設(shè){an}首項為a1,公差為d,
則 a1+d=8
10(2a1+9d)/2=185,
解得
a1=5
d=3
an=5+3(n-1),
∴an=3n+2
設(shè)b1=a2,b2=a4,b3=a8,
則bn=a2^n = 3×2^n+2
∴Sbn=(3×2+2)+(3×2^2+2)+…+(3×2^n+2)
=3×(2+2^2+…+2^n)+2n
=3×2(2^n-1)/(2-1) +2n
= 6×2^n+2n -6
= 3×2^(n+1)+2n -6