1. The problem is to prove a mathematical statement or formula. Since the user did not specify which proof, I will demonstrate a classic proof: the sum of the first $n$ natural numbers is given by the formula $$S = \frac{n(n+1)}{2}.$$ 2. The formula states that if you add all natural numbers from 1 to $n$, the result is $$S = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$$ 3. To prove this, consider the sum written forwards and backwards: $$S = 1 + 2 + 3 + \cdots + (n-1) + n,$$ $$S = n + (n-1) + (n-2) + \cdots + 2 + 1.$$ 4. Adding these two equations term by term, we get: $$2S = (1+n) + (2 + (n-1)) + (3 + (n-2)) + \cdots + ((n-1) + 2) + (n + 1).$$ 5. Each pair sums to $n+1$, and there are $n$ such pairs, so: $$2S = n(n+1).$$ 6. Dividing both sides by 2, we get: $$S = \frac{n(n+1)}{2}.$$ 7. This completes the proof that the sum of the first $n$ natural numbers is $$\frac{n(n+1)}{2}.$$