Question

Demonstrați ca {2ridicat la puterea {3n+1}x3 ridicat la puterea {n+1} + 8 ridicat la puterea {n}x3 ridicat la puterea {n+3}+4 ridicat la puterea {n}x 6 ridicat la puterea {n+1}) este divizibil cu 13. Va rog ajutați-mă ! Dau 50 de puncte !

Answer 1

$2^{3n+1} *3 ^{n+1} +8 ^{n} *3 ^{n+3} +4 ^{n} *6 ^{n+1} =$
$2*2^{3n} *3 ^{n+1} +(2 ^{3}) ^{n} *3 ^{n+3} + (2^{2}) ^{n} *(2*3) ^{n+1} =$
$2*2^{3n} *3 ^{n+1} +2 ^{3} ^{n} *3 ^{n+3} + 2^{3} ^{n} *2*3 ^{n+1} =$
$2^{3n} *3 ^{n+1}*( 2 +3 ^{2} + 2) =2^{3n} *3 ^{n+1}*13$