科学记数法与误差估算
s an integer. It simplifies handling very large or very small measurements, such as Planck' mal and scientific notation, and operations like multiplication and division in this form.
科学记数法将数字表示为 a × 10ᵏ，其中1 ≤ a < 10，k为整数。它简化了极大或极小量的处理，例如普朗克常数6.63 × 10⁻³⁴ J·s。IB要求能在十进制与科学记数法之间转换，并进行乘除运算。
length is (5.0 ± 0.2) cm, the absolute error is 0.2 cm, percentage error = (0.2/5.0) × 100% = 4%. In calculations, errors propagate: when multiplying, relative errors add.
在合并测量值时，绝对误差和百分误差用于评估精度。若长度为(5.0 ± 0.2) cm，绝对误差为0.2 cm，百分误差 = (0.2/5.0) × 100% = 4%。计算时误差会传播：乘法中相对误差相加。
9. Counting Principles, Permutations and Combinations (HL) | 计数原理、排列与组合（HL）
rn arrangements where order matters: the number of ways to arrange n distinct objects is n!; to arrange r from n is P(n,r) = n!/(n–r)!.
基本计数原理指出，若一事件有m种发生方式，另一独立事件有n种，则总结果数为 m × n。排列关注顺序重要的情形：n个不同物体的排列数为 n!；从n个中选r个排列为 P(n,r) = n!/(n–r)!。
Combinations ignore order: choosing r objects from n yields C(n,r) = n!/(r!(n–r)! ions with identical items or circular arrangements, and combinations with constraints like 'at least one'.
组合不考虑顺序：从n个中选r个的方式数为 C(n,r) = n!/(r!(n–r)!)。它们出现在二项式系数和概率中。IB HL题目常将排列与相同物体或环形排列，以及将组合与“至少”等约束条件混合考查。
10. Introduction to Complex Numbers (HL) | 复数入门（HL）
Complex numbers extend the real number system by defining i² = – real numbers. The real part is Re(z) = a, the imaginary part Im(z) = b. The conjugate is z̅ = a – bi, and the modulus is |z| = √a² + b². These tools allow solving equations like x² + 4 = 0.
复数通过定义 i² = –1 扩展了实数系。复数形式为 z = a + bi，其中a和b为实数。实部Re(z) = a，虚部Im(z) = b。共轭为 z̅ = a – bi，模为 |z| = √a² + b²。这些工具可解如 x² + 4 = 0 的方程。
addition by combining real and imaginary parts; multiplication by expanding and using i² = – am represents complex numbers as points or vectors, linking algebraic and geometric views.
运算简单明了：加法合并实部与虚部；乘法展开并利用 i² = –1。除法通过将分子分母同乘以分母的共轭来实现。阿根图将复数表示为点或向量，连接代数与几何视角。
11. Proof by Mathematical Induction (HL) | 数学归纳法（HL）
Mathematical induction proves statements P(n) for all positive integers n. It involves a base step (verify P(1) is true) and an inductive step (assume P(k) true and prove P(k+1) true). IB HL questions commonly ask for induction proofs of series sums, divisibility, and matrix or complex number properties.
数学归纳法用于证明对所有正整数n成立的命题P(n)。它包括基础步骤（验证P(1)为真）和归纳步骤（假设P(k)为真，证明P(k+1)为真）。IB HL常考查级数求和、整除性以及矩阵或复数性质的归纳证明。
des and simplify to (k+1)(k+2)/2, completing the induction. Always write a clear conclusion: “Since the base case is true and the inductive step holds, P(n) is true for all n ∈ ℤ⁺.”
例如，证明 Σᵢ₌₁ⁿ i = n(n+1)/2，先验证n=1成立，再假设 Σᵢ₌₁ᵏ i = k(k+1)/2。两边加上(k+1)并化简为(k+1)(k+2)/2，即完成归纳。务必写出清晰结论：“由于基础情形为真且归纳步骤成立，因此对所有n ∈ ℤ⁺，P(n)为真。”
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