设数列{An}的前n项和Sn=2A(n)-1(n=1,2,3…),数列{bn}满足条件b1=3,bk+1=Ak+bk(k=1,2,

问题描述:

设数列{An}的前n项和Sn=2A(n)-1(n=1,2,3…),数列{bn}满足条件b1=3,bk+1=Ak+bk(k=1,2,
求数列bn的前n项和sn’
(2)设cn=3/(n+2)(Sn’-Sn),且数列cn的前n项和为Tn,求证Tn重点是第二问Tn怎么求

Sn=2^n-1,Sn'=2n+2^n-1,所以Sn'-Sn=2n.
这里的n+2与(Sn’-Sn)应该都是分母上吧.Cn=3/[2n(n+2)=3/2×1/2×(1/n-1/(n+2)),相加,两两抵消,得Tn=3/4×[1+1/2-1/(n+1)-1/(n+2)]<3/4×(1+1/2)=9/8