Closed Form for Triangular Numbers/Proof using Binomial Coefficients

Theorem

The closed-form expression for the $n$th triangular number is:

$\displaystyle T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof

$\forall k \in \Z, k > 0: \dbinom k 1 = k$

Thus:

 $\displaystyle \sum_{k \mathop = 1}^n k$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \binom k 1$ Binomial Coefficient with One $\displaystyle$ $=$ $\displaystyle \binom {n + 1} 2$ Sum of k Choose m up to n‎ $\displaystyle$ $=$ $\displaystyle \frac {\left({n+1}\right) n} 2$ Definition of Binomial Coefficient

$\blacksquare$