1. **State the problem:** We want to prove a statement $P(n)$ is true for all natural numbers $n$ using mathematical induction. 2. **Induction principle:** To prove $P(n)$ for all $n \geq n_0$, we do two steps: - **Base case:** Show $P(n_0)$ is true. - **Inductive step:** Assume $P(k)$ is true for some arbitrary $k \geq n_0$ (inductive hypothesis), then prove $P(k+1)$ is true. 3. **Example:** Prove that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$. 4. **Base case:** For $n=1$, sum is $1$. Formula gives $\frac{1(1+1)}{2} = 1$. So $P(1)$ is true. 5. **Inductive hypothesis:** Assume for $n=k$, sum is $\frac{k(k+1)}{2}$. 6. **Inductive step:** For $n=k+1$, sum is $$\sum_{i=1}^{k+1} i = \left(\sum_{i=1}^k i\right) + (k+1) = \frac{k(k+1)}{2} + (k+1)$$ 7. Simplify: $$\frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$$ 8. This matches the formula for $n=k+1$, so $P(k+1)$ is true. 9. **Conclusion:** By induction, the formula holds for all natural numbers $n$. This is the general method to solve problems by mathematical induction.