# Talk:Closed Form for Triangular Numbers

I've put the proof together for Telescoping Series/Example 1 but I reckon Telescoping Sum would be a stronger result (in that it has wider direct applications and the result for series follows from it) and if we felt real ambitious we could prove it for rings. --prime mover (talk) 21:32, 24 December 2008 (UTC)

I put up the recursive proof for triangular numbers. I couldn't find it anywhere else on the internet so I'm not really sure if it's a quality proof, but I put it up anyway :P. 20:19, 24 May 2010 (UTC)

 $\displaystyle S(n)$ $=$ $\displaystyle n + (n-1) + (n-2) + \cdots + 2 + 1$ $\displaystyle$ $=$ $\displaystyle n + (n-1) + (n-2) + \cdots + (n-(n-2)) + (n-(n-1))$ $\displaystyle$ $=$ $\displaystyle n^2 - (1 + 2 + \cdots + (n-1))$
... can you elaborate? --prime mover (talk) 20:47, 24 May 2010 (UTC)

Well... the first eqn you stated should be obvious right? For example, $S(5) = 5 + 4 + 3 + 2 + 1$. For the second eqn, I'm just restating the constants in terms of $n$. $n-(n-2) = 2$ and $n-(n-1) = 1$; there's no arguing that. For the third eqn I'm just using the associative/commutative rule for addition/subtraction. The right side of the second eqn can be intermediately expressed as:

 $\displaystyle$ $=$ $\displaystyle (n_1 + n_2 + \cdots + n_n) - 1 - 2 - \cdots - (n-1)$

And further simplifying that results in the third eqn.

Oh yes of course. Thanks. --prime mover (talk) 22:12, 25 May 2010 (UTC)

## Landmark

The point has been made that this was the first proof. However, technically speaking, that was held by Closed Form for Triangular Numbers/Direct Proof (which already has the landmark flag on it). The other proofs came later. --prime mover 16:52, 14 March 2012 (EDT)