(1)证明:∵3tSn-(2t+3)Sn-1=3t,∴3tsn+1-(2t+3)Sn =3t (n≥2),两式相减得3tan+1-(2t+3)an =0.
又t>0,∴
an+1
an=
2t+3
3t (n≥2),又当n=2时,3ts2-(2t+3)s1=3t,
即3t (a1+a2)-(2t+3)a1=3t,得 a2=
2t+3
3t,即
a2
a1=
2t+3
3t,∴
an+1
an=
2t+3
3t (n≥1),∴数列{an}为等比数列.
由已知得f(n)=
2t+3
3t,∴bn=f(
1
bn−1)=
2
bn−1+3
3
bn−1=bn-1+
2
3 (n≥2).
∴数列{bn}是以 b1=1为首项,以d=
2
3为公差的等差数列,故 bn=
2
3n+
1
3.
(2)Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)=b2(b1-b3)+b4(b3-b5)+…+b2n(b2n-1-b2n+1)
=2d (b2+b4+…+b2n)=2×
2
3[n•
5
3+
n(n−1)
2•
4
3]=-
8
9n2-
4n
3.