1. **Prove by mathematical induction that** $$4 + 7 + 10 + \dots + (3n + 1) = n \left(\frac{3n + 5}{2}\right)$$ Step 1: **Base case** ($n=1$): Left side: $4$ (since $3(1)+1=4$) Right side: $1 \times \frac{3(1)+5}{2} = 1 \times \frac{8}{2} = 4$ Both sides equal, so base case holds. Step 2: **Inductive hypothesis**: Assume true for $n=k$: $$4 + 7 + 10 + \dots + (3k + 1) = k \left(\frac{3k + 5}{2}\right)$$ Step 3: **Inductive step**: Prove for $n=k+1$: Add the next term $3(k+1)+1 = 3k + 4$ to both sides: $$4 + 7 + \dots + (3k + 1) + (3k + 4) = k \left(\frac{3k + 5}{2}\right) + (3k + 4)$$ Simplify right side: $$= \frac{k(3k + 5)}{2} + \frac{2(3k + 4)}{2} = \frac{3k^2 + 5k + 6k + 8}{2} = \frac{3k^2 + 11k + 8}{2}$$ Step 4: Express right side for $n=k+1$: $$ (k+1) \left(\frac{3(k+1) + 5}{2}\right) = (k+1) \left(\frac{3k + 3 + 5}{2}\right) = (k+1) \left(\frac{3k + 8}{2}\right) = \frac{(k+1)(3k + 8)}{2}$$ Expand: $$= \frac{3k^2 + 8k + 3k + 8}{2} = \frac{3k^2 + 11k + 8}{2}$$ Step 5: Both expressions match, so the formula holds for $n=k+1$. **Conclusion:** By induction, the formula is true for all natural numbers $n$. 2. **Show that the sum of a rational and an irrational number is irrational.** Step 1: Let $r$ be rational and $i$ be irrational. Step 2: Suppose $r + i$ is rational, call it $q$. Step 3: Then $i = q - r$. Step 4: Since $q$ and $r$ are rational, their difference $q - r$ is rational. Step 5: This contradicts that $i$ is irrational. **Conclusion:** The sum of a rational and an irrational number must be irrational. 3. **Disprove by counterexample that if $x$ is prime, then $x^2 - 2$ is prime.** Step 1: Choose $x=2$ (prime). Step 2: Compute $x^2 - 2 = 2^2 - 2 = 4 - 2 = 2$ (prime). Step 3: Choose $x=3$ (prime). Step 4: Compute $x^2 - 2 = 9 - 2 = 7$ (prime). Step 5: Choose $x=5$ (prime). Step 6: Compute $x^2 - 2 = 25 - 2 = 23$ (prime). Step 7: Choose $x=7$ (prime). Step 8: Compute $x^2 - 2 = 49 - 2 = 47$ (prime). Step 9: Choose $x=11$ (prime). Step 10: Compute $x^2 - 2 = 121 - 2 = 119$. Step 11: Check if 119 is prime: $119 = 7 \times 17$, so not prime. **Conclusion:** The statement is false; $x=11$ is a counterexample. 4. **Solve the equation** $$z^2 = 1 + i$$ Step 1: Let $z = a + bi$, where $a,b \in \mathbb{R}$. Step 2: Compute $z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2ab i$. Step 3: Equate real and imaginary parts: $$a^2 - b^2 = 1$$ $$2ab = 1$$ Step 4: From $2ab=1$, solve for $b$: $$b = \frac{1}{2a}$$ Step 5: Substitute into $a^2 - b^2 = 1$: $$a^2 - \left(\frac{1}{2a}\right)^2 = 1$$ $$a^2 - \frac{1}{4a^2} = 1$$ Step 6: Multiply both sides by $4a^2$: $$4a^4 - 1 = 4a^2$$ Step 7: Rearrange: $$4a^4 - 4a^2 - 1 = 0$$ Step 8: Let $x = a^2$, then: $$4x^2 - 4x - 1 = 0$$ Step 9: Solve quadratic for $x$: $$x = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm \sqrt{32}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$$ Step 10: Since $a^2 = x$, $a^2$ must be positive. Step 11: Take $a^2 = \frac{1 + \sqrt{2}}{2}$ (positive), then $$a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}}$$ Step 12: Find $b$: $$b = \frac{1}{2a} = \pm \frac{1}{2 \sqrt{\frac{1 + \sqrt{2}}{2}}}$$ Step 13: Similarly for $a^2 = \frac{1 - \sqrt{2}}{2}$, which is negative, discard. **Final solutions:** $$z = \pm \sqrt{\frac{1 + \sqrt{2}}{2}} + i \times \pm \frac{1}{2 \sqrt{\frac{1 + \sqrt{2}}{2}}}$$ 5. **In a medical examination of 100 persons:** - 20 are short-sighted - 10 are diabetic - 5 are both short-sighted and diabetic Find number of persons neither diabetic nor short-sighted. Step 1: Use inclusion-exclusion principle: $$|S \cup D| = |S| + |D| - |S \cap D| = 20 + 10 - 5 = 25$$ Step 2: Number neither diabetic nor short-sighted: $$100 - 25 = 75$$ **Answer:** 75 persons.