（bn,an是数列,{an}=3n-5,an是首项为-2,公差为3的等差数列）bn=2^an,求数列bn的前n项S

bn=2^an =2^(3n-5) =2^(3n)*2^(-5)b(n-1)=2^[3(n-1)]*2^(-5)=2^(3n)*2^(-3)*2^(-5)bn/b(n-1)=[2^(3n)*2^(-5)]/[2^(3n)*2^(-3)*2^(-5)] =2^3∴bn是以公比q=2^3的等比数列b1=2^(3*1)*2^(-5)...