# Thread: Factorials & Other Stuff

1. ## 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.

2. 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,...

3. The site is under repair thus the equation editor (LaTeX) is not avaliable. You will need to understand this mess.
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

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

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)!

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

Thus you get,
(n-1)!

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

4. Thanks a lot, I really appreciate it.