已知数列{An},Sn是其前n项和,且满足3An=2Sn+n,n为正整数,求证数列{An+1/2}为等比数列

问题描述:

已知数列{An},Sn是其前n项和,且满足3An=2Sn+n,n为正整数,求证数列{An+1/2}为等比数列
1、已知数列{An},Sn是其前n项和,且满足3An=2Sn+n,n为正整数
(1)、求证数列{An+1/2}为等比数列;
(2)、记Tn=S1+S2+S3+.+Sn,求Tn的表达式.
2、已知数列{An}满足A1=1.A2=2,A(n+2)=An+A(n+1)/2,n为正整数
(1)、令Bn=A(n+1)-An,证明:{Bn}是等比数列.
(2)、求{An}的通项式.
求第二题的答案~

1.
证:
Sn=(3an-n)/2
Sn-1=[3a(n-1)-(n-1)]/2
an=Sn-Sn-1=[3an-3a(n-1)-1]/2
an=3a(n-1)+1
an+1/2=3a(n-1)+3/2=3[a(n-1)+1/2]
(an+1/2)/[a(n-1)+1/2]=3,为定值,因此
{An+1/2}为等比数列.
令n=1
3a1=2a1+1
a1=1
a1+1/2=3/2
Tn=S1+S2+...+Sn
=(1/2)(2S1+2S2+...+2Sn)
=(1/2)(3a1-1+3a2-2+...+3an-n)
=[-n(n+1)/4]+(3/2)(a1+a2+...+an)
=[-n(n+1)/4]+(3/2)[a1+1/2+a2+1/2+...+an+1/2-n/2]
=[-n(n+1)/4]-3n/4+(3/2)(3/2)(3^n-1)/2
=[9(3^n-1)-2n(n+4)]/8