# Factorial, but with addition [duplicate]

Is there a notation for addition form of factorial?

$$5! = 5\times4\times3\times2\times1$$

That's pretty obvious. But I'm wondering what I'd need to use to describe

$$5+4+3+2+1$$

like the factorial $5!$ way.

EDIT: I know about the formula. I want to know if there's a short notation.

## marked as duplicate by Niel de Beaudrap, AlexR, user61527, copper.hat, ncmathsadistDec 5 '13 at 1:50

• $1+2+\dots+n=\dfrac{n(n+1)}{2}$; there's no need for a special notation. – egreg Dec 4 '13 at 23:28
• Duplicate of: math.stackexchange.com/q/60578/439 – Niel de Beaudrap Dec 4 '13 at 23:33
• @NieldeBeaudrap I'm asking about it's notation... – akinuri Dec 4 '13 at 23:36
• The sigma notation is a notation for it. – user112167 Dec 4 '13 at 23:37
• So, you didn't see the answer which described that Knuth suggested the notation "$n?$" ? – Niel de Beaudrap Dec 4 '13 at 23:54

It is called the $n$th triangle number and it can be written as $\binom{n+1}2$.

• why would it be called a "triangle number"? – khaverim Jun 10 '16 at 17:01
• @khaverim Check this image. This is something I've come up with some time ago to visualize and understand how the calculation works. I've literally spent an hour or so to think that, becasuse I had nowhere to look then. And I'm guessing it's called triangle number(s) becase you can treat the number set as the half of a rectangle, a triangle. – akinuri Jul 18 '16 at 21:45
• I assumed it was a triangle number due to the obvious relationship to Pascal's Triangle. – James Antill Aug 11 '16 at 21:41
• Unless I'm misunderstanding the notation, this is not a correct answer. I believe it should be ( ( n ( n + 1 ) ) / 2 ), not ( ( n + 1 ) / 2 ). – Oliver Nicholls Aug 25 '17 at 9:37
• Berci just skipped some details. The notation is called binomial coefficient. – akinuri Apr 21 '18 at 20:11

That can be done with the formula $\frac{n^2+n}{2}$

We should also note that the factorial function has a similar look to it as the sigma summation notation; as $$\frac{n(n+1)}{2}=1+2+3+...+n=\sum_{k=1}^nk$$ $$n!=1 \cdot 2 \cdot 3 \cdot ... \cdot n=\prod_{k=1}^nk$$

$\sum_{n=1}^{k} n = 1 +2+3+\ldots+k$. Is a nice notation for it. So $$1 + 2 + 3 + 4 + 5 = \sum_{n=1}^{5} n$$.