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Unit A5 · SolutionsUnit A5 · 解析

Proof & Algebraic Manipulation — Solutions证明与代数运算 —— 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

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

PART I  ·  PAPER 1 SECTION A — SOLUTIONS第一部分  ·  第一卷 A 节 —— 解析No calculator · 21 marks不可使用计算器 · 21 分

Section A — Worked SolutionsA 节 —— 详细解析

Q1EASYPaper 1A1.6 Direct Proof[4 marks]

Prove 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。

Result:结论：  For all $n \in \mathbb{Z}$, $n + (n+1) + (n+2) = 3(n+1)$ — divisible by $3$.对所有 $n \in \mathbb{Z}$，$n + (n+1) + (n+2) = 3(n+1)$ —— 能被 $3$ 整除。

(a) General form A1

Let $n \in \mathbb{Z}$. Three consecutive integers are $n,\; n+1,\; n+2$.

(b) Sum and factor M1·A1

$$ n + (n+1) + (n+2) \;=\; 3n + 3 \;=\; 3(n+1). $$

(c) Conclusion R1

Since $n + 1 \in \mathbb{Z}$, the sum equals $3$ times an integer. $\therefore$ for all $n \in \mathbb{Z}$, the sum of three consecutive integers is divisible by $3$. $\blacksquare$
Why "let $n \in \mathbb{Z}$" matters. An IB direct proof loses marks if the universal quantifier is implicit. The proof works for every integer $n$, so the "let $n$ be any integer" line is what makes the algebra a proof rather than a single calculation. Likewise, the closing "$n+1 \in \mathbb{Z}$" step is the bridge from "equals $3 \times \text{something}$" to "divisible by $3$" — graders look for it.

(a) 一般形式 A1

设 $n \in \mathbb{Z}$。三个连续整数为 $n,\; n+1,\; n+2$。

(b) 求和并提因 M1·A1

$$ n + (n+1) + (n+2) \;=\; 3n + 3 \;=\; 3(n+1). $$

(c) 结论 R1

由 $n + 1 \in \mathbb{Z}$，该和等于 $3$ 的整数倍。$\therefore$ 对所有 $n \in \mathbb{Z}$，三个连续整数之和能被 $3$ 整除。$\blacksquare$
"设 $n \in \mathbb{Z}$" 为何关键。IB 直接证明若隐含全称量词会扣分。该证明对每个整数 $n$ 成立，故"设 $n$ 为任意整数"这一句正是将代数运算升级为"证明"的关键。同理，结尾"$n+1 \in \mathbb{Z}$"是从"等于 $3 \times$ 某物"过渡到"能被 $3$ 整除"的桥梁 —— 评卷人会专门找这一步。
Q2MEDIUMPaper 1A1.6 Proof by Contradiction[5 marks]

Prove $\sqrt{3}$ is irrational. Given: $3 \mid k^{2} \Rightarrow 3 \mid k$.证明 $\sqrt{3}$ 为无理数。给定：$3 \mid k^{2} \Rightarrow 3 \mid k$。

Result:结论：  $\sqrt{3} \notin \mathbb{Q}$.$\sqrt{3} \notin \mathbb{Q}$。

(a) Negation R1

Assume for contradiction $\sqrt{3} \in \mathbb{Q}$. Then $\sqrt{3} = \dfrac{p}{q}$ for some $p, q \in \mathbb{Z}$ with $q \ne 0$ and $\gcd(p, q) = 1$ (write the fraction in lowest terms).

(b) Derive $3 \mid p$ and $3 \mid q$ M1·A1·A1

Square both sides and clear the denominator: $$ 3 \;=\; \frac{p^{2}}{q^{2}} \;\Longrightarrow\; p^{2} \;=\; 3q^{2}. $$ So $3 \mid p^{2}$, hence by the given fact $3 \mid p$. Write $p = 3m$ with $m \in \mathbb{Z}$. Substitute: $$ (3m)^{2} \;=\; 3q^{2} \;\Longrightarrow\; 9m^{2} \;=\; 3q^{2} \;\Longrightarrow\; q^{2} \;=\; 3m^{2}. $$ So $3 \mid q^{2}$, hence $3 \mid q$.

(c) Contradiction R1

But then $3 \mid \gcd(p, q)$, contradicting $\gcd(p, q) = 1$. The assumption must be false, so $\sqrt{3} \notin \mathbb{Q}$. $\blacksquare$
Why "lowest terms" is load-bearing. Without insisting $\gcd(p, q) = 1$ at the start, "$3 \mid p$ and $3 \mid q$" is not a contradiction — every rational has infinitely many such representations. The lowest-terms condition is what turns "both divisible by $3$" into a logical impossibility. IB markers reliably drop the R1 if the condition isn't stated upfront.

(a) 否定 R1

反证假设 $\sqrt{3} \in \mathbb{Q}$。则存在 $p, q \in \mathbb{Z}$、$q \ne 0$、$\gcd(p, q) = 1$（即写为最简分数），使得 $\sqrt{3} = \dfrac{p}{q}$。

(b) 推出 $3 \mid p$ 与 $3 \mid q$ M1·A1·A1

两边平方并去分母： $$ 3 \;=\; \frac{p^{2}}{q^{2}} \;\Longrightarrow\; p^{2} \;=\; 3q^{2}. $$ 故 $3 \mid p^{2}$，由所给事实得 $3 \mid p$。设 $p = 3m$（$m \in \mathbb{Z}$），代入： $$ (3m)^{2} \;=\; 3q^{2} \;\Longrightarrow\; 9m^{2} \;=\; 3q^{2} \;\Longrightarrow\; q^{2} \;=\; 3m^{2}. $$ 故 $3 \mid q^{2}$，进而 $3 \mid q$。

(c) 矛盾 R1

但这意味着 $3 \mid \gcd(p, q)$，与 $\gcd(p, q) = 1$ 矛盾。假设不成立，故 $\sqrt{3} \notin \mathbb{Q}$。$\blacksquare$
"最简分数"为何不可省。若开头不强调 $\gcd(p, q) = 1$，"$3 \mid p$ 且 $3 \mid q$"并不构成矛盾 —— 每个有理数都有无穷多种这样的表示。最简条件正是把"都能被 $3$ 整除"提升为"逻辑不可能"的关键。IB 评卷人若没看到这一前提，几乎一定会扣掉 R1。
Q3MEDIUMPaper 1A1.15 Induction — Sum Identity (HL)[6 marks]

Prove by induction $\sum_{r=1}^{n} r(r+1) = \dfrac{n(n+1)(n+2)}{3}$.用归纳法证明 $\sum_{r=1}^{n} r(r+1) = \dfrac{n(n+1)(n+2)}{3}$。

Result:结论：  $P(n)$ holds for all $n \in \mathbb{Z}^{+}$ by induction.由归纳法，$P(n)$ 对所有 $n \in \mathbb{Z}^{+}$ 成立。

(a) Base case $n = 1$ A1

LHS $= 1 \cdot 2 = 2$. RHS $= \dfrac{1 \cdot 2 \cdot 3}{3} = 2$. LHS $=$ RHS, so $P(1)$ holds.

(b) Inductive hypothesis R1

Assume $P(k)$ holds for some fixed $k \in \mathbb{Z}^{+}$: $$ \sum_{r=1}^{k} r(r+1) \;=\; \frac{k(k+1)(k+2)}{3}. $$

(c) Inductive step — show $P(k+1)$ M1·A1·A1

\begin{aligned} \sum_{r=1}^{k+1} r(r+1) &= \left[\sum_{r=1}^{k} r(r+1)\right] + (k+1)(k+2) \\ &\stackrel{\text{IH}}{=} \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \\ &= (k+1)(k+2)\left[\frac{k}{3} + 1\right] \\ &= (k+1)(k+2)\cdot\frac{k+3}{3} \\ &= \frac{(k+1)(k+2)(k+3)}{3}. \end{aligned} This matches the formula with $n \to k+1$, so $P(k+1)$ holds.

(d) Conclusion R1

$P(1)$ holds and $P(k) \Rightarrow P(k+1)$. By the principle of mathematical induction, $P(n)$ holds for all $n \in \mathbb{Z}^{+}$. $\blacksquare$
The "factor before substitute" pattern. In the inductive step, factor $(k+1)(k+2)$ out before simplifying the bracket. Students who multiply everything out and then refactor lose time and risk arithmetic errors — IB problems are engineered so the common factor falls out naturally. Spot it, lift it, simplify the remaining linear bracket. This template ports verbatim to most "$\sum$ closed-form" induction problems.

(a) 基底 $n = 1$ A1

LHS $= 1 \cdot 2 = 2$。RHS $= \dfrac{1 \cdot 2 \cdot 3}{3} = 2$。LHS $=$ RHS，故 $P(1)$ 成立。

(b) 归纳假设 R1

对某固定 $k \in \mathbb{Z}^{+}$，假设 $P(k)$ 成立： $$ \sum_{r=1}^{k} r(r+1) \;=\; \frac{k(k+1)(k+2)}{3}. $$

(c) 归纳步骤 —— 证明 $P(k+1)$ M1·A1·A1

\begin{aligned} \sum_{r=1}^{k+1} r(r+1) &= \left[\sum_{r=1}^{k} r(r+1)\right] + (k+1)(k+2) \\ &\stackrel{\text{IH}}{=} \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \\ &= (k+1)(k+2)\left[\frac{k}{3} + 1\right] \\ &= (k+1)(k+2)\cdot\frac{k+3}{3} \\ &= \frac{(k+1)(k+2)(k+3)}{3}. \end{aligned} 与目标公式 $n \to k+1$ 一致，故 $P(k+1)$ 成立。

(d) 结论 R1

$P(1)$ 成立且 $P(k) \Rightarrow P(k+1)$。由数学归纳原理，$P(n)$ 对所有 $n \in \mathbb{Z}^{+}$ 成立。$\blacksquare$
"先提因再代入"。归纳步骤中应先把 $(k+1)(k+2)$ 提出，再化简方括号。把所有项展开后再重新因式分解既耗时又易出错 —— IB 题目就是设计成公因式自然出现。看到、提出、化简剩余的一次因式。该模板对绝大多数"$\sum$ 闭式"归纳题都适用。
Q4HARDPaper 1A1.15 Induction — Divisibility (HL)[6 marks]

$P(n)$: $3 \mid (5^{n} + 2 \cdot 11^{n})$.$P(n)$：$3 \mid (5^{n} + 2 \cdot 11^{n})$。

Result:结论：  $P(n)$ holds for all $n \in \mathbb{Z}^{+}$ by induction.由归纳法，$P(n)$ 对所有 $n \in \mathbb{Z}^{+}$ 成立。

(a) Base $n = 1$ A1

$5^{1} + 2 \cdot 11^{1} = 5 + 22 = 27 = 3 \cdot 9$. So $3 \mid 27$, and $P(1)$ holds.

(b) Inductive step R1·M1·A1·A1

Assume $P(k)$: $5^{k} + 2 \cdot 11^{k} = 3m$ for some $m \in \mathbb{Z}$. Then \begin{aligned} 5^{k+1} + 2 \cdot 11^{k+1} &= 5 \cdot 5^{k} + 22 \cdot 11^{k} \\ &= 5 \cdot 5^{k} + 5 \cdot 2 \cdot 11^{k} + 12 \cdot 11^{k} \\ &= 5\bigl(5^{k} + 2 \cdot 11^{k}\bigr) + 12 \cdot 11^{k} \\ &\stackrel{\text{IH}}{=} 5 \cdot 3m + 12 \cdot 11^{k} \\ &= 3\bigl(5m + 4 \cdot 11^{k}\bigr). \end{aligned} Since $5m + 4 \cdot 11^{k} \in \mathbb{Z}$, $3 \mid \bigl(5^{k+1} + 2 \cdot 11^{k+1}\bigr)$, so $P(k+1)$ holds.

(c) Conclusion R1

$P(1)$ holds and $P(k) \Rightarrow P(k+1)$. By induction, $P(n)$ holds for all $n \in \mathbb{Z}^{+}$. $\blacksquare$
The "force-the-IH-block" template. Every divisibility induction follows the same recipe: write the $(k+1)$-expression so the $P(k)$ block appears verbatim, then everything else must independently be divisible by the modulus. Here we wrote $22 \cdot 11^{k}$ as $5 \cdot 2 \cdot 11^{k} + 12 \cdot 11^{k}$ specifically to expose the $\bigl(5^{k} + 2 \cdot 11^{k}\bigr)$ chunk. The leftover $12 \cdot 11^{k} = 3 \cdot 4 \cdot 11^{k}$ then makes the divisibility manifest. Train this split-the-bigger-coefficient move — it cracks $90\%$ of HL divisibility inductions.

(a) 基底 $n = 1$ A1

$5^{1} + 2 \cdot 11^{1} = 5 + 22 = 27 = 3 \cdot 9$。故 $3 \mid 27$，$P(1)$ 成立。

(b) 归纳步骤 R1·M1·A1·A1

假设 $P(k)$：$5^{k} + 2 \cdot 11^{k} = 3m$（$m \in \mathbb{Z}$）。则 \begin{aligned} 5^{k+1} + 2 \cdot 11^{k+1} &= 5 \cdot 5^{k} + 22 \cdot 11^{k} \\ &= 5 \cdot 5^{k} + 5 \cdot 2 \cdot 11^{k} + 12 \cdot 11^{k} \\ &= 5\bigl(5^{k} + 2 \cdot 11^{k}\bigr) + 12 \cdot 11^{k} \\ &\stackrel{\text{IH}}{=} 5 \cdot 3m + 12 \cdot 11^{k} \\ &= 3\bigl(5m + 4 \cdot 11^{k}\bigr). \end{aligned} 由 $5m + 4 \cdot 11^{k} \in \mathbb{Z}$，$3 \mid \bigl(5^{k+1} + 2 \cdot 11^{k+1}\bigr)$，故 $P(k+1)$ 成立。

(c) 结论 R1

$P(1)$ 成立且 $P(k) \Rightarrow P(k+1)$。由归纳法，$P(n)$ 对所有 $n \in \mathbb{Z}^{+}$ 成立。$\blacksquare$
"强行凑出归纳块"模板。每道整除归纳题都遵循同一配方：把 $(k+1)$ 的表达式改写到能原封不动出现 $P(k)$ 块，然后剩余部分必须自身被模数整除。本题把 $22 \cdot 11^{k}$ 写成 $5 \cdot 2 \cdot 11^{k} + 12 \cdot 11^{k}$，正是为了暴露 $\bigl(5^{k} + 2 \cdot 11^{k}\bigr)$ 这一块。剩下的 $12 \cdot 11^{k} = 3 \cdot 4 \cdot 11^{k}$ 让整除性一目了然。把这种"拆大系数"的动作练成肌肉记忆 —— 它能解决 $90\%$ 的 HL 整除归纳题。
PART II  ·  PAPER 1 SECTION B — SOLUTIONS第二部分  ·  第一卷 B 节 —— 解析No calculator · 11 marks不可使用计算器 · 11 分

Section B — Worked SolutionsB 节 —— 详细解析

Q5HARDPaper 1B1.11 Partial Fractions + Telescoping (HL)[11 marks]

$\dfrac{1}{r(r+1)(r+2)}$ — partial fractions, telescoping sum, limit.$\dfrac{1}{r(r+1)(r+2)}$ —— 部分分式、望远镜求和、求极限。

Answers:答案：  (a) $A = \tfrac{1}{2},\; B = -1,\; C = \tfrac{1}{2}$  ·  (c) $S_{n} = \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)}$  ·  (d) $\displaystyle\lim_{n \to \infty} S_{n} = \dfrac{1}{4}$

(a) Partial fractions M1·A1·A1·A1

Clear denominators in $\dfrac{1}{r(r+1)(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+1} + \dfrac{C}{r+2}$: $$ 1 \;=\; A(r+1)(r+2) + B \cdot r(r+2) + C \cdot r(r+1). $$ Substitute strategic values:
  • $r = 0$: $\;1 = A \cdot 1 \cdot 2 \Rightarrow A = \tfrac{1}{2}$.
  • $r = -1$: $\;1 = B \cdot (-1)(1) \Rightarrow B = -1$.
  • $r = -2$: $\;1 = C \cdot (-2)(-1) \Rightarrow C = \tfrac{1}{2}$.
Therefore $\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2r} - \dfrac{1}{r+1} + \dfrac{1}{2(r+2)}.$

(b) Pair-form identity M1·A1

$$ \frac{1}{2}\left[\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right] \;=\; \frac{1}{2} \cdot \frac{(r+2) - r}{r(r+1)(r+2)} \;=\; \frac{1}{2} \cdot \frac{2}{r(r+1)(r+2)} \;=\; \frac{1}{r(r+1)(r+2)}. \;\checkmark $$

(c) Telescoping sum M1·A1·A1

Let $u_{r} = \dfrac{1}{r(r+1)}$. Part (b) gives $\dfrac{1}{r(r+1)(r+2)} = \tfrac{1}{2}(u_{r} - u_{r+1})$. So \begin{aligned} S_{n} &= \sum_{r=1}^{n} \tfrac{1}{2}(u_{r} - u_{r+1}) \\ &= \tfrac{1}{2}\bigl[(u_{1} - u_{2}) + (u_{2} - u_{3}) + \cdots + (u_{n} - u_{n+1})\bigr] \\ &= \tfrac{1}{2}\bigl(u_{1} - u_{n+1}\bigr) \\ &= \tfrac{1}{2}\left[\frac{1}{1 \cdot 2} - \frac{1}{(n+1)(n+2)}\right] \;=\; \frac{1}{4} - \frac{1}{2(n+1)(n+2)}. \end{aligned}

(d) Limit M1·A1

As $n \to \infty$, $\dfrac{1}{2(n+1)(n+2)} \to 0$, so $\displaystyle\lim_{n \to \infty} S_{n} = \dfrac{1}{4}.$
Pair-form > raw three-term form for telescoping. The raw partial fractions in (a) — $\tfrac{1}{2r} - \tfrac{1}{r+1} + \tfrac{1}{2(r+2)}$ — also telescopes, but with a three-term offset: each row has a positive $\tfrac{1}{2r}$ that cancels with two rows ahead, leaving four boundary terms instead of two. Part (b)'s pair-form identity $u_{r} - u_{r+1}$ is the canonical way to collapse a "rising-product" denominator $1/(r(r+1) \cdots (r+m))$ to a clean two-term telescope. This trick generalises: $\dfrac{1}{r(r+1)\cdots(r+m)} = \dfrac{1}{m}\bigl[\dfrac{1}{r(r+1)\cdots(r+m-1)} - \dfrac{1}{(r+1)(r+2)\cdots(r+m)}\bigr]$. Worth memorising for Paper 3.

(a) 部分分式 M1·A1·A1·A1

在 $\dfrac{1}{r(r+1)(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+1} + \dfrac{C}{r+2}$ 中去分母： $$ 1 \;=\; A(r+1)(r+2) + B \cdot r(r+2) + C \cdot r(r+1). $$ 代入策略性数值：
  • $r = 0$：$\;1 = A \cdot 1 \cdot 2 \Rightarrow A = \tfrac{1}{2}$。
  • $r = -1$：$\;1 = B \cdot (-1)(1) \Rightarrow B = -1$。
  • $r = -2$：$\;1 = C \cdot (-2)(-1) \Rightarrow C = \tfrac{1}{2}$。
故 $\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2r} - \dfrac{1}{r+1} + \dfrac{1}{2(r+2)}$。

(b) 成对形式恒等式 M1·A1

$$ \frac{1}{2}\left[\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right] \;=\; \frac{1}{2} \cdot \frac{(r+2) - r}{r(r+1)(r+2)} \;=\; \frac{1}{2} \cdot \frac{2}{r(r+1)(r+2)} \;=\; \frac{1}{r(r+1)(r+2)}. \;\checkmark $$

(c) 望远镜求和 M1·A1·A1

令 $u_{r} = \dfrac{1}{r(r+1)}$。由 (b)，$\dfrac{1}{r(r+1)(r+2)} = \tfrac{1}{2}(u_{r} - u_{r+1})$。故 \begin{aligned} S_{n} &= \sum_{r=1}^{n} \tfrac{1}{2}(u_{r} - u_{r+1}) \\ &= \tfrac{1}{2}\bigl[(u_{1} - u_{2}) + (u_{2} - u_{3}) + \cdots + (u_{n} - u_{n+1})\bigr] \\ &= \tfrac{1}{2}\bigl(u_{1} - u_{n+1}\bigr) \\ &= \tfrac{1}{2}\left[\frac{1}{1 \cdot 2} - \frac{1}{(n+1)(n+2)}\right] \;=\; \frac{1}{4} - \frac{1}{2(n+1)(n+2)}. \end{aligned}

(d) 极限 M1·A1

当 $n \to \infty$，$\dfrac{1}{2(n+1)(n+2)} \to 0$，故 $\displaystyle\lim_{n \to \infty} S_{n} = \dfrac{1}{4}$。
成对形式 > 三项原形（用于望远镜求和）。(a) 中的原始三项分解 $\tfrac{1}{2r} - \tfrac{1}{r+1} + \tfrac{1}{2(r+2)}$ 也能望远镜求和，但需要跨三行抵消：每行正项 $\tfrac{1}{2r}$ 要与往后两行抵消，最终剩四个边界项而非两个。(b) 的成对恒等式 $u_{r} - u_{r+1}$ 是把"升序乘积"分母 $1/(r(r+1) \cdots (r+m))$ 折叠为干净两项望远镜的标准做法。该恒等式可推广：$\dfrac{1}{r(r+1)\cdots(r+m)} = \dfrac{1}{m}\bigl[\dfrac{1}{r(r+1)\cdots(r+m-1)} - \dfrac{1}{(r+1)(r+2)\cdots(r+m)}\bigr]$。Paper 3 值得记下。
PART III  ·  PAPER 2 — SOLUTIONS第三部分  ·  第二卷 —— 解析Calculator · 16 marks可使用计算器 · 16 分

Paper 2 — Worked Solutions第二卷 —— 详细解析

Q6MEDIUMPaper 21.16 $3 \times 3$ System (HL)[7 marks]

Solve $x+y+z=6$, $2x-y+z=3$, $x+2y-z=2$.解 $x+y+z=6$、$2x-y+z=3$、$x+2y-z=2$。

Answer:答案：  $x = 1,\; y = 2,\; z = 3$

(a) Augmented matrix A1

$$ \left[\begin{array}{rrr|r} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right]. $$

(b) Solve M1·M1·A1·A1

Using the GDC's rref (or by hand: $R_{2} \to R_{2} - 2R_{1}$, $R_{3} \to R_{3} - R_{1}$, then $R_{3} \to R_{3} + \tfrac{1}{3}R_{2}$ etc.): $$ \text{rref}\;=\; \left[\begin{array}{rrr|r} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{array}\right]. $$ Therefore $x = 1$, $y = 2$, $z = 3$.

(c) Verify A1·A1

  • Eq 1: $1 + 2 + 3 = 6$ $\checkmark$
  • Eq 2: $2(1) - 2 + 3 = 3$ $\checkmark$
  • Eq 3: $1 + 2(2) - 3 = 2$ $\checkmark$
Always verify on Paper 2. The rref command is fast but a single mistyped coefficient gives a wrong "unique" answer with no warning. The verification step costs 30 seconds and catches data-entry errors that would otherwise lose all of (b). When the system has a unique solution, this check is mandatory — when it has no solution or infinite solutions, the diagnostic is different (an all-zero row or a $0 = \text{nonzero}$ row), and you describe that interpretation instead.

(a) 增广矩阵 A1

$$ \left[\begin{array}{rrr|r} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 2 \end{array}\right]. $$

(b) 求解 M1·M1·A1·A1

用 GDC 的 rref（或手算：$R_{2} \to R_{2} - 2R_{1}$、$R_{3} \to R_{3} - R_{1}$，再 $R_{3} \to R_{3} + \tfrac{1}{3}R_{2}$ 等）： $$ \text{rref}\;=\; \left[\begin{array}{rrr|r} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{array}\right]. $$ 故 $x = 1$、$y = 2$、$z = 3$。

(c) 验证 A1·A1

  • 方程 1：$1 + 2 + 3 = 6$ $\checkmark$
  • 方程 2：$2(1) - 2 + 3 = 3$ $\checkmark$
  • 方程 3：$1 + 2(2) - 3 = 2$ $\checkmark$
Paper 2 一律要验证。rref 命令很快，但一个系数输错就会给出错误的"唯一解"且无任何提示。验证只需 30 秒，却能挡住所有录入错误（否则 (b) 全失）。当方程组有唯一解时，验证是必做步骤；当无解或无穷多解时，诊断信号不同（一整行全 $0$ 或 $0 = (\text{nonzero})$），改为描述该解释。
Q7HARDPaper 21.16 Parametric $3 \times 3$ — Consistency (HL)[9 marks]

$x + 2y + 3z = 4$; $2x - y + z = 3$; $3x + y + 4z = k$. Consistency, solution set, geometry.$x + 2y + 3z = 4$；$2x - y + z = 3$；$3x + y + 4z = k$。相容性、解集、几何意义。

Answers:答案：  (b) $k = 7$  ·  (c) $(x, y, z) = (2 - t,\; 1 - t,\; t),\; t \in \mathbb{R}$  ·  (d) a line in $\mathbb{R}^{3}$.$\mathbb{R}^{3}$ 中的一条直线。

(a) Row dependency M1·A1

Add equation 1 to equation 2: $$ (x + 2y + 3z) + (2x - y + z) \;=\; 3x + y + 4z. $$ This is exactly the LHS of equation 3.

(b) Consistency requires $k = 7$ M1·A1·R1

Adding the RHSs of equations 1 and 2 gives $4 + 3 = 7$. So if any solution exists, equation 3's RHS must equal $7$, i.e. $k = 7$.
  • $k = 7$: equation 3 is the sum of equations 1 and 2 — redundant, so the system reduces to a $2 \times 3$ consistent system with $3 - 2 = 1$ free parameter, giving infinitely many solutions.
  • $k \ne 7$: equation 3 contradicts $(\text{eq 1}) + (\text{eq 2})$, yielding $0 = k - 7 \ne 0$ — no solution.

(c) Parameterise (with $k = 7$) M1·A1·A1

Drop the redundant equation 3 and set $z = t$:
  • From eq 1: $x + 2y = 4 - 3t$.
  • From eq 2: $2x - y = 3 - t$.
Solve this $2 \times 2$ system. From eq 2: $y = 2x - 3 + t$. Substitute into eq 1: $$ x + 2(2x - 3 + t) \;=\; 4 - 3t \;\Longrightarrow\; 5x = 10 - 5t \;\Longrightarrow\; x = 2 - t. $$ Then $y = 2(2 - t) - 3 + t = 1 - t$. So $(x, y, z) = (2 - t,\; 1 - t,\; t).$

(d) Geometric meaning A1

A line in $\mathbb{R}^{3}$ — specifically the line through the point $(2, 1, 0)$ with direction vector $(-1, -1, 1)$ (the coefficients of $t$).
Three planes, three geometries. A $3 \times 3$ system whose coefficient matrix has rank $r$ and augmented matrix has rank $r'$ falls into exactly four geometric cases. Rank $3$ everywhere: unique point (three planes meet at one point). Rank $2$ everywhere ($r = r' = 2$): line (three planes share a common line — what this problem produced for $k = 7$). Rank $1$ everywhere: plane (all three equations describe the same plane). $r \ne r'$: no solution (planes form a prism, or two are parallel). On Paper 2, after running rref, count the all-zero rows: $0$ zero rows $\to$ unique, $1$ zero row + consistent $\to$ line, $2$ zero rows $\to$ plane, $0 = \text{nonzero}$ anywhere $\to$ no solution.

(a) 行依赖 M1·A1

把方程 1 与方程 2 相加： $$ (x + 2y + 3z) + (2x - y + z) \;=\; 3x + y + 4z. $$ 正是方程 3 的左侧。

(b) 相容性要求 $k = 7$ M1·A1·R1

方程 1 与方程 2 右侧之和为 $4 + 3 = 7$。故若存在解，方程 3 的右侧必等于 $7$，即 $k = 7$。
  • $k = 7$：方程 3 等于方程 1 加方程 2 —— 冗余，系统降为相容的 $2 \times 3$，有 $3 - 2 = 1$ 个自由参数，故无穷多解。
  • $k \ne 7$：方程 3 与 $(\text{eq 1}) + (\text{eq 2})$ 矛盾，得 $0 = k - 7 \ne 0$ —— 无解。

(c) 参数化（取 $k = 7$）M1·A1·A1

去掉冗余的方程 3，并令 $z = t$：
  • 由方程 1：$x + 2y = 4 - 3t$。
  • 由方程 2：$2x - y = 3 - t$。
解此 $2 \times 2$ 系统。由方程 2：$y = 2x - 3 + t$。代入方程 1： $$ x + 2(2x - 3 + t) \;=\; 4 - 3t \;\Longrightarrow\; 5x = 10 - 5t \;\Longrightarrow\; x = 2 - t. $$ 则 $y = 2(2 - t) - 3 + t = 1 - t$。故 $(x, y, z) = (2 - t,\; 1 - t,\; t)$。

(d) 几何意义 A1

$\mathbb{R}^{3}$ 中的一条直线 —— 即过点 $(2, 1, 0)$、方向向量为 $(-1, -1, 1)$（即 $t$ 的系数）的直线。
三平面 · 三种几何。$3 \times 3$ 系统的系数矩阵秩 $r$ 与增广矩阵秩 $r'$ 决定恰好四种几何情形。处处秩 $3$：唯一点（三平面交于一点）。处处秩 $2$（$r = r' = 2$）：直线（三平面共一线 —— 本题 $k = 7$ 的情形）。处处秩 $1$：平面（三方程描述同一平面）。$r \ne r'$：无解（平面形成三棱柱，或其中两平行）。Paper 2 中跑完 rref 后，数全 $0$ 行数：$0$ 行 $\to$ 唯一；$1$ 行且相容 $\to$ 直线；$2$ 行 $\to$ 平面；出现 $0 = (\text{nonzero})$ $\to$ 无解。
PART IV  ·  PAPER 3 — SOLUTIONS第四部分  ·  第三卷 —— 解析Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 — Worked Solutions第三卷 —— 详细解析

Q8HARDPaper 31.15 Fibonacci — Strong Induction + Binet (HL)[15 marks]

$F_{1} = F_{2} = 1$, $F_{n} = F_{n-1} + F_{n-2}$. (a) Compute first terms. (b) Strong induction $F_{n} < (7/4)^{n}$. (c) Induction $\sum F_{r} = F_{n+2} - 1$. (d) Verify Binet.$F_{1} = F_{2} = 1$，$F_{n} = F_{n-1} + F_{n-2}$。(a) 求前几项；(b) 强归纳 $F_{n} < (7/4)^{n}$；(c) 归纳 $\sum F_{r} = F_{n+2} - 1$；(d) 验证 Binet 公式。

Answers:答案：  (a) $F_{3} = 2,\, F_{4} = 3,\, F_{5} = 5,\, F_{6} = 8,\, F_{7} = 13$

(a) First terms A1·A1

$F_{3} = F_{2} + F_{1} = 1 + 1 = 2$. $F_{4} = F_{3} + F_{2} = 2 + 1 = 3$. $F_{5} = F_{4} + F_{3} = 3 + 2 = 5$. $F_{6} = F_{5} + F_{4} = 5 + 3 = 8$. $F_{7} = F_{6} + F_{5} = 8 + 5 = 13$.

(b) Strong induction $F_{n} < (7/4)^{n}$ M1·A1·A1·M1·A1·R1

Two base cases (needed because the recurrence reaches back two steps).
  • $n = 1$: $F_{1} = 1 < \tfrac{7}{4}$. $\checkmark$
  • $n = 2$: $F_{2} = 1 < \bigl(\tfrac{7}{4}\bigr)^{2} = \tfrac{49}{16} \approx 3.06$. $\checkmark$
Inductive hypothesis. Assume $P(j)$ holds for every $j$ with $1 \le j \le k$ (where $k \ge 2$). That is, $F_{j} < (7/4)^{j}$ for $j = 1, \dots, k$. Inductive step — show $P(k+1)$: \begin{aligned} F_{k+1} &= F_{k} + F_{k-1} \\ &\stackrel{\text{IH}}{<} \bigl(\tfrac{7}{4}\bigr)^{k} + \bigl(\tfrac{7}{4}\bigr)^{k-1} \\ &= \bigl(\tfrac{7}{4}\bigr)^{k-1}\!\left[\tfrac{7}{4} + 1\right] \\ &= \bigl(\tfrac{7}{4}\bigr)^{k-1} \cdot \tfrac{11}{4}. \end{aligned} Need $\tfrac{11}{4} \le \bigl(\tfrac{7}{4}\bigr)^{2} = \tfrac{49}{16}$. Compute: $\tfrac{11}{4} = \tfrac{44}{16} < \tfrac{49}{16}$. $\checkmark$ Therefore $$ F_{k+1} \;<\; \bigl(\tfrac{7}{4}\bigr)^{k-1} \cdot \bigl(\tfrac{7}{4}\bigr)^{2} \;=\; \bigl(\tfrac{7}{4}\bigr)^{k+1}, $$ so $P(k+1)$ holds. By strong induction, $F_{n} < (7/4)^{n}$ for all $n \in \mathbb{Z}^{+}$. $\blacksquare$

(c) Sum identity $\sum_{r=1}^{n} F_{r} = F_{n+2} - 1$ A1·M1·A1·R1

Base $n = 1$: LHS $= F_{1} = 1$; RHS $= F_{3} - 1 = 2 - 1 = 1$. $\checkmark$ Assume $\sum_{r=1}^{k} F_{r} = F_{k+2} - 1$. Then \begin{aligned} \sum_{r=1}^{k+1} F_{r} &= \left[\sum_{r=1}^{k} F_{r}\right] + F_{k+1} \\ &\stackrel{\text{IH}}{=} (F_{k+2} - 1) + F_{k+1} \\ &= (F_{k+1} + F_{k+2}) - 1 \\ &\stackrel{\text{def}}{=} F_{k+3} - 1 \\ &= F_{(k+1)+2} - 1. \end{aligned} So $P(k+1)$ holds. By induction, the identity holds for all $n \in \mathbb{Z}^{+}$. $\blacksquare$

(d) Verify Binet at $n = 1, 2$ M1·A1·A1

First note that $\varphi$ and $\psi$ are the roots of $x^{2} - x - 1 = 0$, so by Vieta's formulas $\varphi + \psi = 1$ and $\varphi \psi = -1$ (the latter given; the former is read off directly from $\varphi + \psi = \tfrac{1 + \sqrt{5}}{2} + \tfrac{1 - \sqrt{5}}{2} = 1$). $n = 1$: $\dfrac{\varphi - \psi}{\sqrt{5}} = \dfrac{\sqrt{5}}{\sqrt{5}} = 1 = F_{1}$. $\checkmark$ $n = 2$: Factor the difference of squares. $$ \frac{\varphi^{2} - \psi^{2}}{\sqrt{5}} \;=\; \frac{(\varphi - \psi)(\varphi + \psi)}{\sqrt{5}} \;=\; \frac{\sqrt{5} \cdot 1}{\sqrt{5}} \;=\; 1 \;=\; F_{2}. \;\checkmark $$
Why strong induction needs two base cases. The Fibonacci recurrence $F_{n} = F_{n-1} + F_{n-2}$ reaches back two steps. If you only check $P(1)$, then proving $P(3)$ requires $P(2)$ — which you never verified. Two base cases are the minimum needed to "bootstrap" the recurrence into the inductive engine. More generally, an order-$d$ linear recurrence needs $d$ base cases for strong induction. The IB markscheme docks marks on Fibonacci-style proofs that verify only $P(1)$ then jump.

Why Binet's formula is "magic." The roots $\varphi, \psi$ of $x^{2} - x - 1 = 0$ are precisely the values for which $x^{n+1} = x^{n} + x^{n-1}$. Any linear combination $A\varphi^{n} + B\psi^{n}$ also satisfies the Fibonacci recurrence, so the closed form is unique up to the two initial conditions $F_{1} = F_{2} = 1$. Solving for $A, B$ gives $A = 1/\sqrt{5}$, $B = -1/\sqrt{5}$. The "$\sqrt{5}$ in the denominator" appears because that's the gap $\varphi - \psi$. This is the prototype for the entire theory of linear recurrences — Paper 3 explorations sometimes ask you to mimic this derivation for a different recurrence (Lucas numbers, tribonacci, etc.).

(a) 前几项 A1·A1

$F_{3} = F_{2} + F_{1} = 1 + 1 = 2$。$F_{4} = F_{3} + F_{2} = 2 + 1 = 3$。$F_{5} = F_{4} + F_{3} = 3 + 2 = 5$。$F_{6} = F_{5} + F_{4} = 5 + 3 = 8$。$F_{7} = F_{6} + F_{5} = 8 + 5 = 13$。

(b) 强归纳 $F_{n} < (7/4)^{n}$ M1·A1·A1·M1·A1·R1

双基底（因递推回溯两步）。
  • $n = 1$：$F_{1} = 1 < \tfrac{7}{4}$。$\checkmark$
  • $n = 2$：$F_{2} = 1 < \bigl(\tfrac{7}{4}\bigr)^{2} = \tfrac{49}{16} \approx 3.06$。$\checkmark$
归纳假设。对任意 $j$（$1 \le j \le k$，$k \ge 2$）假设 $P(j)$ 成立，即 $F_{j} < (7/4)^{j}$，$j = 1, \dots, k$。 归纳步骤 —— 证 $P(k+1)$： \begin{aligned} F_{k+1} &= F_{k} + F_{k-1} \\ &\stackrel{\text{IH}}{<} \bigl(\tfrac{7}{4}\bigr)^{k} + \bigl(\tfrac{7}{4}\bigr)^{k-1} \\ &= \bigl(\tfrac{7}{4}\bigr)^{k-1}\!\left[\tfrac{7}{4} + 1\right] \\ &= \bigl(\tfrac{7}{4}\bigr)^{k-1} \cdot \tfrac{11}{4}. \end{aligned} 需 $\tfrac{11}{4} \le \bigl(\tfrac{7}{4}\bigr)^{2} = \tfrac{49}{16}$。化为公分母：$\tfrac{11}{4} = \tfrac{44}{16} < \tfrac{49}{16}$。$\checkmark$ 故 $$ F_{k+1} \;<\; \bigl(\tfrac{7}{4}\bigr)^{k-1} \cdot \bigl(\tfrac{7}{4}\bigr)^{2} \;=\; \bigl(\tfrac{7}{4}\bigr)^{k+1}, $$ 即 $P(k+1)$ 成立。由强归纳，对所有 $n \in \mathbb{Z}^{+}$，$F_{n} < (7/4)^{n}$。$\blacksquare$

(c) 求和恒等式 $\sum_{r=1}^{n} F_{r} = F_{n+2} - 1$ A1·M1·A1·R1

基底 $n = 1$：LHS $= F_{1} = 1$；RHS $= F_{3} - 1 = 2 - 1 = 1$。$\checkmark$ 假设 $\sum_{r=1}^{k} F_{r} = F_{k+2} - 1$。则 \begin{aligned} \sum_{r=1}^{k+1} F_{r} &= \left[\sum_{r=1}^{k} F_{r}\right] + F_{k+1} \\ &\stackrel{\text{IH}}{=} (F_{k+2} - 1) + F_{k+1} \\ &= (F_{k+1} + F_{k+2}) - 1 \\ &\stackrel{\text{def}}{=} F_{k+3} - 1 \\ &= F_{(k+1)+2} - 1. \end{aligned} 故 $P(k+1)$ 成立。由归纳，恒等式对所有 $n \in \mathbb{Z}^{+}$ 成立。$\blacksquare$

(d) 在 $n = 1, 2$ 处验证 Binet M1·A1·A1

先注意 $\varphi$、$\psi$ 是 $x^{2} - x - 1 = 0$ 的根，由韦达定理 $\varphi + \psi = 1$、$\varphi \psi = -1$（后者已给；前者由 $\varphi + \psi = \tfrac{1 + \sqrt{5}}{2} + \tfrac{1 - \sqrt{5}}{2} = 1$ 直接读出）。 $n = 1$：$\dfrac{\varphi - \psi}{\sqrt{5}} = \dfrac{\sqrt{5}}{\sqrt{5}} = 1 = F_{1}$。$\checkmark$ $n = 2$：用平方差分解。 $$ \frac{\varphi^{2} - \psi^{2}}{\sqrt{5}} \;=\; \frac{(\varphi - \psi)(\varphi + \psi)}{\sqrt{5}} \;=\; \frac{\sqrt{5} \cdot 1}{\sqrt{5}} \;=\; 1 \;=\; F_{2}. \;\checkmark $$
为何强归纳需要两个基底。斐波那契递推 $F_{n} = F_{n-1} + F_{n-2}$ 回溯两步。若仅检验 $P(1)$，则证 $P(3)$ 需要 $P(2)$ —— 而 $P(2)$ 从未验证。两个基底是把递推"启动"为归纳机器所需的最小数量。一般地，$d$ 阶线性递推强归纳需 $d$ 个基底。IB 评分对只验 $P(1)$ 就跳的斐波那契式证明会扣分。

Binet 公式为何"神奇"。方程 $x^{2} - x - 1 = 0$ 的根 $\varphi, \psi$ 正是使 $x^{n+1} = x^{n} + x^{n-1}$ 成立的值。任何线性组合 $A\varphi^{n} + B\psi^{n}$ 也满足斐波那契递推，故闭式在两个初值 $F_{1} = F_{2} = 1$ 下唯一。解出 $A = 1/\sqrt{5}$、$B = -1/\sqrt{5}$。"$\sqrt{5}$ 出现在分母"是因为它就是 $\varphi - \psi$ 这个间距。这是整套线性递推理论的原型 —— Paper 3 探究题有时会让你对另一种递推（如 Lucas 数、tribonacci）复制这套推导。