（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

n=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)
=2^(-2)
Sn=b1(q^n-1)/(q-1)
=2^(-2)[(2^3)^n-1]/(2^3-1)
=2^(-2)(2^3n-1)/7
=[2^(3n-2)-2^(-2)]/7

对sn=2^-2+2^1+2^3+.....+2^（3n-5），首项为14,公比为8，可得出答案。good luck to you

s=-1/28（1-8的n次方）