$n\geq5$ icin $2^n+2\times2^n>2^n+2n^2$ oldugundan, problem $3^n>2^n+2\times2^n$ oldugunu gostermeye esdeger. Bu da $3^{n-1}>2^n$ in dogoru oldugunu gostermeye esdeger. Bunu gostermek ise kolay.
$1)\, n=5$ icin dogru.
$2)\, n=k$ icin dogru olsun. Yani $3^{k-1}>2^k$ olsun
$3)\, 3^{k}>2^{k+1}$ oldugunu gosterelim..
$3^{k-1}>2^k\implies 3^k>3\times2^k\implies 3^k>3\times2^k>2^{k+1}\implies$ $3^{k}>2^{k+1}$ olur.
$ 3\times2^k>2^{k+1}$ oldugunu gormek kolay, $3>2, \forall k$ dir cunku..