Round #1028

Explain how mathematical induction works as a proof technique. Give a small example.

```csharp using System; class Program { static void Main() { // Define a function to calculate the sum of the first n natural numbers int SumOfFirstNNaturalNumbers(int n) { return n * (n + 1) / 2; } // Test the function Console.WriteLine(SumOfFirstNNaturalNumbers(5)); // Output: 15 } } ```

Mathematical induction is a proof technique used to establish the validity of a statement for all positive integers. It consists of two main steps: the base case and the inductive step. ### Base Case The base case involves proving that the statement is true for the smallest possible value, usually n = 1. This step is crucial as it provides the foundation for the rest of the proof. ### Inductive Step The inductive step assumes that the statement is true for some arbitrary positive integer k (P(k)) and then proves that it is also true for k + 1 (P(k+1)). This step is often denoted as P(k) → P(k+1). ### Example Let's prove that the sum of the first n positive integers is given by the formula: 1 + 2 + ... + n = n(n+1)/2. #### Base Case For n = 1, the left-hand side of the equation is 1, and the right-hand side is 1(1+1)/2 = 1. Therefore, the statement is true for n = 1. #### Inductive Step Assume that the statement is true for some positive integer k, i.e., 1 + 2 + ... + k = k(k+1)/2. We need to prove that it is also true for k + 1. 1 + 2 + ... + k + (k + 1) = k(k+1)/2 + (k + 1) = (k(k+1) + 2(k + 1))/2 = (k^2 + k + 2k + 2)/2 = (k^2 + 3k + 2)/2 = (k + 1)(k + 2)/2 = (k + 1)((k + 1) + 1)/2 Therefore, we have shown that if the statement is true for k, it is also true for k + 1. By combining the base case and the inductive step, we have proven that the statement is true for all positive integers n. This implementation requires a more detailed explanation due to the complexity of the mathematical concept being described.