(1)S3=3a2=a2^2
a2=0(舍去)或3
a5=a2^2=9
d=(a5-a2)/3=2
所以an=2n-1
(2)bn-b(n-1)=2^a(n-1)=1/2*4^(n-1)
bn=(bn-b(n-1))+(b(n-1)-b(n-2))+…+(b2-b1)+b1
=1+1/2*4+…+1/2*4^(n-1)
=1+1/2*(1-4^(n-1))/(1-4)=1/6*4^(n-1)+5/6
(1)S3=3a2=a2^2
a2=0(舍去)或3
a5=a2^2=9
d=(a5-a2)/3=2
所以an=2n-1
(2)bn-b(n-1)=2^a(n-1)=1/2*4^(n-1)
bn=(bn-b(n-1))+(b(n-1)-b(n-2))+…+(b2-b1)+b1
=1+1/2*4+…+1/2*4^(n-1)
=1+1/2*(1-4^(n-1))/(1-4)=1/6*4^(n-1)+5/6