Question

simplifique: a) (n-2)!sobre(n+1)! b) (n-2)! sobre n! c) (n+3)! sobre (n+1)! d) (2n)! sobre 2(2n-2)! e) (n-2)!(n+2)! sobre (n+3)!(n-1) ....espero q vcs tenham entendido....me ajudem pf

Answer 1

a)

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

b)

$\frac{(n-2)!}{n!}=\frac{(n-2)!}{n(n-1)(n-2)!}=\frac{1}{n(n-1)}=\frac{1}{n^2-n}$

c)

$\frac{(n+3)!}{(n+1)!}=\frac{(n+3)(n+2)(n+1)!}{(n+1)!}=(n+3)(n+2)=n^2+5n+6$

d)

$\frac{(2n)!}{2(2n-2)!}=\frac{2n(2n-1)(2n-2)!}{2(2n-2)!}=2n-1$

e)

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