Question

Cate numere naturale 'n' exista,astfel incat fractia: $\frac{2^{n+3}+ 2^{n+1}}{ 3^{n+2}+ 3^{n}}$

Answer 1

    
$\displaystyle \frac{2^{n+3}+ 2^{n+1}}{ 3^{n+2}+ 3^{n}} =\frac{ 2^n(2^{3}+ 2^{1})}{3^n(3^{2}+ 3^{0})} = \\ \\ =\frac{ 2^n(8+ 2)}{3^n(9+ 1)}=\frac{ 2^n \times 10}{3^n \times 10)} = \boxed{\frac{ 2^n }{3^n }} \\ \\ Fractia ~ \frac{ 2^n }{3^n } ~este~ subunitara~daca:~~ \boxed{n \in N^*} \\ \\ Fractia ~ \frac{ 2^n }{3^n } ~este~ echiunitara~daca:~~ \boxed{n \in \{0\}} \\ \\ Fractia ~ \frac{ 2^n }{3^n } ~este~ supraunitara~daca:~~ \boxed{n \in \emptyset ~~(multime ~vida)}$

$\displaystyle \Longrightarrow ~~ fractia~~ \frac{2^n}{3^n} ~nu ~poate~ fi ~supraunitara~pentru \\ nicio~valoare~\underline{naturala}~a~lui~~n.$