Correct Cancelation

[Silvanoshei]

Let me see if this is right...

$\displaystyle \frac{(n+1)!}{(n+1)^{550}}*\frac{n^{550}}{n!}$

Would the Factorials cancel because it's only a + 1 ? Or does n+1! not cancel n!?

 

[JeffM]

Let me see if this is right...

$\displaystyle \frac{n+1!}{(n+1)^{550}}*\frac{n^{550}}{n!}$

Would the Factorials cancel because it's only a + 1 ? Or does n+1! not cancel n!?

One way to define the factorials is this

$\displaystyle 0! = 1.$

$\displaystyle 0 < n \in \mathbb Z \implies n! = n * (n - 1)!.$

So $\displaystyle (n + 1)! = (n + 1) * n!.$

Given the definitions, try simplifying again.

 

[Silvanoshei]

So you'd be left with $\displaystyle \frac{n+1}{n+1}$?

Since the n! cancel and ^500 cancel?

Hmmm... that doesn't work because of the factorial. I suppose another way to do go about would be $\displaystyle (n+1)\frac{n^{550}}{(n+1)^{550}}$

then...

$\displaystyle \frac{n^{550}}{(n+1)^{549}}$?

 

[lookagain]

I suppose another way to do go about would be $\displaystyle (n+1)\frac{n^{550}}{(n+1)^{550}}$

then...

$\displaystyle \frac{n^{550}}{(n+1)^{549}}$? $\displaystyle \ \$ <---- Yes.

.

 

[JeffM]

So you'd be left with $\displaystyle \frac{n+1}{n+1}$?

Since the n! cancel and ^500 cancel?

Hmmm... that doesn't work because of the factorial.

The factorials have nothing to do with why it does not work. It doesn't work because you cannot cancel exponents when what is being exponentiated is different:

$\displaystyle \dfrac{4^3}{3^3} = \dfrac{64}{27} \ne \dfrac{4}{3}.$


I suppose another way to do go about would be $\displaystyle (n+1)\frac{n^{550}}{(n+1)^{550}}$

then...

$\displaystyle \frac{n^{550}}{(n+1)^{549}}$?

This is correct, but perhaps not as simple as possible

$\displaystyle \dfrac{(n + 1)!}{(n + 1)^{550}} * \dfrac{n^{550}}{n!} = \dfrac{(n + 1)!}{n!} * \dfrac{n^{550}}{(n + 1)^{550}} = \dfrac{(n + 1) * n!}{n!} * \dfrac{n^{550}}{(n + 1)^{550}} = \dfrac{n^{550}}{(n + 1)^{549}} = n\left(\dfrac{n}{n + 1}\right)^{549}.$