1、高考资源网( ),您身边的高考专家一、选择题1已知数列an的前n项和为Sn,且Sn2(an1),则a2等于()A4B2C1 D2解析:由题可知Sn2(an1),所以S1a12(a11),解得a12.又S2a1a22(a21),解得a2a124.答案:A2已知数列an满足a10,则数列an是()A递增数列 B递减数列C常数列 D不确定解析:0,则an0,an10,解得n6或n1(舍),从第7项起各项都是正数9已知数列an的前n项和Sn2n22n,数列bn的前n项和Tn2bn.求数列an与bn的通项公式解:(1)当n2时,anSnSn1(2n22n)2(n1)22(n1)4n,当n1时,a1S14
2、也适合,an的通项公式是an4n(nN*)Tn2bn,当n1时,b12b1,b11.当n2时,bnTnTn1(2bn)(2bn1),2bnbn1.数列bn是公比为,首项为1的等比数列bnn1.10已知数列an满足a11,anan13n2(n2)(1)求a2,a3;(2)求数列an的通项公式解:(1)由已知:an满足a11,anan13n2(n2),a2a145,a3a2712.(2)由已知:anan13n2(n2)得:anan13n2,由递推关系,得an1an23n5,a3a27,a2a14,叠加得:ana1473n2,an(n2)当n1时,1a11,数列an的通项公式an.欢迎广大教师踊跃来稿,稿酬丰厚。