1. Let's state the problem: We want to prove a mathematical statement or theorem. 2. A common example is the proof that the sum of the first $n$ natural numbers is given by the formula $$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$ 3. We will use mathematical induction to prove this formula. 4. Base case: For $n=1$, the sum is $1$, and the formula gives $\frac{1(1+1)}{2} = 1$, so the base case holds. 5. Inductive step: Assume the formula holds for some $n = m$, i.e., $$\sum_{k=1}^m k = \frac{m(m+1)}{2}.$$ 6. We need to prove it holds for $n = m+1$: $$\sum_{k=1}^{m+1} k = \left(\sum_{k=1}^m k\right) + (m+1) = \frac{m(m+1)}{2} + (m+1).$$ 7. Simplify the right side: $$\frac{m(m+1)}{2} + (m+1) = \frac{m(m+1) + 2(m+1)}{2} = \frac{(m+1)(m+2)}{2}.$$ 8. This matches the formula with $n = m+1$, so the inductive step holds. 9. By mathematical induction, the formula is true for all natural numbers $n$. This completes the proof.