已知数列{an}满足a1=1,a(n+1)=an+√((an)^2+1),令an=tanθn(0

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已知数列{an}满足a1=1,a(n+1)=an+√((an)^2+1),令an=tanθn(0<θn<π/2),求证(1)
已知数列{an}满足a1=1,a(n+1)=an+√((an)^2+1),令an=tanθn(0<θn<π/2),求证（1）数列{θn-π/2}是等比数列（2）a1+a2+…+an>(n-1)π/2

(1)
a(n+1)=an+√((an)^2+1)
a(n+1)=tan(θ(n+1))
an+√((an)^2+1)=tan(θn)+√(tan^2(θn)+1)=tan(θn)+1/(cos(θn))
=(sin(θn)+1)/(cos(θn))
=(sin(θn)+sin^2(θn/2)+cos^2(θn/2))/(cos(θn))
=(2*sin(θn/2)*cos(θn/2)+sin^2(θn/2)+cos^2(θn/2))/(cos^2(θn/2)-sin^2(θn/2))
=(sin(θn/2)+cos(θn/2))^2/((sin(θn/2)+cos(θn/2))(cos(θn/2)-sin(θn/2)))
=(sin(θn/2)+cos(θn/2))/(cos(θn/2)-sin(θn/2)))
=(tan(θn/2)+1)/(1-tan(θn/2))
=tan(θn/2+π/4)
即θ(n+1)=θn/2+π/4
θ(n+1)-π/2=(1/2)*(θn-π/2)
故{θn-π/2}是等比数列
(2)
a1=tan(θ1)=1
0(n-1)*π/2

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