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Unit A5 · Number & AlgebraUnit A5 · 数与代数

Proof & Algebraic Manipulation证明与代数运算

IB-Style Practice Questions — Paper 1A · Paper 1B · Paper 2 · Paper 3IB 风格练习题 —— 第一卷 A 节 · 第一卷 B 节 · 第二卷 · 第三卷

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus 1.6, 1.11, 1.15, 1.16考纲 1.6、1.11、1.15、1.16AA HL

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v1 · 8 Qs · 63 marks · Papers 1A/1B/2/3 · EASY–HARD · Topics 1.6, 1.11, 1.15, 1.16 · solutions: Solutions/Unit_A5_Proof_and_Algebraic_Manipulation_Solutions.html8 题 · 63 分 · 第一卷 A/B 节 & 第二/三卷 · 简单–较难 · 考点 1.6、1.11、1.15、1.16 · 解析：Solutions/Unit_A5_Proof_and_Algebraic_Manipulation_Solutions.html

PART I  ·  PAPER 1 SECTION A第一部分  ·  第一卷 A 节No calculator · short response · 21 marks不可使用计算器 · 简答题 · 21 分

Section A — Short ResponseA 节 —— 简答题

Write full proofs. Each step must follow from a stated law, definition, or hypothesis. Mark every assumption with "assume", every conclusion with "therefore"/"hence", and label the inductive hypothesis explicitly. No calculator permitted.写出完整证明。每一步都要给出所依据的法则、定义或假设。"假设"用 "assume"、"故"用 "therefore" / "hence" 明示；归纳假设必须显式标出。不可使用计算器。

Q1EASY Paper 1A 1.6 Direct Proof [4 marks]

Prove that the sum of any three consecutive integers is divisible by $3$.证明：任意三个连续整数之和能被 $3$ 整除。

(a) Write a general expression for three consecutive integers in terms of a single integer $n$.用单个整数 $n$ 写出三个连续整数的一般表达式。 [1]
(b) Compute their sum and factor out $3$.求和并把 $3$ 提出。 [2]
(c) State the conclusion in the form "$\therefore$ for all $n \in \mathbb{Z}$, the sum is divisible by $3$."以 "$\therefore$ 对所有 $n \in \mathbb{Z}$，该和能被 $3$ 整除" 的形式写出结论。 [1]
Q2MEDIUM Paper 1A 1.6 Proof by Contradiction [5 marks]

Prove by contradiction that $\sqrt{3}$ is irrational. You may use without proof that if $3 \mid k^{2}$ then $3 \mid k$, for $k \in \mathbb{Z}$.用反证法证明 $\sqrt{3}$ 是无理数。可不证地使用：若 $3 \mid k^{2}$ 则 $3 \mid k$（$k \in \mathbb{Z}$）。

(a) State the negation of "$\sqrt{3}$ is irrational" in the form "$\sqrt{3} = p/q$ for some integers $p, q$" — and state any condition you impose on $p, q$.写出 "$\sqrt{3}$ 是无理数" 的否定，形式为 "$\sqrt{3} = p/q$，$p, q$ 为整数"；并写出对 $p, q$ 施加的条件。 [1]
(b) Square, multiply through, and use the given fact to deduce that $3 \mid p$ and $3 \mid q$.两边平方、整理，并用上述事实推出 $3 \mid p$ 与 $3 \mid q$。 [3]
(c) Show how this contradicts the condition stated in (a) and conclude.说明该结论如何与 (a) 中的条件矛盾，从而得出结论。 [1]
Q3MEDIUM Paper 1A 1.15 Induction — Sum Identity (HL) [6 marks]

Prove by mathematical induction that, for all $n \in \mathbb{Z}^{+}$,用数学归纳法证明：对所有 $n \in \mathbb{Z}^{+}$，

$$ \sum_{r=1}^{n} r(r+1) \;=\; \frac{n(n+1)(n+2)}{3}. $$
(a) Verify the base case $n = 1$.验证基底 $n = 1$。 [1]
(b) State the inductive hypothesis explicitly.显式写出归纳假设。 [1]
(c) Carry out the inductive step: starting from $\sum_{r=1}^{k+1} r(r+1) = \left[\sum_{r=1}^{k} r(r+1)\right] + (k+1)(k+2)$, simplify to the required form.完成归纳步骤：从 $\sum_{r=1}^{k+1} r(r+1) = \left[\sum_{r=1}^{k} r(r+1)\right] + (k+1)(k+2)$ 出发，化简至目标形式。 [3]
(d) State the conclusion citing the principle of mathematical induction.引用数学归纳原理写出结论。 [1]
Q4HARD Paper 1A 1.15 Induction — Divisibility (HL) [6 marks]

Let $P(n)$ be the statement: $3 \mid \bigl(5^{n} + 2 \cdot 11^{n}\bigr)$ for $n \in \mathbb{Z}^{+}$.设命题 $P(n)$：$3 \mid \bigl(5^{n} + 2 \cdot 11^{n}\bigr)$，$n \in \mathbb{Z}^{+}$。

(a) Verify $P(1)$.验证 $P(1)$。 [1]
(b) Assume $P(k)$ holds, so $5^{k} + 2 \cdot 11^{k} = 3m$ for some $m \in \mathbb{Z}$. Express $5^{k+1} + 2 \cdot 11^{k+1}$ in a form that makes the factor of $3$ explicit. (Hint: $11 = 5 + 6$, or use $5(5^{k} + 2 \cdot 11^{k}) + 12 \cdot 11^{k}$.)假设 $P(k)$ 成立，即 $5^{k} + 2 \cdot 11^{k} = 3m$（$m \in \mathbb{Z}$）。将 $5^{k+1} + 2 \cdot 11^{k+1}$ 改写为能显式提出 $3$ 的形式。（提示：$11 = 5 + 6$，或用 $5(5^{k} + 2 \cdot 11^{k}) + 12 \cdot 11^{k}$。） [4]
(c) State the conclusion.写出结论。 [1]
PART II  ·  PAPER 1 SECTION B第二部分  ·  第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分

Section B — Extended ResponseB 节 —— 长答题

Show every algebraic step. In partial-fraction setups, clear denominators and equate coefficients (or substitute strategic values) — both methods must be shown until the unknowns are found. Telescoping sums must list at least the first two and last two terms explicitly.写出每一步代数运算。部分分式分解须先去分母，再比较系数（或代入策略性值）—— 两种方法都要写直到求出待定系数。望远镜（telescoping）求和必须显式列出至少前两项和后两项。

Q5HARD Paper 1B 1.11 Partial Fractions + Telescoping (HL) [11 marks]

Consider the rational expression $\dfrac{1}{r(r+1)(r+2)}$ for $r \in \mathbb{Z}^{+}$.考虑有理式 $\dfrac{1}{r(r+1)(r+2)}$，$r \in \mathbb{Z}^{+}$。

(a) Find constants $A, B, C$ such that $\dfrac{1}{r(r+1)(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+1} + \dfrac{C}{r+2}.$求常数 $A, B, C$ 使得 $\dfrac{1}{r(r+1)(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+1} + \dfrac{C}{r+2}$。 [4]
(b) Hence show that $\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2}\left[\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right]$.由此证明 $\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2}\left[\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right]$。 [2]
(c) Using (b), find a closed form for $S_{n} = \displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)(r+2)}$.利用 (b) 求 $S_{n} = \displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)(r+2)}$ 的闭式。 [3]
(d) Deduce $\displaystyle\lim_{n \to \infty} S_{n}$.求 $\displaystyle\lim_{n \to \infty} S_{n}$。 [2]
PART III  ·  PAPER 2第三部分  ·  第二卷Calculator · mixed response · 16 marks可使用计算器 · 混合题型 · 16 分

Paper 2 — Calculator Permitted第二卷 —— 允许使用计算器

A graphing calculator is required. You may use the calculator's rref command directly in Paper 2, but show the augmented matrix you fed it and interpret the row-echelon output in words ("unique" / "no" / "infinite" with parameter count). Give exact answers where reasonable.需要图形计算器（GDC）。第二卷允许直接调用 rref 命令，但必须写出输入的增广矩阵，并用文字解释行阶梯输出（"唯一" / "无解" / "无穷多（参数数量）"）。能精确求解的题目就给精确答案。

Q6MEDIUM Paper 2 1.16 $3 \times 3$ System (HL) [7 marks]

Solve the system求解下列方程组

$$ \begin{cases} \;\;x + \;\;y + \;\;z = 6 \\ 2x - \;\;y + \;\;z = 3 \\ \;\;x + 2y - \;\;z = 2 \end{cases} $$
(a) Write the augmented matrix.写出增广矩阵。 [1]
(b) Solve the system. State $x$, $y$, $z$ as integers.求解。给出 $x$、$y$、$z$ 的整数值。 [4]
(c) Verify your solution by substituting back into all three equations.将解代入三个方程进行验证。 [2]
Q7HARD Paper 2 1.16 Parametric $3 \times 3$ — Consistency (HL) [9 marks]

Consider the system in $x, y, z$ depending on the real parameter $k$:考虑依赖实参数 $k$ 的方程组（未知数为 $x, y, z$）：

$$ \begin{cases} \;\;x + 2y + 3z = 4 \\ 2x - \;\;y + \;\;z = 3 \\ 3x + \;\;y + 4z = k \end{cases} $$
(a) Show that the LHS of the third equation equals the sum of the LHSs of the first two.证明：第三个方程的左侧等于前两个方程左侧之和。 [2]
(b) Hence determine the value of $k$ for which the system is consistent, and explain why no other value of $k$ gives a solution.由此求使方程组相容的 $k$ 值，并解释为何其他 $k$ 值无解。 [3]
(c) For the value of $k$ found in (b), parameterise the solution set by setting $z = t$ ($t \in \mathbb{R}$) and expressing $x, y$ in terms of $t$.对 (b) 中求得的 $k$，令 $z = t$（$t \in \mathbb{R}$），将 $x, y$ 用 $t$ 表示，给出参数化解集。 [3]
(d) State the geometric meaning of the solution set in $\mathbb{R}^{3}$.用一句话说明该解集在 $\mathbb{R}^{3}$ 中的几何意义。 [1]
PART IV  ·  PAPER 3第四部分  ·  第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 — HL Extended Problem第三卷 —— HL 长题探究

A graphing calculator is required. Method marks dominate. For strong induction, you must state the hypothesis "$P(j)$ holds for every $j$ with $1 \le j \le k$" before the inductive step.需要图形计算器（GDC）。方法分占主导。对强归纳法（strong induction），归纳步骤之前必须显式写出假设："$P(j)$ 对 $1 \le j \le k$ 内所有 $j$ 成立"。

Q8HARD Paper 3 1.15 Fibonacci — Strong Induction + Binet (HL) [15 marks]

The Fibonacci sequence $(F_{n})$ is defined by $F_{1} = F_{2} = 1$ and $F_{n} = F_{n-1} + F_{n-2}$ for $n \ge 3$.斐波那契数列 $(F_{n})$ 定义为 $F_{1} = F_{2} = 1$，$F_{n} = F_{n-1} + F_{n-2}$（$n \ge 3$）。

(a) Compute $F_{3}, F_{4}, F_{5}, F_{6}, F_{7}$.计算 $F_{3}, F_{4}, F_{5}, F_{6}, F_{7}$。 [2]
(b) Prove by strong induction that $F_{n} < \left(\dfrac{7}{4}\right)^{n}$ for all $n \in \mathbb{Z}^{+}$. (You will need to verify two base cases and use the inequality $\dfrac{7}{4} + 1 \le \left(\dfrac{7}{4}\right)^{2}$ in the inductive step.)用强归纳法证明：对所有 $n \in \mathbb{Z}^{+}$，$F_{n} < \left(\dfrac{7}{4}\right)^{n}$。（需验证两个基底，并在归纳步骤中使用 $\dfrac{7}{4} + 1 \le \left(\dfrac{7}{4}\right)^{2}$。） [6]
(c) Prove by induction that $\displaystyle\sum_{r=1}^{n} F_{r} = F_{n+2} - 1$ for all $n \in \mathbb{Z}^{+}$.用归纳法证明：对所有 $n \in \mathbb{Z}^{+}$，$\displaystyle\sum_{r=1}^{n} F_{r} = F_{n+2} - 1$。 [4]
(d) Let $\varphi = \dfrac{1 + \sqrt{5}}{2}$ and $\psi = \dfrac{1 - \sqrt{5}}{2}$. Binet's formula states $F_{n} = \dfrac{\varphi^{n} - \psi^{n}}{\sqrt{5}}$. Verify Binet's formula gives $F_{1} = F_{2} = 1$ by direct calculation. (You may use $\varphi - \psi = \sqrt{5}$ and $\varphi \psi = -1$.)设 $\varphi = \dfrac{1 + \sqrt{5}}{2}$、$\psi = \dfrac{1 - \sqrt{5}}{2}$。Binet 公式（Binet's formula）给出 $F_{n} = \dfrac{\varphi^{n} - \psi^{n}}{\sqrt{5}}$。通过直接计算验证 Binet 公式给出 $F_{1} = F_{2} = 1$。（可使用 $\varphi - \psi = \sqrt{5}$ 与 $\varphi \psi = -1$。） [3]