help solving (3(n+1)!) / 5n!?

Question

hi guys... i was wondering if any of y'all can help me solve:
3(n+1)! / 5n

Note: 3(n+1)! is the numerator && 5n is the denominator.
i need to know how to solvee it.... pleease~
THanks in advance :D

sorry i don't think i was very clear..
the question was
3(n+1)! / 5n!

Answer 1

Well, recalling that factorial means, for example,
4! = 4*3*2*1

Then, we can write n! as n*(n-1)*(n-2)*...*1
Or, (n+1)! as (n+1)*n*(n-1)*...*1
Noticing the "n*(n+1)*..." part of the (n+1)! is the same as n!, we can rewrite:
(n+1)! = (n+1)*n!

Substitute that in and solve:

3(n+1)! / 5n!
3(n+1)n! / 5n! (n!'s cancel)
3(n+1)/5 [ = (3/5)(n+1) ]

Answer 2

The factorial of n = n * (n - 1) * (n - 2) * ... * (2) * (1)

So, you can rewrite the problem to look like this:
3 (n+1) (n)!
------------------
5n!

Now, the n! in the numerator cancels out with the n! in the denominator, leaving:
3 (n+1)
-----------
5

Answer 3

3(n+1)! / 5n! = 3(n + 1)/5

since

(n + 1)! = (n + 1) * n!

So the n! cancels
.

Answer 4

n! = n(n-1)(n-2)...1*0! where 0! is defined to be 1. So we have

3(n+1)!/5n! = 3(n+1)n!/5n! = 3(n+1)/5

Math Rules!

Answer 5

3 (n + 1).(n).(n - 1)-------1
_____________________
5 (n).(n-1)---------1

= (3/5).(n + 1)