(Ⅰ)由S1=a1=1,S2=1+a2,得3t(1+a2)-(2t+3)=3t,,又3tSn-(2t+3)Sn-1=3t,3tSn-1-(2t+3)Sn-2=3t(n=3,4,)两式相减,得:3tan-(2t+3)an-1=0,由此能够证明数列{an}为等比数列.
(Ⅱ)由,得,所以,由此能求出(b1-b3)b2+(b3-b5)b4+…+(b2n-1-b2n+1)b2n之和.
【解析】
(Ⅰ)由S1=a1=1,S2=1+a2,得3t(1+a2)-(2t+3)=3t,∴
又3tSn-(2t+3)Sn-1=3t,3tSn-1-(2t+3)Sn-2=3t(n=3,4,)两式相减,
得:3tan-(2t+3)an-1=0,
∴(n=3,4,)
综上,数列{an}为首项为1,公比为的等比数列
(Ⅱ)由,得,
所以{bn}是首项为1,,公差为的等差数列,b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1=(b1-b3)b2+(b3-b5)b4+…+(b2n-1-b2n+1)b2n==