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