Closed Form for Triangular Numbers/Proof using Binomial Coefficients

Theorem

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

$\displaystyle T_n = \sum_{i=1}^{n} i = \frac {n \left({n+1}\right)} {2}$

Proof

From Properties of Binomial Coefficients: Particular Values, we have that:

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

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{k=1}^n k\)	\(=\)	\(\displaystyle \sum_{k=1}^n \binom k 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Properties of Binomial Coefficients: Particular Values
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {n + 1} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of k Choose m up to n‎
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({n+1}\right) n} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of binomial coefficient

$\blacksquare$