资源描述
第一节 数列的概念与简单表示一、选择题(65分30分)1(2011平顶山模拟)数列1,2,2,3,3,3,4,4,4,4,5,的第100项是()A14B12C13 D15解析:易知数字为n时共有n个,到数字n时,总共的数字的个数为123n.易得n13时,最后一项为第91项,n14共有14个,故第100项为14.答案:A2(2011商丘一模)已知数列an中,a1b(b为任意正数),an1(n1,2,3,),能使anb的n的数值是()A14 B15C16 D17解析:a1b,a2,a3,a4b,此数列的周期为3,能使anb的n的数值满足n3k2(kN*)答案:C3(2011珠海月考)在数列an中,a11,anan1an1(1)n(n2,nN*),则的值是()A. B.C. D.解析:由已知得a21(1)22,a3a2a2(1)3,a3,a4(1)4,a43,3a53(1)5,a5,.答案:C4(2011北京石景山高三模拟)已知数列an的前n项和Snn3,则a5a6的值为()A91 B152C218 D279解析:a5a6S6S46343152.答案:B5已知数列an满足a01,ana0a1an1(n1),则当n1时,an等于()A2n B.C2n1 D2n1解析:由ana0a1a2an1(n1),得an1a0a1a2an1an2an,an(nN*)是公比为2的等比数列,a1a01,an2n1.答案:C6(2011清远阶段测试)已知数列an的前n项和Snn29n,第k项满足5ak8,则k等于()A9 B8C7 D6解析:Snn29n,n2时,anSnSn12n10,a1S18适合上式,an2n10(nN*),52k108,得7.5k0,an0,解得n6或n6(nN*)时,an0.令n2n300,解得5n6.又nN*,0n6.当0n6(nN*)时,an0.(3)由ann2n30(n)230,nN*,知an是递增数列,且a1a2a5a60a7a8a9,故Sn存在最小值S5S6,Sn不存在最大值
展开阅读全文