Binomial coefficient as a summation series proof?

Question

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.

Fantastic, thanks!
I assume, then, that this is a common proof? — Leonidas Lanier, Aug 18 '15 at 00:27
Here's another one: $\sum_{i=1}^{n-1} i^3 = {n \choose 2}^2$. Can you prove it? — , Aug 18 '15 at 00:33
I just did it on pencil and paper. It was identical, only I had two of everything.
Considering I'm new to $ LaTeX $ , I don't feel like showing the work on here, haha. Thanks for the example, though! — Leonidas Lanier, Aug 18 '15 at 00:50
I did do some proofs with higher trial successes. I stopped at n choose 7 because I was multiplying 4 summations on one side and had a fraction with two sets of product notation in the numerator and three in the denominator on the other side.
Speaking of which, what do you call multiplying a series of integers with $ \prod $ ? I would think production, but I'm pretty sure it's something else. — Leonidas Lanier, Aug 18 '15 at 00:55
The sums of consecutive integer powers gets complicated quickly (take a look at Faulhaber's formula to see what I mean). Far more approachable are sums of the form $\sum_{i=1}^n i(i-1)\cdots (i-k+1)$---see what you can make of those! — Semiclassical, Aug 18 '15 at 01:31
Sure thing!
Yeah, in my comment to the only reply on this thread, I talked about something similar where you put n choose 3, n choose 4, etc. in terms of summations.

But yeah, I'll definitely mess around with the summation you just posted. — Leonidas Lanier, Aug 18 '15 at 02:10
Congratulations! You have rediscovered the case $r=1$ of the "hockey stick " formula: http://www.artofproblemsolving.com/wiki/index.php/Combinatorial_identity — , Aug 18 '15 at 13:47
That's a fascinating page. When I used to solve combinatorics problems by writing out possibilities, I used the hockey-stick method (?) to fund sums of sums of sums... — Leonidas Lanier, Aug 18 '15 at 16:13

score 1 · Answer 1 · answered Aug 18 '15 at 00:34

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:
1. Take the first 2 elements. You can do that in 1 way.
2. 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.
3. 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.
4. 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 $\sum_{i=1}^{n-1}i = {n \choose 2}$.

I saw that (and personally thought it was the coolest part, haha). My last step was supposed to show that $ (n-2)!(n-1)n = n! $ , hence $ n! = n! $
Yeah, the way I had noticed the identity was that I was writing down combinations of 2 with 6 possible values and noticed the stairstep you mentioned (5 possibilities involving 1, 4 involving 2, 3 involving 3, etc.)that is indicative of summation. It wasn't until I began looking at probability for my AP Stat class that I realized binomial coefficients were the standard method for solving similar problems. — Leonidas Lanier, Aug 18 '15 at 00:42

score 1 · Accepted Answer · answered Aug 18 '15 at 08:18

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.

Binomial coefficient as a summation series proof?

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