Question

aratati ca ( 2 la puterea n+2) divide pe ( 4 la puterea n+1 + 4 la puterea n), pentru orice n apartine N.

Answer 1

$2^{n+1}$+$2^{n}$ Ι $4^{n+1}$+$4^{n}$
$4^{n+1} + 4^{n} = 4^{n} * 4^{1} + 4^{n} = 4^{n}$(4+1)=$4^{n} * 5$
$2^{n+2}+ 2^{n}= 2^{n}* 2^{2}+ 2^{n}= 2^{n}$(4+1)=$2^{n} * 5$
$4^{n}*5= 2^{n}* 2^{n}*5$