题目

等差数列{an}的首项为a1=1，公差d≠0，已知、、、…、、…成等比数列，且k1=1，k2=2，k3=5．(Ⅰ)求{an}、{kn}的通项公式；(Ⅱ)求数列{}的前n项和Sn． 答案：解：(1)∵=a1·a5  ∴(1+d)2=1·(1+4d)，∵d≠0，∴d=2，    ∴an=2n-1，=2kn-1,又{}的公比q==3，∴=3n-1，∴2kn-1=3n-1，∴kn=．(Ⅱ)由(Ⅰ)知，设Sn=，∴Sn=，∴Sn=1+=1+2×，∴Sn=3-．The school days are busy enough, yet the Taylors try hard to fit as much as possible _____ their kids’ lives． A． in        B． into         C． on        D． at