从第二问开始吧,不要用“观察法”……(1) b1=π/3 ,b2=π/6(2)bn=(π/3)(1/2)^(n-1)(3

(2) a1=√3 b1= arctana1=π/3
a(n+1)=[√（1+an²)-1]/an
=[√(1+tanbn²)-1]/tanbn
=[√(cos²bn+sin²bn)/cos²bn-1]/(sinbn/cosbn)
=(1-cosbn)/sinbn
=tanb(n+1)
∴ tanb(n+1)=sinb(n+1)/cosb(n+1)=(1-cosbn)/sinbn
∴sinbnsinb(n+1)+cosbncosb(n+1)=cosb(n+1)
∴cos[b(n+1)-bn]=cosb(n+1)
∵0