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.