{an}是等差数列,bn={1/2}^an,已知b1+b2+b3=21/8,b1b2b3=1/8,(1)求an (2)求bn(3)求Sn=b1+b2+…+bn

题目

{an}是等差数列,bn={1/2}^an,已知b1+b2+b3=21/8,b1b2b3=1/8,(1)求an (2)求bn(3)求Sn=b1+b2+…+bn
{an}是等差数列,bn={1/2}^an,已知b1+b2+b3=21/8,b1b2b3=1/8,(1)求an (2)求bn(3)求Sn（Sn=b1+b2+…+bn)

答案

b1b2b3=(1/2)^(a1+a2+a3)=1/8所以a1+a2+a3=3设an公差为d,则3a2=3,a2=1,b2=1/2bn/b(n-1)=(1/2)^[an-a(n-1)]=(1/2)^d所以bn是等比数列b1+b3=17/8,b1b3=1/4所以b1=2,b3=1/8或b1=1/8,b3=2b1=(1/2)^a1,所以a1=-1或a1=3所...