# Math Help - Simplify ratios of factorials

1. ## Simplify ratios of factorials

$\frac{(n+1)!}{n!} = \frac{n!(n+1)}{n!} = (n+1)$

I understand canceling, but I don't see how the initial fraction became $\frac{n!(n+1)}{n!}$.

Same with $\frac{(2n-1)!}{(2n+1)!} = \frac{(2n-1)!}{(2n-1)!(2n)(2n+1)} = \frac{1}{2n(2n+1)}$.

2. Originally Posted by cinder
$\frac{(n+1)!}{n!} = \frac{n!(n+1)}{n!} = (n+1)$

I understand canceling, but I don't see how the initial fraction became $\frac{n!(n+1)}{n!}$.

Same with $\frac{(2n-1)!}{(2n+1)!} = \frac{(2n-1)!}{(2n-1)!(2n)(2n+1)} = \frac{1}{2n(2n+1)}$.
do you know what factorial means?

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

but $n! = n(n - 1)(n - 2)(n - 3)...(3)(2)(1)$

so we have: $(n + 1)! = (n + 1)n!$

similarly, $(2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2)...(3)(2)(1) = (2n + 1)(2n)(2n - 1)!$

3. Originally Posted by Jhevon
do you know what factorial means?

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

but $n! = n(n - 1)(n - 2)(n - 3)...(3)(2)(1)$

so we have: $(n + 1)! = (n + 1)n!$

similarly, $(2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2)...(3)(2)(1) = (2n + 1)(2n)(2n - 1)!$
I know what factorial means, but I guess I don't understand when there are variables involved.

4. Originally Posted by cinder
I know what factorial means, but I guess I don't understand when there are variables involved.
well, did my illustration help?

5. Originally Posted by Jhevon
well, did my illustration help?
No.

6. Originally Posted by cinder
No.
ok, let's start from the basics and build our way up.

the factorial of an integer is defined as the product of all integers starting from 1 going up to the integer.

example, 5! = 1*2*3*4*5 and 6! = 1*2*3*4*5*6

so the integers go up in order until we get to the integer we are finding the factorial of. but how do we go from one integer to the other? we simply add 1 to the current integer to get the next integer. equivalently, this means we can subtract one from an integer to get the previous integer and keep doing so until we hit 1 and stop, which is what we do here. take 5! again as an example.

we can say that 5! = 5*4*3*2*1, the same thing, just going backwards. since we are going backwards, we subtract 1 from the first integer to get the second, and then subtract 1 from the second integer to get the third and so on. so we can say,

5! = 5*(5 - 1)*((5 - 1) - 1)*(((5 - 1) - 1) - 1)*((((5 - 1) - 1) - 1) - 1)

which is just simpler to write as

5! = 5*(5 - 1)(5 - 2)(5 - 3)(5 - 4)

if the factorial is very long or indefinite, we can omit some of the terms in the middle, again, let's use 5! as an illustration

5! = 5*(5 - 1)*...*(5 - 3)(5 - 4)

but we usually simplify the end, so instead of 5 - 4 we write 1 and instead of 5 - 3 we write 2, so we get:

5! = 5*(5 - 1)*...*2*1

but this representation can work for any integer. if we want to have an arbitrary integer factorial, we can use n. so in general:

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

there is no difference if we want (n + 1)!, we just start at n + 1 and subtract 1 until we get to 1

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

so here is where the algebraic manipulation comes in.

recall that n! = n(n - 1)(n - 2)(n - 3)....(3)(2)(1)

now let's write (n + 1)! out.

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

notice anything? yes, what's in red is n! so we can just replace it

thus we get, (n + 1)! = (n + 1)n!

do you get it now?

7. I'm 99% sure I got it (I'll reread to verify my thinking). Thank you very much for the explanation!

8. Originally Posted by cinder
I'm 99% sure I got it (I'll reread to verify my thinking). Thank you very much for the explanation!
well, if you have any doubts, come back