How would you simplify "(3n-2)!"?

[JohnCU]

Originally posted by: Kyteland

Originally posted by: JohnCU
Um yes (3n-2)! does = (3n-2)(3n-1)(3n)!

Bullsh!t

n=1

(3n-2)! = (3*1-2)! = 1! = 1

(3n-2)(3n-1)(3n)! = (3*1-2)(3*1-1)(3*1)! = 1*2*6 = 12

Copnsidered NOT proven.

True, I was thinking of how (n+1)! = (n+1)n!

 

[Kyteland]

Originally posted by: NYSTrooper

Originally posted by: MCrusty

Originally posted by: Kyteland

Originally posted by: MCrusty

Originally posted by: Kyteland

Originally posted by: MCrusty
(3n-2)(3n-1)(3n)!

Um, no.

(3n-2)! = (3n)!/((3n)(3n-1))

Uhmm...wtf are you talking about...

What you put up above is not equal to (3n-2)!

(3n-2)! != (3n-2)(3n-1)(3n)!

(3n-2)! = (3n)!/((3n)(3n-1))
edit: I edited my post with better math.

Heh..i thought they were (3n+2)!

Way to go guys, now you've got him really confused.

I thought that was the goal?

 

[NYSTrooper]

Originally posted by: JohnCU

Originally posted by: Kyteland

Originally posted by: JohnCU
Um yes (3n-2)! does = (3n-2)(3n-1)(3n)!

Bullsh!t

n=1

(3n-2)! = (3*1-2)! = 1! = 1

(3n-2)(3n-1)(3n)! = (3*1-2)(3*1-1)(3*1)! = 1*2*6 = 12

Copnsidered NOT proven.

True, I was thinking of how (n+1)! = (n+1)n!

CU?

 

[SuPrEIVIE]

Originally posted by: JohnCU

Originally posted by: Kyteland

Originally posted by: MCrusty

Originally posted by: Kyteland

Originally posted by: MCrusty
(3n-2)(3n-1)(3n)!

Um, no.

(3n-2)! = (3n)!/((3n)(3n-1))

Uhmm...wtf are you talking about...

What you put up above is not equal to (3n-2)!

(3n-2)! != (3n-2)(3n-1)(3n)!

(3n-2)! = (3n)!/((3n)(3n-1))

edit: I edited my post with the correct math.

Um yes (3n-2)! does = (3n-2)(3n-1)(3n)!


according to zee mathematical conztituzion zis procedure of zimplifying zis againzt zee lawz!

 

[JohnCU]

Originally posted by: NYSTrooper

CU?

CU, my name? that stands for Clemson University...

 

[LordMorpheus]

You COULD do a lot of things. I'm guessing you are supposed to do something like:

((3n)!)
-----------------
((3n)(3n-1))

 

[JohnCU]

Originally posted by: SuPrEIVIE
according to zee mathematical conztituzion zis procedure of zimplifying zis againzt zee lawz!

How does (n+1)! = (n+1)n! then?

 

[NYSTrooper]

Originally posted by: JohnCU

Originally posted by: NYSTrooper

CU?

CU, my name? that stands for Clemson University...

Booo.

CU = University of Colorado at Boulder.

 

[LordMorpheus]

Originally posted by: JohnCU

Originally posted by: SuPrEIVIE
according to zee mathematical conztituzion zis procedure of zimplifying zis againzt zee lawz!

How does (n+1)! = (n+1)n! then?

That works.

(3n)! =(3n)(3n-1)(3n-2)(3n-3) . . . . .
(n+1)! = (n+1)(n)(n-1)(n-2)(n-3) . . . .
n! = (n)(n-1)(n-2)(n-3) . . . .

therefore, (n+1)! = (n+1)(n!)

 

[JohnCU]

I guess you can subtract from it, but can't add.

 

[Chaotic42]

Originally posted by: CrackaLackaZe

Originally posted by: Gibson486
i will give you a hint.....start plugging in numbers from 0 to whenever....

So just go from n=0 to n=5, look for a pattern, and put down a series for it?

Here are n=1 to n=15

1=1
2=24
3=5040
4=3628800
5=6.227*10^9
6=2.092*10^13
7=1.216*10^17
8=1.124*10^21
9=1.551*10^25
10=3.049*10^29
11=8.223*10^33
12=2.952*10^38
13=1.376*10^43
14=8.159*10^47
15=6.042*10^52

 

[DrPizza]

Originally posted by: LordMorpheus

Originally posted by: JohnCU

Originally posted by: SuPrEIVIE
according to zee mathematical conztituzion zis procedure of zimplifying zis againzt zee lawz!

How does (n+1)! = (n+1)n! then?

That works.

(3n)! =(3n)(3n-1)(3n-2)(3n-3) . . . . .
(n+1)! = (n+1)(n)(n-1)(n-2)(n-3) . . . .
n! = (n)(n-1)(n-2)(n-3) . . . .

therefore, (n+1)! = (n+1)(n!)

Yes, the pattern is to subtract, not add...
10! equals 10*9*8!
not 10*11*12! (which is what the (3n-2)(3n-1)(3n)! people were mistakenly doing above.)


Anyway...
take (3n-2)! , put it over 1

Multiply the numerator and the denominator by (3n-1)(3n)
The numerator becomes (3n)(3n-1)(3n-2)! which is (3n)!
So, the resulting fraction is

(3n)! / [(3n)(3n-1)]

which is the result posted earlier from a TI calculator.
I'm not sure I'd call that "simplified", but it's probably useful when working with series in Calculus (comparison test, ratio test, etc)

 

[GhettoFob]

This thread gives me a headache.