已知正项数列{an}满足:an2-nan-(n+1)=0,数列{bn}的前n项和为Sn,且Sn=2bn-2.

1个回答

  • 解题思路:(Ⅰ)解方程an2-nan-(n+1)=0,得an,由Sn=2bn-2,得n≥2时,Sn-1=2bn-1-2,两式相减得bn的递推式,根据递推式可判断{bn}为等比数列,进而可求得bn

    (Ⅱ)由(Ⅰ)可得

    1

    a

    n

    lo

    g

    2

    b

    n

    ,拆项后利用裂项相消法可求得Tn

    (Ⅰ)由an2-nan-(n+1)=0,得an=n+1,或an=-1(舍去),

    ∴an=n+1;

    又Sn=2bn-2,∴n≥2时,Sn-1=2bn-1-2,

    两式相减,得bn=Sn-Sn-1=2bn-2bn-1

    ∴bn=2bn-1(n≥2),

    ∴{bn}为等比数列,公比q=2,

    又∵S1=b1=2b1-2,∴b1=2,

    ∴bn=2×2n−1=2n.

    (Ⅱ)由(Ⅰ)知,an=n+1,bn=2n,

    ∴[1

    an•log2bn=

    1

    (n+1)log22n=

    1

    n(n+1)=

    1/n−

    1

    n+1],

    ∴Tn=1−

    1

    2+

    1

    2−

    1

    3+…+

    1

    n−

    1

    n+1

    =1-[1/n+1]=[n/n+1].

    点评:

    本题考点: 数列递推式;等差数列的通项公式;等比数列的通项公式;数列的求和.

    考点点评: 本题考查由递推式求数列通项、等差数列等比数列的通项公式、数列求和等知识,裂相消法对数列求和是高考考查的重点内容,要熟练掌握.