1、专题三 数列第1讲数列的通项与求和问题一、选择题1在等差数列an中,若a2a34,a4a56,则a9a10等于()A9B10 C11D12解析设等差数列an的公差为d,则有(a4a5)(a2a3)4d2,所以d.又(a9a10)(a4a5)10d5,所以a9a10(a4a5)511.答案C2(2014嘉兴教学测试)在各项均为正数的等比数列an中,a31,a51,则a2a2a6a3a7()A4B6 C8D84解析在等比数列an中,a3a7a,a2a6a3a5,所以a2a2a6a3a7a2a3a5a(a3a5)2(11)2(2)28.答案C3已知数列1,3,5,7,则其前n项和Sn为()An21B
2、n22 Cn21Dn22解析因为an2n1,则Snn21.答案A4(2014烟台一模)在等差数列an中,a12 012,其前n项和为Sn,若2 002,则S2 014的值等于()A2 011B2 012 C2 014D2 013解析等差数列中,Snna1d,a1(n1),即数列是首项为a12 012,公差为的等差数列;因为2 002,所以,(2 01210)2 002,1,所以,S2 0142 014(2 012)(2 0141)12 014,选C.答案C5(2014合肥质量检测)数列an满足a12,an,其前n项积为Tn,则T2 014()A.B C6D6解析由an,得an1.a12,a23
3、,a3,a4,a52,a63.故数列an具有周期性,周期为4,a1a2a3a41,T2 014T2a1a22(3)6.答案D二、填空题6(2014衡水中学调研)已知数列an满足a1,an1an(n2),则该数列的通项公式an_.解析an1an(n2),1,又a1,3,an.答案7设等差数列an的前n项和为Sn,Sm12,Sm0,Sm13,则m等于_解析由Sm12,Sm0,Sm13,得am2,am13,所以d1,因为Sm0,故ma1d0,故a1,因为amam15,故amam12a1(2m1)d(m1)2m15,即m5.答案58(2014广东卷)若等比数列an的各项均为正数,且a10a11a9a1
4、22e5,则ln a1ln a2ln a20_.解析a10a11a9a122a10a112e5,a10a11e5,ln a1ln a2ln a2010ln(a10a11)10ln e550.答案50三、解答题9(2014北京卷)已知an是等差数列,满足a13,a412,数列bn满足b14,b420,且bnan为等比数列(1)求数列an和bn的通项公式;(2)求数列bn的前n项和解(1)设等差数列an的公差为d,由题意得d3.所以ana1(n1)d3n(n1,2,)设等比数列bnan的公比为q,由题意得q38,解得q2.所以bnan(b1a1)qn12n1.从而bn3n2n1(n1,2,)(2)
5、由(1)知bn3n2n1(n1,2,)数列3n的前n项和为n(n1),数列2n1的前n项和为2n1.所以,数列bn的前n项和为n(n1)2n1.10(2014江西卷)已知首项都是1的两个数列an,bn(bn0,nN*)满足anbn1an1bn2bn1bn0.(1)令cn,求数列cn的通项公式;(2)若bn3n1,求数列an的前n项和Sn.解(1)因为anbn1an1bn2bn1bn0,bn0(nN*),所以2,即cn1cn2.所以数列cn是以首项c11,公差d2的等差数列,故cn2n1.(2)由bn3n1知ancnbn(2n1)3n1,于是数列an前n项和Sn130331532(2n1)3n1
6、,3Sn131332(2n3)3n1(2n1)3n,相减得2Sn12(31323n1)(2n1)3n2(2n2)3n,所以Sn(n1)3n1.11(2014成都诊断)已知数列an前n项和为Sn,首项为a1,且,an,Sn成等差数列(1)求数列an的通项公式;(2)数列bn满足bn(log2a2n1)(log2a2n3),求数列的前n项和解(1),an,Sn成等差数列,2anSn,当n1时,2a1S1,a1,当n2时,Sn2an,Sn12an1,两式相减得:anSnSn12an2an1,2,所以数列an是首项为,公比为2的等比数列,即an2n12n2.(2)bn(log2a2n1)(log2a2n3)(log222n12)(log222n32)(2n1)(2n1),数列的前n项和Tn.