在等差数列{an},S10=185 a2=8 若从数列{an}中,依次取出a2,a4,a8……a2^n构成{bn}求{bn}前n项和

设{an}首项为a1,公差为d,
则 a1+d=8
10(2a1+9d)/2=185,
解得
a1=5
d=3
an=5+3(n-1),
∴an=3n+2
设b1=a2,b2=a4,b3=a8,
则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