(1)由已知3tSn-(2t+3)Sn-1=3t,可得3tsn-1-(2t+3)sn-2=3t,两式相减可得数列an与an-1的递推关系,从而可证.
(2)把f(t)的解析式代入bn,进而可知bn=+bn-1,判断出{bn}是一个首项为1,公差为的等差数列.进而根据等差数列的通项公式求得答案.
(3){bn}是等差数列,用分组法求得数列的b1b2-b2b3+b3b4-…+b2n-1b2n-b2nb2n+1和.
(1)证:∵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
∴(n≥2),
又当n=2时,3tS2-(2t+3)S1=3t,
即3t(a1+a2)-(2t+3)a1=3t,得,
即,
∴(n≥1),
∴{an}为等比数列
(2)【解析】
由已知得,f(t)=
∴bn=f()==+bn-1(n≥2,n∈N*).
∴{bn}是一个首项为1,公差为的等差数列.
于是bn=n+
(3)【解析】
Tn=b1b2-b2b3+b3b4-…+b2n-1b2n-b2nb2n+1
=b2(b1-b3)+b4(b3-b5)+…+b2n(b2n-1-b2n+1)=-2(b2+b4+…+b2n)
=-2d(b2+b4+…+b2n)
=
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