Closed Form for Triangular Numbers/Proof using Binomial Coefficients

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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

From Binomial Coefficient with One:

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

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

$\blacksquare$


Sources