题目内容

等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=a52.

(1)求数列{an}的通项公式;

(2)若数列{bn}满足bn=,求数列{bn}的前99项的和.

解:(1)设数列{an}公差为d(d>0),

∵a1,a3,a9成等比数列,∴a32=a1a9.

(a1+2d)2=a1(a1+8d),d2=a1d.

∵d≠0,∴a1=d.                                                         ①

∵S5=a52,∴5a1+·d=(a1+4d)2.                      ②

由①②得a1=,d=.

∴an=+(n-1)×=n.                                                    

(2)bn=,

∴b1+b2+b3+…+b99

=[99+(1-)+(-)+…+ (-)]

=(100-)=.


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