题目

数列{an}的前n项和Sn满足：t(Sn+1+1)-(2t+1)Sn,n∈N*,t≠0.(1)求证:{an}是等比数列;(2)若{an}的公比为f(t),数列{bn}满足：b1=1,bn+1=f(),求{bn}的通项公式；(3)定义数列{cn}为：cn=,求{cn}的前n项和Tn，并求Tn. 答案：解:(1)由T(Sn+1+1)=(2T+1)Sn,得T(Sn+1)=(2T+1)Sn-1,相减得=2+,∴{an}是等比数列.                                                                     (2)bn+1=f()=2+bn,∴bn+1-bn=2,bn=1,得bn=2n-1.                                                                                    (3)cn===(-),∴Tn=［(1-)+(-)+…+(-)=(1-)］.∴Tn=.

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