Question

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

Answer 1

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

 

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Answer 2

(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 3

To simplify the expression $(n+2)!(n^2-9)/(n+4)!$, let's break it down step by step:

First, let's expand the factorials: $(n+2)! = (n+2)(n+1)n!$ $(n+4)! = (n+4)(n+3)(n+2)!$

Substituting these expressions back into the original expression, we have: $(n+2)!(n^2-9)/(n+4)! = \frac{(n+2)(n+1)n!(n^2-9)}{(n+4)(n+3)(n+2)!}$

Now, we can simplify some terms: $(n+2)!$ cancels out with one of the $(n+2)!$ terms in the denominator: $\frac{(n+2)(n+1)n!(n^2-9)}{(n+4)(n+3)(n+2)!} = \frac{(n+1)n!(n^2-9)}{(n+4)(n+3)}$

Next, we can cancel out the common factors: $n!$ cancels out with $n!$ in the numerator: $\frac{(n+1)n!(n^2-9)}{(n+4)(n+3)} = \frac{(n+1)(n^2-9)}{(n+4)(n+3)}$

Finally, let's expand and simplify the numerator: $(n+1)(n^2-9) = n^3 - 8n - 9$

Therefore, the simplified form of the expression $(n+2)!(n^2-9)/(n+4)!$ is $\frac{n^3 - 8n - 9}{(n+4)(n+3)}$.