First, let's think inductively about this equation. In fact, we know this is true for other reasons (reverse and add comes to mind). But why might induction be applicable? The left-hand side adds up the numbers from 1 to \(n\text{.}\) If we know how to do that, adding just one more term (\(n+1\)) would not be that hard. For example, if \(n = 100\text{,}\) suppose we know that the sum of the first 100 numbers is \(5050\) (so \(1 + 2 + 3 + \cdots + 100 = 5050\text{,}\) which is true). Now to find the sum of the first 101 numbers, it makes more sense to just add 101 to 5050, instead of computing the entire sum again. We would have \(1 + 2 + 3 + \cdots + 100 + 101 = 5050 + 101 = 5151\text{.}\) In fact, it would always be easy to add just one more term. This is why we should use induction.
Now the formal proof:
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 \(n \ge 1\text{.}\)