摘要: 解:(1)由a1=S1=1.S2=1+a2.得a2= 又3tSn-(2t+3)Sn-1=3t ① 3tSn-1-(2t+3)Sn-2=3t ② ①-②得3tan-(2t+3)an-1=0 ∴. 所以{an}是一个首项为1.公比为的等比数列. (2)由f(t)=.得bn=f+bn-1. ∴bn=1+(n-1)= (3)由bn=.可知{b2n-1}和{b2n}是首项分别为1和.公差均为的等差数列于是 b1b2-b2b3+b3b4-b4b5+-+b2n-1b2n-b2nb2n+1 =b2(b1-b3)+b4(b3-b5)+b6(b5-b7)+-+b2n(b2n-1+b2n+1) =-(b2+b4+-+b2n)=-=-(2n2+3n)

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