已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+(1/2)bn=1.

问题描述:

已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+(1/2)bn=1.
(1)求证:数列{bn}是等比数列
(2)记cn=ab*bn,求{cn}的前n项和Sn

a5-a2=3d=18-6d=4a1=a2-d=2an=4n-2Tn+1/2bn=1所以T(n-1)+1/2b(n-1)=1相减,Tn-T(n-1)=bnbn+1/2bn-1/2b(n-1)=03/2bn=1/2n(n-1)bn/b(n-1)=1/3所以是等比数列 T1=b1所以b1+1/2b1=1b1=2/3,q=bn/b(n-1)=1/3bn=2/3*(1/3)^(n...