设函数y=f(x)的定义域与值域均为R,其反函数为y=f-1(x),且对任意实数x都有f(x)+23f−1(x)=53x

1个回答

  • (Ⅰ)∵an+1=f(an),∴an=f-1(an+1

    ∵任意实数x都有f(x)+

    2

    3f−1(x)=

    5

    3x,∴f(an+1)+

    2

    3f−1(an+1)=

    5

    3an+1

    ∵an+1=f(an),即an=f-1(an+1),∴f(an+1)=an+2,f-1(an+1)=an

    ∴an+2+

    2

    3an=

    5

    3an+1,即an+2−an+1=

    2

    3(an+1−an)

    ∵bn=an+1-an(n∈N*),a1=1,a2=

    5

    3]

    ∴数列{bn}是以a2−a1=

    5

    3−1=

    2

    3为首项,以q=

    2

    3为公比的等比数列

    故数列{bn}的通项为bn=(

    2

    3)n

    (Ⅱ)(文)由bn=an+1−an=(

    2

    3)n得an-a1=(an-an-1)+(an-1-an-2)+…+(a2-a1

    =(

    2

    3)n−1+(

    2

    3)n−2+…+(

    2

    3)2+

    2

    3=2[1−(

    2

    3)n−1]

    又∵a1=1,∴an=3−

    2n

    3n−1(n∈N*),即数列{an}是递增数列,且an<3(n∈N*

    ∴满足

    m−1

    2m+1>an对所有n∈N*恒成立的参数m必须满足