解:(1)∵等比數(shù)列a
n的前n項和為f(n)-c,
∴a
1=f(1)-c=

-c,
∴a
2=[f(2)-c]-[f(1)-c]=-

,a
3=[f(3)-c]-[f(2)-c]=-

又數(shù)列{a
n}成等比數(shù)列,

=-

,
∵a
1=

-c
∴-

=

-c,∴c=1
又公比q=

=

所以a
n=-

•

,n∈N;
∵S
n-S
n-1=

=

(n≥2)
又b
n>0,

>0,∴

=1;
∴數(shù)列{

}構(gòu)成一個首項為1公差為1的等差數(shù)列,
∴

=1+(n-1)×1=n,S
n=n
2當(dāng)n≥2,b
n=S
n-S
n-1=n
2-(n-1)
2=2n-1;
又b
1=c=1適合上式,∴b
n=2n-1(n∈N);
(2)由(1)知

=(2n-1)+(2n-1)•(

)
n
設(shè)(2n-1)•(

)
n前n項和為Q
n 設(shè)數(shù)列2n-1的前n項和為S
nQ
n=

+3×(

)
2+5×(

)
3+…+(2n-3)•(

)
n-1+(2n-1)•(

)
n ①

Q
n=(

)
2+3×(

)
3+5×(

)
4+…+(2n-3)•(

)
n+(2n-1)•(

)
n+1 ②
①-②得:

=

∴Q
n=1-(n+1)(

)
n
∴S
n=n
2
∴T
n=S
n+Q
n=n
2+1-(n+1)(
)n分析:(1)由等比數(shù)列{a
n}的前n項和為f(n)-c求出數(shù)列{a
n}的公比和首項,得到數(shù)列{a
n}的通項公式;由數(shù)列{b
n}的前n項和S
n滿足S
n-S
n-1=

可得到數(shù)列{

}構(gòu)成一個首項為1公差為1的等差數(shù)列,進而得到數(shù)列{

}的通項公式,再由b
n=S
n-S
n-1可確定{b
n}的通項公式.
(2)首先寫出數(shù)列

的通項公式,然后利用錯位相減的方法求數(shù)列前n項和.
點評:本題主要考查數(shù)列遞推式和數(shù)列求和的知識點,解答本題的關(guān)鍵是熟練掌握等差數(shù)列和等比數(shù)列的性質(zhì)及其求和公式,此題還要熟練掌握錯位相減法在數(shù)列求和中的應(yīng)用,此題有一定的難度.