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 B70 C75 D100解析因为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)5010150103100.故选B.答案B3数列a12,ak2k,a1020共有十
2、项,且其和为240,则a1aka10的值为()A31 B120 C130 D185解析a1aka10240(22k20)240240110130.答案C4(2015西安质检)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
3、Sn为()A. 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(2015宝鸡测试)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(2014武汉模拟)等比数列an的前n项和Sn2n1,则aaa_解析当n1时,
4、a1S11,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(2014济南模拟)设等差数列an的前n项和为Sn,且S32S24,a536.(1)求an,Sn;(2)设bnSn1(nN),Tn,求Tn.解(1)因为S32S24,所以a1d4,又因为a536,所以a14d36.解得d8,a14,所以an48(n1)8n4,Sn4n2.(2)bn4n21(2n1)(2n1),所以.Tn.10(2015汉中模拟)已知an 是各项均为正数的等比数列,且a1a22,a3a432.(1
5、)求数列an的通项公式;(2)设数列bn的前n项和为Snn2(nN),求数列anbn的前n项和解(1)设等比数列an的公比为q,由已知得又a10,q0,解得an2n1.(2)由Snn2得Sn1(n1)2(n2),当n2时,bnSnSn12n1,当n1时,b11符合上式,bn2n1(nN),anbn(2n1)2n1.Tn1321522(2n1)2n1,2Tn12322523(2n3)2n1(2n1)2n,两式相减得Tn12(2222n1)(2n1)2n(2n3)2n3,Tn(2n3)2n3.能力提升题组(建议用时:25分钟)11(2015西安模拟)数列an满足anan1(nN*),且a11,Sn
6、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(2015长沙模拟)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)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
7、差数列所以a1a2a3a100(a1a3a99)(a2a4a100)503450(5)4100.答案A13设f(x),利用倒序相加法,可求得f f f 的值为_解析当x1x21时,f(x1)f(x2)1.设Sf ff ,倒序相加有2S10,即S5.答案514在数列an中,a15,a22,记A(n)a1a2an,B(n)a2a3an1,C(n)a3a4an2(nN),若对于任意nN,A(n),B(n),C(n)成等差数列(1)求数列an的通项公式;(2)求数列|an|的前n项和解(1)根据题意A(n),B(n),C(n)成等差数列,A(n)C(n)2B(n),整理得an2an1a2a1253,数列an是首项为5,公差为3的等差数列,an53(n1)3n8.(2)|an|记数列|an|的前n项和为Sn.当n2时,Snn;当n3时,Sn7n14,综上,Sn