# Summation analogue of the factorial operation?

1. Jul 14, 2007

### gabee

Is there such a thing?

The factorial is usually defined as
$$n! = \prod_{k=1}^n k$$ if k is a natural number greater than or equal to 1.

Is there an operation that is defined as
$$\sum_{k=0}^n k$$
if one wants to find, for instance, something like 5+4+3+2+1?

I ask because I was thinking about binomial expansions and Pascal's triangle, and I'm just curious as to why the factorial operation (!) exists for products but I've never heard of such a thing for sums.

2. Jul 15, 2007

### Manchot

No, defining that operation would be fairly useless: the summation is equal to n(n+1)/2.

3. Jul 15, 2007

### gabee

Haha, that's why :P

4. Jul 15, 2007

### Gib Z

In case you wish to see why, we wish to sum : $$1, 2, 3, 4, 5, 6, 7 \cdots n$$. Add the last term, and the first term. We get n+1. Add the second last term, and the second term, we still get n+1. Add the third, and third last, we still get n+1. How many pairs of these (n+1)'s are there? Well if n is even we can easily see the number of pairs is n/2.

However is n is an Odd number, then the number of pairs that add up to (n+1) is (n-1)/2. And we still have a term in the middle, and we can see that one is (n+1)/2. Simple algebra gives the same sum : n(n+1)/2

5. Jul 15, 2007

### gabee

Yes, I've heard the legend about Gauss as a child being asked to sum the numbers 1 through 100 and he realized he could do it in this way...that guy was a genius. I am trying to brush up on my Cal I and II this summer so I can commit this sort of thing to memory again. Thanks!

Last edited: Jul 15, 2007
6. Jul 15, 2007

### Curious3141

7. Jul 15, 2007

### ZioX

Last edited: Jul 15, 2007