1、限时规范训练一、选择题1已知数列an满足a15,anan12n,则()A2B4C5 D.解析:因为22,所以令n3,得224,故选B.答案:B2若数列an满足a115,且3an13an2,则使akak10的k值为()A22 B21C24 D23解析:因为3an13an2,所以an1an,所以数列an是首项为15,公差为的等差数列,所以an15(n1)n,令ann0,得n23.5,所以使akak10,an13an,又a12,an是首项为2,公比为3的等比数列,Sn3n1.答案:3n111已知Sn为数列an的前n项和,且满足a11,anan13n(nN*),则S2 014_.解析:由anan13n
2、知,当n2时,anan13n1.所以3,所以数列an所有的奇数项构成以3为公比的等比数列,所有的偶数项也构成以3为公比的等比数列又因为a11,所以a23,a2n13n1,a2n3n.所以S2 014(a1a3a2 013)(a2a4a2 014)4231 0072.答案:231 007212数列an中,a1,an1(nN*),则数列an的通项公式an_.解析:由已知可得(n1)an1,设nanbn,则bn1,所以1,可得122,即是公比为2,首项为3的等比数列,故132n1 an.答案:三、解答题13在数列an中,a18,a42,且满足an22an1an0.(1)求数列an的通项公式;(2)设
3、Sn|a1|a2|an|,求Sn.解析:(1)an22an1an0,an2an1an1an,an1an为常数列,an是以a1为首项的等差数列,设ana1(n1)d,则a4a13d,d2,an102n.(2)由(1)知an102n,令an0,得n5.当n5时,an0;当n5时,an0;当n0.当n5时,Sn|a1|a2|an|a1a2a5(a6a7an)T5(TnT5)2T5Tnn29n40,其中Tna1a2an.当n5时,Sn|a1|a2|an|a1a2an9nn2.Sn.14正项数列an的前n项和为Sn,且a4Sn2an1(nN*)(1)求数列an的通项公式;(2)若bn,数列bn的前n项和
4、为Tn.求证:T2n1.解析:(1)当n1时,a11;当n2时,因为an0,a4Sn2an1,所以a4Sn12an11,两式相减得aa4an2an2an12(anan1),所以anan12,所以数列an是以1为首项,2为公差的等差数列,所以an2n1.(2)证明:bn(1)n1T2n11.T2n1.15已知数列an的前n项和为Sn,a12,且满足an1Sn2n1(nN*)(1)证明:数列为等差数列;(2)求S1S2Sn.解析:(1)证明:由条件可知,Sn1SnSn2n1,即Sn12Sn2n1,整理得1,所以数列是以1为首项,1为公差的等差数列(2)由(1)可知,1n1n,即Snn2n,令TnS1S2Sn,则Tn12222n2n,2Tn122(n1)2nn2n1,Tn2222nn2n1,整理得Tn2(n1)2n1.