Closed Form for Triangular Numbers/Proof by Induction

Theorem
The closed-form expression for the $n$th triangular number is:
 * $\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \left({n+1}\right)} 2$

Proof
Proof by induction:
 * Base Case:

When $n=1$, we have $\displaystyle \sum_{i \mathop = 1}^1 i = 1$.

Also, $\displaystyle \frac{n \left({n+1}\right)} 2 = \frac {1 \cdot 2} 2 = 1$.

This is our base case.


 * Induction Hypothesis:
 * $\forall k \in \N: k \ge 1: \displaystyle \sum_{i \mathop = 1}^k i = \frac {k \left({k+1}\right)} 2$

This is our induction hypothesis.


 * 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:

Thus, the result has been shown by induction.

This is usually the first proof by induction that a student mathematician experiences.