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Gauss's Identity for Arithmetic Progression Mathematical Induction
$$ \sum_{i=1}^n i = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} $$

1. Base Case (\(n = 1\)):
For \(n = 1\), the Left-Hand Side (LHS) is \(1\).
The Right-Hand Side (RHS) is \(\frac{1(1+1)}{2} = \frac{2}{2} = 1\).
Since \(\text{LHS} = \text{RHS} = 1\), \(P(1)\) holds.

2. Inductive Hypothesis:
Assume that the statement \(P(k)\) is true for some positive integer \(k \ge 1\): $$ 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2} $$

3. Inductive Step (\(n = k+1\)):
We must prove that \(P(k+1)\) holds, namely: $$ 1 + 2 + \dots + k + (k+1) = \frac{(k+1)(k+2)}{2} $$ Substituting our hypothesis: $$ \text{LHS} = \left(\sum_{i=1}^k i\right) + (k+1) = \frac{k(k+1)}{2} + (k+1) $$ Factoring out \((k+1)\): $$ = (k+1) \left( \frac{k}{2} + 1 \right) = (k+1) \left( \frac{k+2}{2} \right) = \frac{(k+1)(k+2)}{2} $$ This matches the RHS for \(n = k+1\). Thus, \(P(k) \implies P(k+1)\).

By the Principle of Mathematical Induction, \(P(n)\) holds for all \(n \in \mathbb{N}\). ■