Question

a)Quanto é An,₃ = 6(n-2)
b)Quanto é An,₄ = 12. An,₂


resolução
a) 3
b) 6

Answer 1

AE Fábio,

use o conceito de fatorial, com n ∈ IN..

$\dfrac{n!}{(n-3)!}=6\cdot(n-2)\Rightarrow \dfrac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)!}{(n-3)!}=6\cdot(n-2)\\\\\\ \Rightarrow \dfrac{n\cdot(n-1)\cdot(n-2)\cdot(\not n-3)!}{(\not n-3)!}=6\cdot(n-2)\\\\\\ \Rightarrow n\cdot(n-1)\cdot(\not n-2)=6\cdot(\not n-2)\\\\ \Rightarrow n^2-n=6\\\\ \Rightarrow n^2-n-6=0\\\\\ n_1=-2~~\notin \mathbb{N}~~e~~n_2=3\\\\\\ \Large\boxed{\text{S}=\{3\}}$

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$\dfrac{n!}{(n-4)!}=12\cdot \left\{\dfrac{n!}{(n-2)}\right\}\\\\\\ \Rightarrow \dfrac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)!}{(n-4)!}= 12\cdot\left\{\dfrac{n\cdot(n-1)\cdot(n-2)!}{(n-2)!}\right\}\\\\\\ \Rightarrow n\cdot(n-1)\cdot(n-2)\cdot(n-3)=12\cdot\{n\cdot(n-1)\}\\\\ \Rightarrow (n-2)\cdot(n-3)=12\\ \Rightarrow n^2-3n-2n+6=12\\ \Rightarrow n^2-5n-6=0\\\\ n_1=-1~\notin\mathbb{N}~~e~~n_2=6\\\\\\ \Large\boxed{\text{S}=\{6\}}$