# Simplified Factorial Notation

• Jan 4th 2009, 04:00 PM
ninjahcamp3r
Simplified Factorial Notation
Alright.

So i'm sorry if my typing for my math equations might be off but i'm barely 5 minutes old in these forums so give me some slack. Please? (Happy)

So on about factorial notation. I have no idea what it means.

Key

_ = sub

Givens:

{a_n} n=1 is defined by a_1 = 1 x 1!, a_2 = 2 x 2!,... , a_n = n x n!

C_n = a_n + a_(n + 1)

I need to write it in factorial notation. And Simplified. Any helpers out there today? =)
• Jan 4th 2009, 04:26 PM
chabmgph
Quote:

Originally Posted by ninjahcamp3r
Alright.

So i'm sorry if my typing for my math equations might be off but i'm barely 5 minutes old in these forums so give me some slack. Please? (Happy)

So on about factorial notation. I have no idea what it means.

Key

_ = sub

Givens:

{a_n} n=1 is defined by a_1 = 1 x 1!, a_2 = 2 x 2!,... , a_n = n x n!

C_n = a_n + a_(n + 1)

I need to write it in factorial notation. And Simplified. Any helpers out there today? =)

For any $\displaystyle n \in N$, $\displaystyle n! = n*(n-1)*(n-2)...* 1$
So for example,
$\displaystyle 5! = 5*4*3*2*1$
and $\displaystyle 8! = 8*7*6*5*4*3*2*1$

Hope that helps.
• Jan 4th 2009, 04:26 PM
Chris L T521
Quote:

Originally Posted by ninjahcamp3r
Alright.

So i'm sorry if my typing for my math equations might be off but i'm barely 5 minutes old in these forums so give me some slack. Please? (Happy)

So on about factorial notation. I have no idea what it means.

Key

_ = sub

Givens:

{a_n} n=1 is defined by a_1 = 1 x 1!, a_2 = 2 x 2!,... , a_n = n x n!

C_n = a_n + a_(n + 1)

I need to write it in factorial notation. And Simplified. Any helpers out there today? =)

If $\displaystyle a_n=n\cdot n!$, then it would follow that $\displaystyle a_{n+1}=(n+1)\cdot(n+1)!$

So, $\displaystyle C_n=a_n+a_{n+1}=n\cdot n!+(n+1)\cdot(n+1)!=\left[n+(n+1)^2\right]n!=\color{red}\boxed{(n^2+3n+1)\cdot n!}$

Does this make sense?
• Jan 4th 2009, 04:40 PM
ninjahcamp3r
Quote:

Originally Posted by Chris L T521
If $\displaystyle a_n=n\cdot n!$, then it would follow that $\displaystyle a_{n+1}=(n+1)\cdot(n+1)!$

So, $\displaystyle C_n=a_n+a_{n+1}=n\cdot n!+(n+1)\cdot(n+1)!=\left[n+(n+1)^2\right]n!=\color{red}\boxed{(n^2+3n+1)\cdot n!}$

Does this make sense?

Alright the last part is also what I have but is it all the way simplified.

I also forgot to include that a_n can also be expressed as (n + 1)! - n!.

Could you also give me a link on how you guys right it as neatly as you guys do? It looks so much better then my typing =).
• Jan 4th 2009, 04:48 PM
chabmgph
Quote:

Originally Posted by ninjahcamp3r
Alright the last part is also what I have but is it all the way simplified.

I also forgot to include that a_n can also be expressed as (n + 1)! - n!.

Could you also give me a link on how you guys right it as neatly as you guys do? It looks so much better then my typing =).

These formulas are written using LaTex coding. There is a section in here that may help:
http://www.mathhelpforum.com/math-help/latex-help/