(1)由S1=a1=1,S2=1+a2,得3t(1+a2)-(2t+3)=3t. ∴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 ∴,n=2,3,4…, 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. 可见{bn}是一个首项为1,公差为的等差数列. 于是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b2n=, ∴b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1 =b2(b1-b3)+b4(b3-b5)+…+b2n(b2n-1-b2n+1) =- (b2+b4+…+b2n)=-·n(+)=- (2n2+3n). |