Closed Form for Triangular Numbers/Proof by Recursion

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

Let:

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

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle n + (n-1) + (n-2) + \cdots + 2 + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n + (n-1) + (n-2) + \cdots + (n-(n-2)) + (n-(n-1))\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^2 - (1 + 2 + \cdots + (n-1))\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle n^2 - S(n-1)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Establishes a recursive definition of triangular numbers
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle n + S(n-1)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	The standard recursive definition of triangular numbers
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2 S \left({n}\right)\)	\(=\)	\(\displaystyle n^2 + n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Adding the equations together
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle \frac{n(n+1)} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$