(1)bn,√an,bn+1成等比
所以an=bn*bn+1
所以a1=b1*b2=3 a2=b2*b3=6
所以b1*(b1+d)=3 (b1+d)*(b1+2d)=6
解得：b1=√2 d=√2/2或者b1=-√2 d=-√2/2舍去
所以bn=b1+(n-1)d=√2(n+1)/2
(2)an=bn*bn+1=(n+1)(n+2)/2
所以1/an=2/(n+1)(n+2)=2(1/(n+1)-1/(n+2))
所以Sn=1/a1+1/a2+.+1/an
=2(1/2-1/3+1/3-1/4……+1/(n+1)-1/(n+2))
=1-2/(n+2)