若数列{An}满足前n项之和Sn=2An-4(n∈N*),Bn+1=An+2Bn,且B1=2(1)求数列{An}的通项公式An (2)求证数列{(Bn)/2^n}是等差数列,并求Bn(3)求数列{Bn}的前N项个Tn麻烦写详细过程
问题描述:
若数列{An}满足前n项之和Sn=2An-4(n∈N*),Bn+1=An+2Bn,且B1=2
(1)求数列{An}的通项公式An
(2)求证数列{(Bn)/2^n}是等差数列,并求Bn
(3)求数列{Bn}的前N项个Tn
麻烦写详细过程
答
(1)Sn=2An-4……(1),n=1代入,得:A1=4,Sn+1=(2An+1)-4……(2),(2)-(1),得:An+1=2An+1-2An,An+1=2An,数列{An}是首项为4,公比为2的等比数列,An=2^(n+1)(2)Bn+1=An+2Bn=2Bn+2^(n+1)Bn+1/2^(n+1)=(Bn/2^n)+1即...