Closed Form for Triangular Numbers/Proof by Induction

Proof
Proof by induction:

Basis for the Induction
When $n = 1$, we have:
 * $\displaystyle \sum_{i \mathop = 1}^1 i = 1$

Also:
 * $\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$

This is our base case.

Induction Hypothesis

 * $\forall k \in \N: k \ge 1: \displaystyle \sum_{i \mathop = 1}^k i = \frac {k \paren {k + 1} } 2$

This is our induction hypothesis.

It is to be demonstrated that:
 * $\displaystyle \sum_{i \mathop = 1}^{k + 1} i = \frac {\paren {k + 1} \paren {k + 2} } 2$

Induction Step
This is our induction step:

Consider $n = k + 1$.

By the properties of summation:
 * $\displaystyle \sum_{i \mathop = 1}^{k + 1} i = \sum_{i \mathop = 1}^k i + k + 1$

Using the induction hypothesis this can be simplified to:

Hence the result by induction.

Also see
This is usually the first proof by induction that a student mathematician encounters.

The second one is often Proof by Induction of Sum of Sequence of Squares.