丰台区2013年高三年级第二学期统一练习(一)

\r11\r由a0得ak0,即a,\rk11k(k1)1k1\r11n1\ra(n1)(nN,n2n1).\rnk1k(k1)k(k1)k\r„„„„„„„„\r„8分\r1\rS0\r(Ⅲ)(1)当k=n时,显然成立;„„„„„„„„„„„„„„„„„„„9\rn2\r分\r当k<n时,据条件①得\rSaaa(aaa),\rk12kk1k2n\r即Saaaaaa\rk12kk1k2n,\r2Saaaaaa\rk12kk1k2n\raaaaaa1\r12kk1k2n,\r1\rS(k1,2,3,,n).\rk2\r„„„„„„„„„„„„„„„„„„„„„„„„„„1\r1分\rnaaaaaaa\r(2)i1234n1n\r\ri1234n1n\ri1\rSSSSSSSSSS\rS213243n1n2nn1\r1234n1n\rSSSSSS\r1234n1n\r2233445(n1)nn\r-12-EvaluationWarning:ThedocumentwascreatedwithSpire.PDFfor.NET.\r新东方在线[www.koolearn.com]2013高考-倒计时冲刺命题趋势志愿填报招生简章模拟考试\rSSSSS\r1234n1\r2233445(n1)n\r111111\r\r22233445(n1)n\r111111111111\r.\r22233445n1n22n„„„„„„„„„„„„14\r分\r-13-