1. The problem: We want to understand mathematical induction, a method to prove statements for all natural numbers $n \geq 1$. 2. Explanation: Mathematical induction has two main steps: - Base case: Show the statement is true for $n=1$. - Inductive step: Assume the statement is true for some arbitrary $n=k$, then prove it is true for $n=k+1$. If both steps are successful, the statement holds for all natural numbers $n$. 3. Example: Prove that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$. Step 1 (Base case): For $n=1$, the left side is $1$, the right side is $\frac{1 \cdot 2}{2} = 1$. So the base case holds. Step 2 (Inductive step): Assume for $n=k$, $1 + 2 + ... + k = \frac{k(k+1)}{2}$. We must prove for $n=k+1$: \[1 + 2 + ... + k + (k+1) = \frac{(k+1)((k+1)+1)}{2} \] By the inductive hypothesis, \[\frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2} \] Which matches the right side for $n=k+1$. Hence, the statement is true for $k+1$. 4. Conclusion: By induction, the formula for the sum of the first $n$ natural numbers is true for all natural $n$. 5. Another example: Prove $2^n \geq n+1$ for all $n \geq 1$. Base case ($n=1$): $2^1 = 2 \geq 2$ true. Inductive step: Assume $2^k \geq k+1$, then \[2^{k+1} = 2 \cdot 2^k \geq 2(k+1) \geq (k+1) + 1\]\for $k \geq 1$. So the inequality also holds for $k+1$. Hence, $2^n \geq n+1$ for all $n$. Mathematical induction is a powerful, fundamental tool to prove statements over natural numbers.