Question

4) (PUC) Se (n−1)! = 1 , então n é igual a:
(n+1)!−n! 81
a) 13 b) 11 c) 9 d) 8 e) 6

Answer 1

Resposta: c)

Explicação passo-a-passo:

$\sf \dfrac{(n-1)!}{(n+1)!-n!}=\dfrac{1}{81}\\\\\\ \sf \dfrac{(n-1)!}{(n+1)\cdot n \cdot (n-1)!-n\cdot (n-1)!}=\dfrac{1}{81}\\\\\\ \sf \dfrac{(n-1)!\cdot 1}{(n-1)!\cdot \big[n\cdot(n+1)-n\big]}=\dfrac{1}{81}\\\\\\ \sf \dfrac{1}{n\cdot(n+1)-n}=\dfrac{1}{81}\\\\\\ \sf \dfrac{1}{n\cdot \big[(n+1)-1\big]}=\dfrac{1}{81}\\\\\\ \sf \dfrac{1}{n\cdot n}=\dfrac{1}{9\cdot 9}\\\\\\ \sf \dfrac{1}{n^2}=\dfrac{1}{9^2}\\\\\\ \sf n^2=9^2\\\\ \boxed{\boxed{\sf n=9}}$

n é um número natural, por isso n = 9 é a única solução aceitável.

Answer 2

Resposta: letra c)

Explicação passo-a-passo:

$\dfrac{(n-1)!}{(n+1)!-n!}=\dfrac{1}{81}\\\\\\ \dfrac{(n-1)!}{(n+1)\cdot (n+1-1)\cdot (n+1-1-1)!-n\cdot (n-1)!}=\dfrac{1}{81}\\\\\\ \dfrac{(n-1)!}{(n+1)\cdot n\cdot (n-1)!-n\cdot (n-1)!}=\dfrac{1}{81}\\\\\\ \dfrac{(n-1)!\cdot 1}{(n-1)!\cdot \big[n\cdot(n+1)-n\big]}=\dfrac{1}{81}\\\\\\ \dfrac{1}{n\cdot(n+1)-n}=\dfrac{1}{81}\\\\\\ \dfrac{1}{n\cdot \big[(n+1)-1\big]}=\dfrac{1}{81}\\\\\\ \dfrac{1}{n\cdot n}=\dfrac{1}{9\cdot 9}\\\\\\ \boxed{n=9}$