1. ## simplifying n factorials

can someone explain this process:

the original question is f(n) = [(2^n)(n!)]/(n+2)!] and the solution says it simplifies to f(n) = [(2^n)]/ (k+1)(k+2)] can someone explain to me how the cancelled out the factorials and the addition of the (k+1) in the denominator?

thanks

2. not sure about the k part but

$\displaystyle f(n) = \frac{2^nn!}{(n+2)!} = \frac{2^nn!}{(n+2)(n+1)n!}= \frac{2^n}{(n+2)(n+1)}$

3. Originally Posted by pickslides
not sure about the k part but

$\displaystyle f(n) = \frac{2^nn!}{(n+2)!} = \frac{2^nn!}{(n+2)(n+1)n!}= \frac{2^n}{(n+2)(n+1)}$
wait for the 2nd step how did u get (n+1)n!??

4. Originally Posted by johntuan
wait for the 2nd step how did u get (n+1)n!??
$\displaystyle (n+2)! = (n+2)(n+1)\textcolor{red}{(n)(n-1)(n-2) ...}$

what is that in red?

5. ## Re: simplifying n factorials

1 + [tex] cos [tex] theta [tex] / 2

6. ## Re: simplifying n factorials

Originally Posted by anil86
1 + [tex] cos [tex] theta [tex] / 2
What on Earth?? Did you mean this to go into another thread or something?

-Dan

7. ## Re: simplifying n factorials

Originally Posted by johntuan
wait for the 2nd step how did u get (n+1)n!??
It comes directly from the definition of "factorial". n! is defined as "n(n-1)(n-2)...(3)(2)(1)". That is "n times all positive integers less than n". (n+2)! is "n+ 2 times all positive integers less than n+ 2" or (n+2)(n+1)(n)(n-1(n-2)...(3)(2)(1)= (n+2)(n+1)(n!).