Question

Va rog frumos, nu inteleg ce vrea pb asta si nici nu prea sunt eu cu demonstratiile ... ma poate ajuta cineva? Multumesc mult! ^^ >:]​

Answer 1

$4^{n+1}$·$3^n$·6+$12^{n+1$·3+$2^{2n+1$·$3^n$·4=
4·$4^n$·$3^n$·6+$12^n$·12·3+2·$2^{2n}$·$3^n$·4=
4·6·(4·3$)^n$+$12^n$·36+2·4·$(2^2)^n$·$3^n$=
24·$12^n$+36·$12^n$+8·(4·3$)^n$=
24  ·$12^n$+36·$12^n$+8·$12^n$=
$12^n$(24+36+8)=
$12^n$·68
dar 68=4·17,
deci a=$12^n$·4·17, adica a e divizibil cu 17 pentru orice numar natural n