Closed Form for Triangular Numbers/Proof by Induction

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

Proof
Proof by induction:
 * Base Case''': $n=1$:

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

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

So the base case is true.


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


 * Inductive Step: Consider $n=k+1$.

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

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.