1、第4讲数列求和基础巩固题组(建议用时:40分钟)一、选择题1等差数列an的通项公式为an2n1,其前n项和为Sn,则数列的前10项的和为()A120 B70C75 D100解析由于n2,所以的前10项和为10375.答案C2已知函数f(n)且anf(n)f(n1),则a1a2a3a100等于()A0 B100 C100 D10 200解析由题意,得a1a2a3a1001222223232424252992100210021012(12)(32)(99100)(101100)(1299100)(23100101)1101100.故选B.答案B3数列a12,ak2k,a1020共有十项,且其和为2
2、40,则a1aka10的值为()A31 B120 C130 D185解析a1aka10240(22k20)240240110130.答案C4(2021台州质检)已知数列an满足a11,an1an2n(nN*),则S2 016()A22 0161 B321 0083C321 0081 D321 0072解析a11,a22,又2.2.a1,a3,a5,成等比数列;a2,a4,a6,成等比数列,S2 016a1a2a3a4a5a6a2 015a2 016(a1a3a5a2 015)(a2a4a6a2 016)321 0083.故选B.答案B5已知数列an:,若bn,那么数列bn的前n项和Sn为()A
3、. B. C. D.解析an,bn4,Sn44.答案B二、填空题6在等差数列an中,a10,a10a110,a10a110可知d0,a110,T18a1a10a11a18S10(S18S10)60.答案607(2021湖州测试)在数列an中,a11,an1(1)n(an1),记Sn为an的前n项和,则S2 013_.解析由a11,an1(1)n(an1)可得a11,a22,a31,a40,该数列是周期为4的数列,所以S2 013503(a1a2a3a4)a2 013503(2)11 005.答案1 0058(2022武汉模拟)等比数列an的前n项和Sn2n1,则aaa_.解析当n1时,a1S1
4、1,当n2时,anSnSn12n1(2n11)2n1,又a11适合上式an2n1,a4n1.数列a是以a1为首项,以4为公比的等比数列aaa(4n1)答案(4n1)三、解答题9(2021金华十校联考)已知数列an的前n项和是Sn,且Snan1(nN*)(1)求数列an的通项公式;(2)设bnlog(1Sn1)(nN*),令Tn,求Tn.解(1)当n1时,a1S1,由S1a11,得a1,当n2时,Sn1an,Sn11an1,则SnSn1(an1an),即an(an1an),所以anan1(n2)故数列an是以为首项,为公比的等比数列故ann12n(nN*)(2)由于1Snann.所以bnlog(
5、1Sn1)logn1n1,由于,所以Tn.10(2021山东卷)设等差数列an的前n项和为Sn,且S44S2,a2n2an1.(1)求数列an的通项公式;(2)设数列bn的前n项和为Tn,且Tn(为常数),令cnb2n,(nN*),求数列cn的前n项和Rn.解(1)设等差数列an的首项为a1,公差为d.由S44S2,a2n2an1,得解得a11,d2.因此an2n1,nN*.(2)由题意知Tn,所以n2时,bnTnTn1.故cnb2n(n1)n1,nN*,所以Rn00112233(n1)n1,则Rn011223(n2)n1(n1)n,两式相减得Rn123n1(n1)n(n1)nn,整理得Rn.
6、所以数列cn的前n项和Rn.力量提升题组(建议用时:35分钟)11(2021西安模拟)数列an满足anan1(nN*),且a11,Sn是数列an的前n项和,则S21()A. B6 C10 D11解析依题意得anan1an1an2,则an2an,即数列an中的奇数项、偶数项分别相等,则a21a11,S21(a1a2)(a3a4)(a19a20)a2110(a1a2)a211016,故选B.答案B12(2021温州十校联考)已知函数f(n)n2cos(n),且anf(n)f(n1),则a1a2a3a100()A100 B0 C100 D10 200解析若n为偶数,则anf(n)f(n1)n2(n1
7、)2(2n1),为首项为a25,公差为4的等差数列;若n为奇数,则anf(n)f(n1)n2(n1)22n1,为首项为a13,公差为4的等差数列所以a1a2a3a100(a1a3a99)(a2a4a100)503450(5)4100.答案A13数列an的前n项和为Sn,a11,a22,an2an1(1)n(nN*),则S100_.解析由an2an1(1)n,知a2k2a2k2,a2k1a2k10,a1a3a5a2n11,数列a2k是等差数列,a2k2k.S100(a1a3a5a99)(a2a4a6a100)50(246100)502 600.答案2 60014(2021湖南卷)设Sn为数列an
8、的前n项和,Sn(1)nan,nN*,则:(1)a3_;(2)S1S2S100_.解析n2时,anSnSn1(1)nan(1)n1an1,an(1)nan(1)n1an1.当n为偶数时,an1,当n为奇数时,2anan1,当n4时,a3.依据以上an的关系式及递推式可求a1,a3,a5,a7,a2,a4,a6,a8.a2a1,a4a3,a6a5,S1S2S100(a2a1)(a4a3)(a100a99).答案(1)(2)15若Sn是公差不为0的等差数列an的前n项和,且S1,S2,S4成等比数列(1)求等比数列S1,S2,S4的公比;(2)若S24,求数列an的通项公式;(3)在(2)的条件下
9、,设bn,Tn是数列bn的前n项和,求使得Tn对全部nN*都成立的最小正整数m.解(1)由于an为等差数列,设an的公差为d(d0),所以S1a1,S22a1d,S44a16d.由于S1,S2,S4成等比数列且设其公比为q,所以S1S4S.所以a1(4a16d)(2a1d)2.所以2a1dd2.由于公差d0.所以d2a1.所以q4.(2)由于S24,所以2a1d4.又d2a1,所以a11,d2.所以an2n1.(3)由于bn,所以Tn.要使Tn对全部nN*都成立,则有,即m30.由于mN*,所以m的最小值为30.16已知数列an与bn满足bnanan1bn1an20,bn,nN*,且a12,a
10、24.(1)求a3,a4,a5的值;(2)设cna2n1a2n1,nN*,证明cn是等比数列;(3)设Ska2a4a2k,kN*,证明 1 (nN*)(1)解由bn,nN*,可得bn又bnanan1bn1an20,当n1时,a1a22a30,由a12,a24,可得a33;当n2时,2a2a3a40,可得a45;当n3时,a3a42a50,可得a54.(2)证明对任意nN*,a2n1a2n2a2n10,2a2na2n1a2n20,a2n1a2n22a2n30,得a2na2n3,将代入,可得a2n1a2n3(a2n1a2n1),即cn1cn(nN*)又c1a1a31,故cn0,因此1.所以cn是等比数列(3)证明由(2)可得a2k1 a2k1(1)k,于是,对任意kN*且k2,有a1a31,(a3a5)1,a5a71,(1)k(a2k3a2k1)1.将以上各式相加,得a1(1)ka2k1(k1),即a2k1(1)k1(k1),此式当k1时也成立由式得a2k(1)k1(k3)从而S2k(a2a4)(a6a8)(a4k2a4k)k,S2k1S2ka4kk3,所以,对任意nN*,n2, .对于n1,不等式明显成立.特殊提示:老师配赠习题、课件、视频、图片、文档等各种电子资源见创新设计高考总复习光盘中内容.