### Exercise 03: Prove using mathematical induction #### 1. Prove that for all $n \in \mathbb{N}$, $\sum_{k=0}^{n-1} (2k+1) = n^2$. 1. **Base case** ($n=1$): $$\sum_{k=0}^{0} (2k+1) = 2\cdot0 + 1 = 1$$ Right-hand side (RHS): $$1^2 = 1$$ Base case holds. 2. **Induction hypothesis:** Assume the formula is true for some $n$: $$\sum_{k=0}^{n-1} (2k+1) = n^2$$ 3. **Inductive step:** Show it holds for $n+1$: $$\sum_{k=0}^{n} (2k+1) = \left(\sum_{k=0}^{n-1} (2k+1)\right) + (2n+1)$$ By induction hypothesis: $$= n^2 + (2n+1) = n^2 + 2n + 1 = (n+1)^2$$ Therefore, formula holds for all $n$ by induction. --- #### 2. Prove that $6^n -1$ is divisible by 5 for all $n \in \mathbb{N}$. 1. **Base case** ($n=1$): $$6^1 - 1 = 6 - 1 = 5$$ which is divisible by 5. 2. **Induction hypothesis:** Assume $6^n - 1$ is divisible by 5. 3. **Inductive step:** Consider: $$6^{n+1} - 1 = 6 \cdot 6^n - 1 = 6 (6^n) - 1$$ Rewrite as: $$= 6 (6^n - 1) + 6 - 1 = 6(6^n - 1) + 5$$ By hypothesis, $6^n - 1$ is divisible by 5, so $6(6^n - 1)$ is divisible by 5. Since 5 is divisible by 5, their sum is divisible by 5. Thus, true for $n+1$. --- #### 3. Prove that $ forall n \geq 7, n! > 3^2 = 9$ 1. For $n=7$, $$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$$ Clearly, $5040 > 9$. 2. For all $n > 7$, $n!$ grows even larger, so the inequality holds. Induction is trivial from 7 upwards. --- ### Exercise 04: Solve equations and inequalities #### 1) Solve $|2x - 1| = 3$ This means: $$2x - 1 = 3 \quad \text{or} \quad 2x - 1 = -3$$ Solving both: - $2x = 4 \Rightarrow x=2$ - $2x = -2 \Rightarrow x = -1$ Solutions: $x = 2, -1$ --- #### 2) Solve $|2x - 4| \leq |x + 2|$ Consider two cases: - Case 1: $x+2 \geq 0 \Rightarrow x \geq -2$ Here: $$|2x - 4| \leq x + 2$$ Solve subcases: - If $2x - 4 \geq 0 \Rightarrow x \geq 2$, then $2x -4 \leq x + 2 \Rightarrow x \leq 6$ So for $x \geq 2$, $2 \leq x \leq 6$ - If $2x - 4 < 0 \Rightarrow x < 2$, then $-(2x -4) \leq x + 2 \Rightarrow -2x +4 \leq x + 2 \Rightarrow 4 - 2 \leq x + 2x \Rightarrow 2 \leq 3x \Rightarrow x \geq \frac{2}{3}$ Combined with $x \geq -2$ and $x < 2$, valid $\frac{2}{3} \leq x < 2$ - Case 2: $x + 2 < 0 \Rightarrow x < -2$ Then: $$|2x - 4| \leq -(x + 2) = -x - 2$$ - Subcase 1: If $2x -4 \geq 0 \Rightarrow x \geq 2$, conflict with $x < -2$ discard. - Subcase 2: If $2x -4 < 0$, then $-(2x -4) \leq -x -2 \Rightarrow -2x + 4 \leq -x - 2 \Rightarrow 4 + 2 \leq - x + 2x \Rightarrow 6 \leq x$ Conflict with $x < -2$, no solution here. **Final solution:** $$\boxed{\left[ \frac{2}{3}, 6 \right]}$$ --- #### 3) Solve $\left| \frac{1}{x} - 2 \right| \leq 3$ Rewrite: $$-3 \leq \frac{1}{x} - 2 \leq 3$$ Add 2 to each part: $$-1 \leq \frac{1}{x} \leq 5$$ Consider $\frac{1}{x}$: - $\frac{1}{x} \geq -1$ and $\frac{1}{x} \leq 5$ Split into intervals based on sign of $x$: - If $x > 0$: $$-1 \leq \frac{1}{x} \leq 5$$ Means $$\frac{1}{x} \geq -1 \Rightarrow \text{always true since } \frac{1}{x} > 0 > -1$$ $$\frac{1}{x} \leq 5 \Rightarrow x \geq \frac{1}{5}$$ - If $x < 0$: $$-1 \leq \frac{1}{x} \leq 5$$ means $$\frac{1}{x} \leq 5$$ always true since $\frac{1}{x} < 0$ $$\frac{1}{x} \geq -1 \Rightarrow x \leq -1$$ - $x=0$ not allowed **Solution:** $$x \in (-\infty, -1] \cup [\frac{1}{5}, \infty)$$