1. **State the problem:** We want to prove a statement using the principle of mathematical induction, which is a loop variant proof technique. 2. **Explain the principle of mathematical induction:** To prove a statement $P(n)$ for all integers $n \geq k$ (where $k$ is the base case), we do two steps: - **Base case:** Show that $P(k)$ is true. - **Inductive step:** Assume $P(m)$ is true for some arbitrary $m \geq k$, then prove $P(m+1)$ is true. 3. **Example statement:** Prove that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$. That is, prove: $$P(n): 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$ 4. **Base case ($n=1$):** $$1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$$ So $P(1)$ is true. 5. **Inductive step:** Assume $P(m)$ is true: $$1 + 2 + \cdots + m = \frac{m(m+1)}{2}$$ We need to prove $P(m+1)$: $$1 + 2 + \cdots + m + (m+1) = \frac{(m+1)(m+2)}{2}$$ 6. **Proof:** Starting from the assumption, $$1 + 2 + \cdots + m + (m+1) = \frac{m(m+1)}{2} + (m+1)$$ Factor out $(m+1)$: $$= (m+1)\left(\frac{m}{2} + 1\right) = (m+1)\frac{m+2}{2} = \frac{(m+1)(m+2)}{2}$$ 7. **Conclusion:** Since the base case is true and the inductive step holds, by mathematical induction, the formula is true for all $n \geq 1$. This completes the loop variant proof with an example.