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

(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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