Is there a way to denote the calculation $1+2+3+\dots+n$?

Question

Since $n!$ represents $$1\cdot2\cdot3\cdots n,$$ I am wondering if there is a way to represent $$1+2+3+\dots+n?$$ What are some usual notations for the computation of some common sequences? Any other examples?

Answer 1

One way to write it would be simply by using the sumation notation, meaning $$1+2+3+\dots+n=\sum_{k=1}^n k.$$ Of course, that is equivalent to writing the factorials with the product notation, meaning $$n!=\prod_{k=1}^n k,$$ so I don't think that is what you were asking.

If you already know that $1+2+\dots+n=\frac{n(n+1)}{2}$, then you can just write $\frac{n(n+1)}{2}$ instead of the sum. The fact that the sum can be expressed as this rather short fraction is in my oppinion the real reason why a shortened notation does not exist. Unlike in the case of $n!$, which cannot be expressed by a polynomial in $n$, this one can be, thus shortening the notation is not needed.

EDIT: I was of course proven wrong by Chris Culter in his answer. However, I would like to add that the notation $T_n$, where $T$ stands for triangular, is not as common as $n!$. Any mathematitian in the world will know that if you write $5!$, you really mean $5\cdot 4\cdot 3\cdot 2\cdot 1$, however, if you say $T_n$, that notation can be used for other things, such as Chebyshev polynomials.

Answer 2

$$\sum_{n=1}^{k} n = 1+2+3+...+ k$$

Also, $$k! = \prod\limits_{n=1}^k n$$

Answer 3

$T_n$, where the letter T stands for Triangular.

Answer 4

Another way to write it would be ${n+1 \choose 2}$:

${n \choose 0}=1$ for all $n$, and ${n \choose k+1} = \sum_{k=0}^{n-1}{n \choose k}$ (with ${0 \choose k} = 0$ for $k>0$). Therefore ${n\choose 1}=n$ and thus ${n+1 \choose 2}=\sum_{k=0}^n k$. Note that this actually is $0+1+\dots+n$, but obviously the additional term $0$ doesn't change the sum.