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.