Question

Simplifique:

$\frac{(n+3)!}{(n-2)!}. \frac{(n-1)!}{(n+2)!}$

Answer 1

Observe que, $(n+3)!=(n+3)(n+2)!$ e $(n-1)!=(n-1)(n-2)!$.

Assim:

$\dfrac{(n+3)!}{(n-2)!}\cdot\dfrac{(n-1)!}{(n+2!}=\dfrac{(n+3)(n+2)!\cdot(n-1)(n-2)!}{(n-2)!\cdot(n+2)!}=(n+3)(n-1)$

Answer 2

$\frac{(n+3)!}{(n-2)!}.\frac{(n-1)!}{(n+2)!}=\\ \\ \frac{(n+3)(n+2)!}{(n-2)!}.\frac{(n-1)(n-2)!}{(n+2)!}=(n+3)(n-1)=\boxed{n^2+2 n-3}$