Proof

Let \(P(n)\) be the statement \(1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\text{.}\) We will show that \(P(n)\) is true for all natural numbers \(n \ge 1\text{.}\)

Base case: \(P(1)\) is the statement \(1 = \frac{1(1+1)}{2}\) which is clearly true.

Inductive case: Let \(k \ge 1\) be a natural number. Assume (for induction) that \(P(k)\) is true. That means \(1 + 2 + 3 + \cdots + k = \frac{k(k+1)}{2}\text{.}\) We will prove that \(P(k+1)\) is true as well. That is, we must prove that \(1 + 2 + 3 + \cdots + k + (k+1) = \frac{(k+1)(k+2)}{2}\text{.}\) To prove this equation, start by adding \(k+1\) to both sides of the inductive hypothesis:

\begin{equation*} 1 + 2 + 3 + \cdots + k + (k+1) = \frac{k(k+1)}{2} + (k+1)\text{.} \end{equation*}

Now, simplifying the right side we get:

\begin{align*} \frac{k(k+1)}{2} + k+1 \amp = \frac{k(k+1)}{2} + \frac{2(k+1)}{2}\\ \amp = \frac{k(k+1) + 2(k+1)}{2}\\ \amp = \frac{(k+2)(k+1)}{2}\text{.} \end{align*}

Thus \(P(k+1)\) is true, so by the principle of mathematical induction \(P(n)\) is true for all natural numbers \(n \ge 1\text{.}\)