公差为d的等差数列{an},d不等于0,a1=0,bn=2^an,Sn是{bn}的前n项和,Tn=Sn/bn

因为是等差数列所以An=a1+（n-1）d则有：a2=a1+da4=a1+3da8=a1+7d又因为A2、A4、A8成等比数列所以a2/a4=a4/a8（a4）2=a2×a8（a1+3d）2=（a1+d）（a1+7d）（a1）2+2（3a1d）+（3d）2=（a1）2+7a1d+a1d+7d22d2=2a1da1...