I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it I'm writing some software that takes a group of users and compares each user with every other user in the group. I need to display the amount of comparisons needed for a countdown type feature. 
For example, this group [1,2,3,4,5] would be analysed like this:
1-2, 1-3, 1-4, 1-5
2-3, 2-4, 2-5
3-4, 3-5
4-5

By creating little diagrams like this I've figured out the pattern which is as follows:
Users - Comparisons
2     -   1
3     -   3 (+2)
4     -   6 (+3)
5     -   10 (+4)
6     -   15 (+5)
7     -   21 (+6)
8     -   28 (+7)
9     -   36 (+8)

I need to be able to take any number of users, and calculate how many comparisons it will take to compare every user with every other user.
Can someone please tell me what the formula for this is?
 A: The following way to getting the solution is beautiful and said to have been found by young Gauss in school. The idea is that the order of adding $1+2+\cdots+n=S_n$ does not change the value of the sum. Therefore:
$$1 + 2 + \ldots + (n-1) + n=S_n$$
$$n + (n-1) + \ldots + 2 + 1=S_n$$
Adding the two equations term by term gives
$$(n+1)+(n+1)+\ldots+(n+1)=2S_n$$
so $n(n+1)=2S_n$. For $n$ persons, there are $S_{n-1}$ possibilities, as others answers have shown already nicely.
A: You want to know how many ways there are to choose $2$ users from
a set of $n$ users.
Generally, the number of ways to choose $k$ elements from a set of
order $n$ (that is, all elements in the set are distinct) is denoted
by 
$$
\binom{n}{k}
$$
and is equivalent to 
$$
\frac{n!}{(n-k)!k!}
$$
In the case of $k=2$ the latter equals to 
$$
\frac{n!}{(n-2)!2!}=\frac{n(n-1)}{2}
$$
which is also the sum of $1+2+...+n-1$.
For more information see Binomial coefficient and Arithmetic progression
A: The sum of $0+\cdots + n-1$ is $$\frac12(n-1)n.$$
Here $n$ is the number of users; there are 0 comparisons needed for the first user alone, 1 for the second user (to compare them to the first), 2 for the third user, and so on, up to the $n$th user who must be compared with the $n-1$ previous users.
For example, for $9$ people you are adding up $0+1+2+3+4+5+6+7+8$, which is equal to $$\frac12\cdot 8\cdot 9= \frac{72}{2} = 36$$ and for $10$ people you may compute $$\frac12\cdot9\cdot10 = \frac{90}2 = 45.$$
A: The discrete sum up to a finite value $N$ is given by,
$$\sum_{n=1}^{N} n = \frac{1}{2}N(N+1)$$
Proof:
The proof by induction roughly boils down to:
$$S_N = 1+ 2 +\dots+N$$
$$S_{N+1}= 1+ 2 + \dots + N + (N+1) = \underbrace{\frac{1}{2}N(N+1)}_{S_N} + (N+1)$$
assuming that the induction hypothesis is true. The right hand side:
$$\frac{N(N+1)}{2}+(N+1)=\frac{(N+1)(N+2)}{2}$$
which is precisely the induction hypothesis applied to $S_{N+1}$.

Just for your own curiosity, the case $N=\infty$ is of course divergent. However, with the use of the zeta function, it may be regularized to yield,
$$\sum_{n=1}^{\infty}n = \zeta(-1)=-\frac{1}{12}$$
A: $$N=2:\ 1 + 2 = (1 + 2) = 1\times3$$
$$N=4:\ 1 + 2 + 3 + 4 = (1 + 4) + (2 + 3) = 2\times5$$
$$N=6:\ 1 + 2 + 3 + 4 + 5 + 6 = (1 + 6) + (2 + 5) + (3 + 4) = 3\times7$$
$$N=8:\ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8= (1 + 8) + (2 + 7) + (3 + 6)+ (4 + 5) = 4\times9$$
More generally, $N/2\times(N + 1)$.
For odd $N$, sum the $N-1$ first terms (using the even formula) together with $N$, giving $(N-1)/2\times N+N=N\times(N+1)/2$.
A: Here is another way to find the sum of the first $n$ squares that generalizes to sums of higher powers.
$(k+1)^2 - k^2 = 2k+1$
$\sum_{k=1}^n ( (k+1)^2 - k^2 ) = \sum_{k=1}^n (2k+1)$
$(n+1)^2 - 1^2 = 2 \sum_{k=1}^n k + n$
$\sum_{k=1}^n k = \frac{ (n+1)^2 - 1 - n }{2} = \frac{n^2+n}{2}$
A: This is kind of a pseudo code:
Say you have n number of people, and you labeled them.
for i in (1,2,3,...,n), person i need to compare with all the people who has a number larger (strictly), so person i need to compare (n-i) times.
so adding up would be
(n-1) + (n-2) + ... + 3 + 2 + 1...
which would be the sum from 1 to (n-1)
