设数列{a n }的各项都是正数,且对任意n∈N * ,都有(a n -1)(a n +3)=4S n ,其中S n 为

1个回答

  • (Ⅰ)∵(a n-1)(a n+3)=4S n,当n≥2时,(a n-1-1)(a n-1+3)=4S n-1

    两式相减,得

    a 2n -

    a 2n-1 +2 a n -2 a n-1 =4 a n ,即(a n+a n-1)(a n-a n-1-2)=0,又a n>0,∴a n-a n-1=2.

    当n=1时,(a 1-1)(a 1+3)=4a 1,∴(a 1+1)(a 1-3)=0,又a 1>0,∴a 1=3.

    所以,数是以3为首项,2为公差的等差数列.

    (Ⅱ)由(Ⅰ),a 1=3,d=2,∴a n=2n+1.

    设 b n =

    4

    a n 2 -1 ,n∈N *;∵a n=2n+1,∴ a n 2 -1=4n(n+1)))

    ∴ b n =

    4

    4n(n+1) =

    1

    n(n+1) =

    1

    n -

    1

    n+1

    ∴T n=b 1+b 2+b 3+…+b n= (1-

    1

    2 )+(

    1

    2 -

    1

    3 )+…+(

    1

    n -

    1

    n+1 ) = 1-

    1

    n+1 <1 .

    又∵ T n+1 - T n =

    n+1

    n+2 -

    n

    n+1 =

    1

    (n+2)(n+1) >0 ,∴ T n+1 > T n > T n-1 >…> T 1 =

    1

    2 ,

    综上所述:不等式

    1

    2 ≤ T n <1 成立.