设数列{an}前n项和为Sn,且Sn=2An-2,令bn=log2an.试求数列{an}的通项公式.设Cn=Bn/an,求证{an}的Tn

Sn=2An-2,S(n+1)=2A(n+1)-2,S(n+1)-Sn=A(n+1)=2A(n+1)-2AnA(n+1)=2An,A(n+1)/An=2S1=A1=2A1-2,A1=1An=2的n次方Cn=Bn/An=n/(2的n次方),是差比数列,即n*(1/2)的－n次方Tn=n*(1/2)的－n 1式（1/2）Tn=n*(1/2)de-(n+1) 2...