Binomial coefficient as a summation series proof? Alright, so I was wondering if the following is a well known identity or if its existence provides any real benefits other than serving as a time-saver when dealing with higher values for combinations.
After screwing around with some basic combinations stuff, I noticed the following:
$$ \sum_{i=1}^{n-1} i = \begin{pmatrix}n\\2\\ \end{pmatrix}$$
To prove this, I used Gauss' method to simplify the summation, and I wrote n choose 2 in terms of factorials to simplify the right side.
$$ \frac{ (n-1) n } {2} = \frac{ n! } { (2!) (n-2)! } $$
$$ 2!(n-2)!(n-1)(n) = 2n! $$
$$ 2(n-2)!(n-1)(n) = 2n! $$
$$ (n-2)!(n-1)(n) = n! $$
$$ n! = n! $$
I did this on lunch break one day over the summer. I'm in high school, so my math skills are very subpar on this forum, but I was hoping some people might discuss it and/or answer my aforementioned questions. I didn't see anything about it on here or Google, for that matter. If you found this banal or rudimentary, just let me know and I'll refrain from posting until I come up with something more interesting. Regardless, I hope you found it worth your time.
 A: That is an identity. Notice that $n!=n(n-1)(n-2)!$, so that the $(n-2)!$ cancel out, so you can shorten your proof!
Another way of seeing the same thing is that, if you need to choose 2 elements from a total of n, you can do it in:


*

*$n \choose 2$ ways (by definition)

*In the following way:


*

*Take the first 2 elements. You can do that in 1 way. 

*Take the first 3 elements. The ways in which you can do that that don't involve the 3rd element are already counted, so you only count the ways of taking two elements out of 3 taking the 3rd number. Because there are 2 other numbers, you can either take 1 or 2. You can do this in 2 ways.

*Repeat this over and over until you get to taking n elements. You take the last one and you're left with n-1 choices.

*You add all the numbers that you got, which is just $1+2+\cdots+(n-1)$, the summation you wrote.
Because we have counted the same in two different ways, the two expressions are the same. This is called double counting and is an important tool in combinatorics. We can say, then, that $\sum_{i=1}^{n-1}i = {n \choose 2}$.
A: 
I was wondering if the following is a well known identity

Not only is it well-known, but it's part of a much larger group. In general, we have 

$$\sum_{k=0}^nk~(k+1)~\cdots~(k+p)~=~(p+1)!~{n+p+1\choose n-1}~=~(p+1)!~{n+p+1\choose p+2}$$

The whole idea is to rewrite the summand as $(p+1)!~\displaystyle{p+k\choose p+1}.~$ See also Faulhaber's formulas.
