Question

arata ca nr 6 la puterea lui n +3 la puterea lui n +2 la puterea lui n +1 este nr compus oricare ar fi n nr natural

Answer 1

$6^{n}+3^{n}+2^{n}+1=$

$=(2*3)^{n}+3^{n}+(2^{n}+1)=$

$=2^{n}*3^{n}+1*3^{n}+(2^{n}+1)=$

$=3^{n}(2^{n}+1)+1*(2^{n}+1)=$

$=(2^{n}+1)(3^{n}+1)$


$n~\in~\math{N}~=>min~n=0$

$=>min~2^{n}+1=2^{0}+1=1+1=2>1$

$=>min~3^{n}+1=3^{0}+1=1+1=2>1$


$=>~numarul~este~compus~oricare~ar~fi~n~numar~natural$