# Factorials & Other Stuff

• Sep 9th 2006, 07:14 PM
StackFryer
Factorials & Other Stuff
Hi there, I need some help with these questions please:
_______________________________________

Find the first 5 terms of the sequence:

tn=2^tn-1, t1=0 (I think I know this one, I just need some confirmation)

Evaluate each of these expressions:

12!
----
5!3!

9!
---- x21
6!3!

Simplify these expressions:

(nē+5n+6)(n+1)!

and

(n+1)!
------
nē+n

Solve for n:

3n(n!)=60n(n-2)!
_______________________________________

Thanks a lot. Oh, and where I put -----, it means divided by, just in case that wasn't clear. :)
• Sep 9th 2006, 07:25 PM
ThePerfectHacker
Quote:

Originally Posted by StackFryer
Hi there, I need some help with these questions please:
_______________________________________

Find the first 5 terms of the sequence:

tn=2^tn-1, t1=0 (I think I know this one, I just need some confirmation)

Looks like,
0,1,2,4,16,...
• Sep 9th 2006, 07:33 PM
ThePerfectHacker
The site is under repair thus the equation editor (LaTeX) is not avaliable. You will need to understand this mess.
Quote:

Originally Posted by StackFryer

12!
----
5!3!

Expand it
12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
---------------------------------------------------
5 x 4 x 3 x 2 x 1 x 3 x 2 x 1

What are you left with?
12 x 11 x 10 x 9 x 8 x 7

Quote:

9!
---- x21
6!3!
Same thing.
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
----------------------------------- x 21
6 x 5 x 4 x 3 x 2 x 1 x 3 x 2 x 1

What are you left with?
2 x 8 x 7 x 21

Quote:

Simplify these expressions:

(n&#178;+5n+6)(n+1)!
Factor,
(n+2)(n+3)(n+1)!
By the definition of factorial this is,
(n+3)!

Quote:

(n+1)!
------
n&#178;+n
Factor the numerator and denominator as,
(n+1)n(n-1)!
--------------
n(n+1)

Thus you get,
(n-1)!

Quote:

3n(n!)=60n(n-2)!
Divide by 3n both sides,
n! = 20 (n-2)!
Express right hand side as,
n(n-1)(n-2)!=20(n-2)!
Divide out thus,
n(n-1)=20
Thus,
n^2-n=20
Thus,
n^2-n-20=0
Thus,
(n-5)(n+4)=0
Thus,
n=5 or n=-4
But!
Factorial cannot be negative
Thus,
n=5
• Sep 9th 2006, 07:57 PM
StackFryer
Thanks a lot, I really appreciate it.