题目内容

已知数列{an}是等差数列,其前n项和为Sn.若a5=11,S4=24.

(1)求数列{an}的通项公式;

(2)设bn=(),求数列{bn}的前n项和Tn.

解:(1)设{an}的首项为a1,公差为d,则                         ?

解得a1=3,d=2.                                                                                                        ?

an=2n+1.                                                                                                             ?

(2)∵an=2n+1,∴bn=()2n+1.                                                                                    ?

∴{bn}是首项b1=,公式q=的等比数列.                                                             ?

Tn==[1-()n].

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