已知数列{a n }中,a 1 =3,a 2 =5,S n 为其前n项和,且满足S n +S n-2 =2S n-1 +

1个回答

  • (1)由 S n + S n-2 =2 S n-1 + 2 n-1 得 a n = a n-1 + 2 n-1 (n≥3,n∈N*) ,

    ∵a 2=5,∴当n≥3时,a n=a 2+(a 3-a 2)+(a 4-a 3)+…+(a n-a n-1)=5+2 2+2 3+…+2 n-1=2 n+1,

    经验证a 1=3,a 2=5也符合上式,

    ∴ a n = 2 n +1(n∈ N * ) ;

    (2)由(1)可得 b n =

    n

    a n -1 =

    n

    2 n ,

    ∴ T n =

    1

    2 +

    2

    2 2 +

    3

    2 3 +…+

    n

    2 n ① ⇒

    1

    2 T n =

    1

    2 2 +

    2

    2 3 +…+

    n-1

    2 n +

    n

    2 n+1 ②,

    ①-②有:

    1

    2 T n =

    1

    2 +

    1

    2 2 +

    1

    2 3 +…+

    1

    2 n -

    n

    2 n+1 =1-

    1

    2 n -

    n

    2 n+1 ,

    ∴ T n =2-

    n+2

    2 n ;

    (3)∵ f(x)= 2 x-1 , c n =

    1

    a n a n+1 ,

    ∴ c n f(n)=

    2 n-1

    ( 2 n +1)( 2 n+1 +1) =

    1

    2 (

    1

    2 n +1 -

    1

    2 n+1 +1 )(n∈N*) ,

    ∴Q n=c 1f(1)+c 2f(2)+…+c nf(n)

    =

    1

    2 [(

    1

    2 1 +1 -

    1

    2 2 +1 )+(

    1

    2 2 +1 -

    1

    2 3 +1 )+…+(

    1

    2 n +1 -

    1

    2 n+1 +1 )]

    =

    1

    2 (

    1

    1+2 -

    1

    2 n+1 +1 )<

    1

    2 ×

    1

    3 =

    1

    6 .