出来てんじゃん。
んじゃゴリ押し出来るよ。

S_{n+1}=3S_n-4n+2^{n+1} (n ≧ 1)
[S_{n+1}-2(n+1)+2]=3[S_n-2n+2]+2^{n+1} (n ≧ 1)
[S_{n+1}-2(n+1)+2]/2^{n+1}=(3/2)[S_n-2n+2]/2^n+1 (n ≧ 1)
{[S_{n+1}-2(n+1)+2]/2^{n+1}+2}=(3/2){[S_n-2n+2]/2^n+2} (n ≧ 1)