Summation analogue of the factorial operation?

[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.

 

[Manchot]

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

 

[gabee]

Haha, that's why :P

 

[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

 

[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!

 

[Curious3141]

Kinda, sorta, they're called triangular numbers : http://en.wikipedia.org/wiki/Triangular_numbers

 

[ZioX]

I've seen S_k(n)=1^k+2^k+...+(n-1)^k used for expositions on Bernoulli's theorem.

Wikipedia sticks to the sigma notation here: http://en.wikipedia.org/wiki/Bernoulli_numbers