题目

已知数列{an}满足a1=,an+1=. (1)证明数列是等差数列,并求{an}的通项公式; (2)若数列{bn}满足bn=,求数列{bn}的前n项和Sn. 答案：(1)证明∵an+1=, =2, 是等差数列, +(n-1)×2=2+2n-2=2n,即an= (2)解∵bn=, ∴Sn=b1+b2+…+bn=1++…+, 则Sn=+…+, 两式相减得Sn=1++…+=2,∴Sn=4-

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