f(x)= x 2 + x,数列{a n }的前n项和为S n ,点(n,S n )(n∈N * )均在函数y=f(x)

1个回答

  • f(x)=

    x 2

    x,数列{a n}的前n项和为S n,点(n,S n)(n∈N *)均在函数y=f(x)的图象上.

    (Ⅰ)求数列{a n}的通项公式a n

    (Ⅱ)令b n

    ,求数列{b n}的前n项和T n

    (Ⅰ)∵点(n,S n)在f(x)的图象上,

    ∴S n

    n 2

    n.

    当n≥2时,a n=S n-S n -1=n+1;当n=1时,a 1=S 1=2,满足上式,∴a n=n+1(n∈N *).

    (Ⅱ)b n

    T n=b 1+b 2+…+b n=2+

    +…+

    ,①

    T n

    +…+

    ,②

    由①-②,得

    T n=2+

    +…+

    =(1+

    +…+

    )+(1-

    )=

    +1-

    =2(1-

    )+1-

    ∴T n=6-

    .