设数列{a n }的前n项和为S n ,且a 1 =1,a n+1 =2S n +1,数列{b n }满足a 1 =b

1个回答

  • (Ⅰ)由a n+1=2S n+1可得a n=2S n-1+1(n≥2),

    两式相减得a n+1-a n=2a n

    a n+1=3a n(n≥2).

    又a 2=2S 1+1=3,

    所以a 2=3a 1

    故{a n}是首项为1,公比为3的等比数列.

    所以a n=3 n-1

    由点P(b n,b n+1)在直线x-y+2=0上,所以b n+1-b n=2.

    则数列{b n}是首项为1,公差为2的等差数列.

    则b n=1+(n-1)•2=2n-1

    (Ⅱ)因为 c n =

    b n

    a n =

    2n-1

    3 n-1 ,所以 T n =

    1

    3 0 +

    3

    3 1 +

    5

    3 2 ++

    2n-1

    3 n-1 .

    1

    3 T n =

    1

    3 1 +

    3

    3 2 +

    5

    3 2 ++

    2n-3

    3 n-1 +

    2n-1

    3 n ,

    两式相减得:

    2

    3 T n =1+

    2

    3 +

    2

    3 2 ++

    2

    3 n-1 -

    2n-1

    3 n .

    所以 T n =3-

    1

    2• 3 n-2 -

    2n-1

    2• 3 n-1 = 3-

    n+1

    3 n-1 .