in105 6 2sin sin30 A??? ? ?或226??c……………10分所以1 1 6 2 2 3 1sin 12 2 2 2 4ABCS ac B???? ?????或413???ABCS………12分(选择②③不能确定三角形)21.(本小题12分)已知数列??na满足:1111, ( 2, )4 ( 1) 2nnnnaa a n n Na???? ? ????,设nnnab)1(1???.(1)求证:数列??b是等比数列,并求数列??na的通项公式;(2)求数列???????nbn23的前n项和nS解:(1)由1111, ( 2, )4 ( 1) 2nnnnaa a n n Na???? ? ????得:??????????????11)1(12)1(1nnnnaa,所以,)2(21????nbbnn又03)1(1111?????ab,所以)2(21????nbbnn所以,数列{}nb是等比数列. )2()2(31?????nbnn,31?b故,)()2(31??????Nnbnn,)()1()2(3111?????????Nnbnnn…………6分(2)1)2(32323??????nnnbn1210)2(323)2(37)2(34)2(31??????????????nnnS?nnnnnS)2(323)2(353)2(37)2(34)2(31211321????????????????????两式相减,nnnnS)2(323)2(1)2(1)2(1)2(131231321???????????????nn)21(????