Question

Resolva as expressões abaixo:
a)n!÷(n-2)!
b)n!÷(n+2)!
c)(n-1)!÷(n+1)!
d)(n+2)÷(n-1)!​​

Answer 1

Vamos utilizar a definição de fatorial para simplificar as expressões dadas.

$n!~=~n\cdot(n-1)\cdot(n-2)\cdot~...~\cdot(n-k)!\\\\ou\\\\n!~=~n\cdot(n-1)\cdot(n-2)\cdot~...~\cdot1$

a)

$\dfrac{n!}{(n-2)!}~=\\\\\\=~\dfrac{n\cdot(n-1)\cdot(n-2)!}{(n-2)!}\\\\\\=~\dfrac{n\cdot(n-1)\cdot1}{1}\\\\\\=~\boxed{n\cdot(n-1)}~~ou~~ \boxed{n^2-n}$

b)

$\dfrac{n!}{(n+2)!}~=\\\\\\=~\dfrac{n!}{(n+2-1)\cdot(n+2-1-1)!}\\\\\\=~\dfrac{n!}{(n+1)\cdot n!}\\\\\\=~\dfrac{1}{(n+1)\cdot1}\\\\\\=~\boxed{\dfrac{1}{n+1}}$

c)

$\dfrac{(n-1)!}{(n+1)!}~=\\\\\\=~\dfrac{(n-1)!}{(n+1)\cdot(n+1-1)\cdot(n+1-1-1)!}\\\\\\=~\dfrac{(n-1)!}{(n+1)\cdot(n)\cdot(n-1)!}\\\\\\=~\dfrac{1}{(n+1)\cdot n\cdot1}\\\\\\=~\boxed{\dfrac{1}{(n+1)\cdot n}}~~ou~~ \boxed{\dfrac{1}{n^2+n}}$

d)

$\dfrac{(n+2)!}{(n-1)!}~=\\\\\\=~\dfrac{(n+2)\cdot(n+2-1)\cdot(n+2-1-1)\cdot(n+2-1-1-1)!}{(n-1)!}\\\\\\=~\dfrac{(n+2)\cdot(n+1)\cdot(n)\cdot(n-1)!}{(n-1)!}\\\\\\=~\dfrac{(n+2)\cdot(n+1)\cdot n\cdot1}{1}\\\\\\=~\boxed{(n+2)\cdot(n+1)\cdot n}~~ou~~\boxed{n^3+3n^2+2n}$