Year 12 CIE Further Mathematics: Quick-Reference Formula & Theorem Handbook | CIE 进阶数学公式定理速查手册

📚 Year 12 CIE Further Mathematics: Quick-Reference Formula & Theorem Handbook | CIE 进阶数学公式定理速查手册

This quick-reference handbook collects the formulas, identities and theorems that carry the most marks in the Year 12 CIE Further Mathematics course (AS Level, 9231). It is written for fast revision of Further Pure Mathematics 1: roots of polynomial equations and symmetric functions, summation of series, matrices and determinants, complex numbers, polar coordinates, vectors, proof by induction and numerical methods. Every result is stated in exact symbolic form with a short note on the condition under which it may be used, so you can revise a whole topic in a few minutes before a mock or a timed paper.

本速查手册汇总了 CIE 进阶数学（AS Level，9231）Year 12 阶段最常考、分值最集中的公式、恒等式与定理，面向进阶纯数学 1 的快速复习：多项式方程的根与对称函数、级数求和、矩阵与行列式、复数、极坐标、向量、数学归纳法与数值方法。每条结论都以精确的符号形式给出，并附上使用条件的中英双语说明，方便你在模考或限时训练前几分钟内完成一个专题的回顾。

1. Roots of Polynomial Equations | 多项式方程的根与系数

⁻¹ + ... + a₁x + a₀ = 0, whose roots are α₁, α₂, ..., αₙ, the coefficient relations (Vieta' as) give the sum of roots Σαᵢ = −aₙ₋₁ / aₙ, the sum of products of two distinct roots Σαᵢαⱼ = aₙ₋₂ / aₙ, and the sum of triple products Σαᵢαⱼ ime, and the pattern closes with the product of all the roots, which equals (−1)ⁿ a₀ / aₙ.

对于多项式方程 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ = 0，其根记为 α₁, α₂, ..., αₙ，系数关系（韦达定理）给出：所有根之和 Σα = −aₙ₋₁ / aₙ，所有两根乘积之和 Σαβ = aₙ₋₂ / aₙ，所有三根乘积之和 Σαβγ = −aₙ₋₃ / aₙ。每多取一根相乘，符号交替变化；最终以所有根的乘积收尾，等于 (−1)ⁿ a₀ / aₙ。

lations are the ones used most often in algebra and in later work on series and complex numbers:

对于二次方程 ax² + bx + c = 0，以下两个关系式使用频率最高，后续级数与复数题目中也经常调用：

α + β = −b / a     αβ = c / a

d = 0 there are three relations, and every cubic question in the paper is built from them:

对于三次方程 ax³ + bx² + cx + d = 0，共有三个关系式，试卷中的三次方程题目几乎都围绕它们展开：

α + β + γ = −b / a    αβ + βγ + γα = c / a    αβγ = −d / a

may also be tackled by noticing that if the coefficients are real, complex roots occur in conjugate pairs.

对于四次方程 ax⁴ + bx³ + cx² + dx + e = 0，四个关系式为 Σα = −b / a，Σαβ = c / a，Σαβγ = −d / a，αβγδ = e / a。若系数均为实数，还可利用共轭复根成对出现这一性质来减少未知量。

A useful identity that converts a sum of squares into sums you already know is:

把平方和转化为已知和式的常用恒等式为：

Σα² = (Σα)² − 2Σαβ

³ through (Σα)³ = Σα³ + 3Σα²β + 6Σαβγ, so always reduce to the three basic symmetric sums.

对于三次方程，该式即 (−b / a)² − 2(c / a)，是考试中求 α² + β² + γ² 的最快路径。类似地，利用 (Σα)³ = Σα³ + 3Σα²β + 6Σαβγ 可以求 Σα³，核心思路始终是化归到三个基本对称和。

2. Symmetric Functions and Transformations of Roots | 对称函数与根的变换

e α² + β² = (α + β)² − 2αβ, α³ + β³ = (α + β)³ − 3αβ(α + β), 1/α + 1/β = (α + β) / αβ, and α² − β² = (α − β)(α + β), where (α − β)² = (α + β)² − 4αβ.

考试题目很少只问 α + β 与 αβ，而常问 α² + β²、α³ + β³、1/α + 1/β 或 (α − β)² 之类的组合。标准化简为：α² + β² = (α + β)² − 2αβ，α³ + β³ = (α + β)³ − 3αβ(α + β)，1/α + 1/β = (α + β) / αβ，α² − β² = (α − β)(α + β)，其中 (α − β)² = (α + β)² − 4αβ。

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When the question asks for the equation whose roots are a transformation of α, β, γ, either substitute directly or work with the new symmetric sums. The common substitutions are:

当题目要求写出以 α, β, γ 的某种变换为根的新方程时，可以直接代换，也可以用新的对称和来构造。常见代换如下：

• New root y = α + k: replace x by (y − k) in the original equation.
  新根 y = α + k：在原方程中用 x = y − k 代换。
• New root y = kα: replace x by y / k, then multiply through by kⁿ.
  新根 y = kα：用 x = y / k 代换，再整体乘以 kⁿ。
• New root y = 1/α: replace x by 1 / y and clear denominators.
  新根 y = 1/α：用 x = 1 / y 代换并通分整理。
• New root y = α²: put x = √y and eliminate the square root, then discard any extraneous roots introduced.
  新根 y = α²：令 x = √y 并消去根号，最后剔除增根。

bic builds the quadratic whose roots are α + β and γ, and dividing gives the required answer. This "sum-of-two-roots" trick appears in almost every series of past papers.

对于以 α, β, γ 为根的三次方程，处理 y = α + β 这类变换时可用 α + β = (Σα) − γ，即 y = S − γ，从而 γ = S − y。把 γ = S − y 代入原三次方程，即得一个以 α + β 和 γ 为根的方程，再做除法即可得到所求结果。这一"两根之和"技巧在历年真题中反复出现。

3. Summation of Series | 级数求和

The standard results for sums of powers of the first n natural numbers are the backbone of this topic, and they must be quoted exactly:

前 n 个自然数的幂次和标准结果是本专题的核心，必须准确引用：

Σᵣ₌₁ⁿ r = ½n(n + 1)   |   Σᵣ₌₁ⁿ r² = ⅙n(n + 1)(2n + 1)   |   Σᵣ₌₁ⁿ r³ = ¼n²(n + 1)²

c = nc, and every summation operator is linear, so a polynomial can be split term by term:

常数 c 连加 n 项之和为 nc，即 Σc = nc；求和运算具有线性性，因此多项式可以逐项拆分：

Σ (ar² + br + c) = a Σr² + b Σr + c Σ1

ecause all intermediate terms cancel in pairs. Partial fractions: expressions such as 1 / (r(r + 1)) are split as 1/r − 1/(r + 1) before summing, which usually produces a telescoping result plus one leftover term at each end.

还有两个必备技巧。其一为差分法：若通项可写成 f(r) − f(r + 1)，则 Σᵣ₌₁ⁿ (f(r) − f(r + 1)) = f(1) − f(n + 1)，中间项两两抵消。其二为部分分式：把 1 / (r(r + 1)) 这类表达式拆成 1/r − 1/(r + 1) 再求和，通常得到"首尾各剩若干项"的望远镜式结果。

4. Matrices: Operations and Determinants | 矩阵运算与行列式

ngular. For products, det(AB) = det A × det B, and in general matrix multiplication is not commutative, so AB ≠ BA.

对于 2×2 矩阵 A = [[a, b], [c, d]]，其行列式为 det A = ad − bc。若 det A = 0，矩阵为奇异矩阵，不存在逆矩阵；若 det A ≠ 0，则为非奇异矩阵。对乘积有 det(AB) = det A × det B；矩阵乘法一般不满足交换律，故 AB ≠ BA。

, and the entry in row i and column j of AB is the scalar product of row i of A with column j of B.

两个矩阵相乘的前提是前者的列数等于后者的行数。若 A 为 m×n，B 为 n×p，则 AB 为 m×p；AB 中第 i 行第 j 列的元素等于 A 的第 i 行与 B 的第 j 列的数量积。

nating signs, or by row operations that create zeros. Key facts worth memorising are collected below.

3×3 矩阵的行列式按任意一行或一列展开：每个元素乘以其余子式（2×2 行列式），符号交替；也可先用行变换制造零元素再展开。下表汇总了必须记住的要点。

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