Question

(n+2)!(n^2-9)/(n+4)!

Answer 1

(n+2)!(n^2-9)/(n+4)!  =

=(n+2)! (n^2-9)  / (n+2)! (n+3)(n+4)=

=(n-3)(n+3) /  (n+3)(n+4)=

=(n-3)/ (n+4)

Answer 2

$frac{(n+2)!(n^2-9)}{(n+4)!}=frac{(n+3)(n-3)}{(n+3)(n+4)}=frac{n-3}{n+4}$

 

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