Question

若数列{An}满足前n项之和Sn=2An-4(n∈N*),Bn+1=An+2Bn,且B1=2

Answer 1

(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(Bn)/2^n是公差为1的等差数列,首项为B1/2^1=2/2=1,通项 (Bn)/2^n=n

(3)(Bn)/2^n=n,Bn=n*2^n

Tn=1*2^1+2*2^2+……+n*2^n

2Tn=1*2^2+2*2^3+……+(n-1)*2^n+n*2^(n+1)

相减得,-Tn=2^1+2^2+2^3+……+2^n-n*2^(n+1)

=2^(n+1)-2-n*2^(n+1)

=(1-n)*2^(n+1)-2

Tn=(n-1)*2^(n+1)+2