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1、考点规范练31数列求和基础巩固1.数列112,314,518,7116,(2n-1)+12n,的前n项和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n2.数列an满足a1=1,且对任意的nN+都有an+1=a1+an+n,则1an的前100项和为()A.100101B.99100C.101100D.2001013.已知函数f(n)=n2cos(n),且an=f(n)+f(n+1),则a1+a2+a3+a100=()A.0B.-100C.100D.10 2004.已知函数f(x)=xa的图象过点(4,2),令an=1f(n+1)+f
2、(n),nN+.记数列an的前n项和为Sn,则S2 016等于()A.2016-1B.2016+1C.2017-1D.2017+15.若数列an满足an+1+(-1)nan=2n-1,则an的前60项和为()A.3 690B.3 660C.1 845D.1 8306.32-1+42-2+52-3+(n+2)2-n=.7.已知数列an满足:a3=15,an-an+1=2anan+1,则数列anan+1前10项的和为.8.已知an是等差数列,bn是等比数列,且b2=3,b3=9,a1=b1,a14=b4.(1)求an的通项公式;(2)设cn=an+bn,求数列cn的前n项和.9.设等差数列an的公
3、差为d,前n项和为Sn,等比数列bn的公比为q,已知b1=a1,b2=2,q=d,S10=100.(1)求数列an,bn的通项公式;(2)当d1时,记cn=anbn,求数列cn的前n项和Tn.10.已知Sn为数列an的前n项和,an0,an2+2an=4Sn+3.(1)求an的通项公式;(2)设bn=1anan+1,求数列bn的前n项和.11.已知各项均为正数的数列an的前n项和为Sn,满足an+12=2Sn+n+4,a2-1,a3,a7恰为等比数列bn的前3项.(1)求数列an,bn的通项公式;(2)若cn=(-1)nlog2bn-1anan+1,求数列cn的前n项和Tn.能力提升12.(2
4、017山东烟台模拟)已知数列an中,a1=1,且an+1=an2an+1,若bn=anan+1,则数列bn的前n项和Sn为()A.2n2n+1B.n2n+1C.2n2n-1D.2n-12n+113.(2017福建龙岩一模)已知Sn为数列an的前n项和,对nN+都有Sn=1-an,若bn=log2an,则1b1b2+1b2b3+1bnbn+1=.14.(2017广西模拟)已知数列an的前n项和为Sn,且Sn=32an-1(nN+).(1)求数列an的通项公式;(2)设bn=2log3an2+1,求1b1b2+1b2b3+1bn-1bn.高考预测15.几位大学生响应国家的创业号召,开发了一款应用软
5、件.为激发大家学习数学的兴趣,他们推出了“解数学题获取软件激活码”的活动.这款软件的激活码为下面数学问题的答案:已知数列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一项是20,接下来的两项是20,21,再接下来的三项是20,21,22,依此类推.求满足如下条件的最小整数N:N100,且该数列的前N项和为2的整数幂.那么该款软件的激活码是()A.440B.330C.220D.11016.已知数列an的前n项和为Sn,且a1=2,Sn=2an+k,等差数列bn的前n项和为Tn,且Tn=n2.(1)求k和Sn;(2)若cn=anbn,求数列cn的前n项和Mn.参考答案考点规范
6、练31数列求和1.A解析该数列的通项公式为an=(2n-1)+12n,则Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.D解析an+1=a1+an+n,an+1-an=1+n.an-an-1=n(n2).an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=n+(n-1)+2+1=n(n+1)2.1an=2n(n+1)=21n-1n+1.1an的前100项和为21-12+12-13+1100-1101=21-1101=200101.故选D.3.B解析f(n)=n2cos(n)=-n2,n为奇数,n2,n为偶数=(-1)nn2,an=f(n)+f(n
7、+1)=(-1)nn2+(-1)n+1(n+1)2=(-1)nn2-(n+1)2=(-1)n+1(2n+1).a1+a2+a3+a100=3+(-5)+7+(-9)+199+(-201)=50(-2)=-100.故选B.4.C解析由f(4)=2,可得4a=2,解得a=12,则f(x)=x12.an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2016=a1+a2+a3+a2016=(2-1)+(3-2)+(4-3)+(2017-2016)=2017-1.5.D解析an+1+(-1)nan=2n-1,当n=2k(kN+)时,a2k+1+a2k=4k-1,当n=2k+1(kN+)时,a2
8、k+2-a2k+1=4k+1,+得:a2k+a2k+2=8k.则a2+a4+a6+a8+a60=(a2+a4)+(a6+a8)+(a58+a60)=8(1+3+29)=815(1+29)2=1800.由得a2k+1=a2k+2-(4k+1),a1+a3+a5+a59=a2+a4+a60-4(0+1+2+29)+30=1800-430292+30=30,a1+a2+a60=1800+30=1830.6.4-n+42n解析设Sn=32-1+42-2+52-3+(n+2)2-n,则12Sn=32-2+42-3+(n+1)2-n+(n+2)2-n-1.-,得12Sn=32-1+2-2+2-3+2-n-
9、(n+2)2-n-1=22-1+2-1+2-2+2-3+2-n-(n+2)2-n-1=1+2-1(1-2-n)1-2-1-(n+2)2-n-1=2-(n+4)2-n-1.故Sn=4-n+42n.7.1021解析an-an+1=2anan+1,an-an+1anan+1=2,即1an+1-1an=2.数列1an是以2为公差的等差数列.1a3=5,1an=5+2(n-3)=2n-1.an=12n-1.anan+1=1(2n-1)(2n+1)=1212n-1-12n+1.数列anan+1前10项的和为121-13+13-15+1210-1-1210+1=121-121=122021=1021.8.解
10、(1)因为等比数列bn的公比q=b3b2=93=3,所以b1=b2q=1,b4=b3q=27.设等差数列an的公差为d.因为a1=b1=1,a14=b4=27,所以1+13d=27,即d=2.所以an=2n-1.(2)由(1)知,an=2n-1,bn=3n-1.因此cn=an+bn=2n-1+3n-1.从而数列cn的前n项和Sn=1+3+(2n-1)+1+3+3n-1=n(1+2n-1)2+1-3n1-3=n2+3n-12.9.解(1)由题意,有10a1+45d=100,a1d=2,即2a1+9d=20,a1d=2,解得a1=1,d=2或a1=9,d=29.故an=2n-1,bn=2n-1或a
11、n=19(2n+79),bn=929n-1.(2)由d1,知an=2n-1,bn=2n-1,故cn=2n-12n-1,于是Tn=1+32+522+723+924+2n-12n-1,12Tn=12+322+523+724+925+2n-12n.-可得12Tn=2+12+122+12n-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.10.解(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.两式相减可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由于an0,可
12、得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去),a1=3.所以an是首项为3,公差为2的等差数列,故an的通项公式为an=2n+1.(2)由an=2n+1可知bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.设数列bn的前n项和为Tn,则Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).11.解(1)因为an+12=2Sn+n+4,所以an2=2Sn-1+n-1+4(n2).两式相减得an+12-an2=2an+1,所以an+12=an2+2an+1=(an+1)2.因为an是各项均为正数的数列
13、,所以an+1-an=1.又a32=(a2-1)a7,所以(a2+1)2=(a2-1)(a2+5),解得a2=3,a1=2,所以an是以2为首项,1为公差的等差数列,所以an=n+1.由题意知b1=2,b2=4,b3=8,故bn=2n.(2)由(1)得cn=(-1)nlog22n-1(n+1)(n+2)=(-1)nn-1(n+1)(n+2),故Tn=c1+c2+cn=-1+2-3+(-1)nn-123+134+1(n+1)(n+2).设Fn=-1+2-3+(-1)nn.则当n为偶数时,Fn=(-1+2)+(-3+4)+-(n-1)+n=n2;当n为奇数时,Fn=Fn-1+(-n)=n-12-n
14、=-(n+1)2.设Gn=123+134+1(n+1)(n+2),则Gn=12-13+13-14+1n+1-1n+2=12-1n+2.所以Tn=n-12+1n+2,n为偶数,-n+22+1n+2,n为奇数.12.B解析由an+1=an2an+1,得1an+1=1an+2,数列1an是以1为首项,2为公差的等差数列,1an=2n-1,又bn=anan+1,bn=1(2n-1)(2n+1)=1212n-1-12n+1,Sn=1211-13+13-15+12n-1-12n+1=n2n+1,故选B.13.nn+1解析对nN+都有Sn=1-an,当n=1时,a1=1-a1,解得a1=12.当n2时,an
15、=Sn-Sn-1=1-an-(1-an-1),化为an=12an-1.数列an是等比数列,公比为12,首项为12.an=12n.bn=log2an=-n.1bnbn+1=1-n(-n-1)=1n-1n+1.则1b1b2+1b2b3+1bnbn+1=1-12+12-13+1n-1n+1=1-1n+1=nn+1.14.解(1)当n=1时,a1=32a1-1,a1=2.当n2时,Sn=32an-1,Sn-1=32an-1-1(n2),-得an=32an-1-32an-1-1,即an=3an-1,数列an是首项为2,公比为3的等比数列,an=23n-1.(2)由(1)得bn=2log3an2+1=2n
16、-1,1b1b2+1b2b3+1bn-1bn=113+135+1(2n-3)(2n-1)=121-13+13-15+12n-3-12n-1=n-12n-1.15.A解析设数列的首项为第1组,接下来两项为第2组,再接下来三项为第3组,以此类推,设第n组的项数为n,则前n组的项数和为n(1+n)2.第n组的和为1-2n1-2=2n-1,前n组总共的和为2(1-2n)1-2-n=2n+1-2-n.由题意,N100,令n(1+n)2100,得n14且nN+,即N出现在第13组之后.若要使最小整数N满足:N100且前N项和为2的整数幂,则SN-Sn(1+n)2应与-2-n互为相反数,即2k-1=2+n(
17、kN+,n14),所以k=log2(n+3),解得n=29,k=5.所以N=29(1+29)2+5=440,故选A.16.解(1)Sn=2an+k,当n=1时,S1=2a1+k.a1=-k=2,即k=-2.Sn=2an-2.当n2时,Sn-1=2an-1-2.an=Sn-Sn-1=2an-2an-1.an=2an-1.数列an是以2为首项,2为公比的等比数列.即an=2n.Sn=2n+1-2.(2)等差数列bn的前n项和为Tn,且Tn=n2,当n2时,bn=Tn-Tn-1=2n-1.又b1=T1=1符合bn=2n-1,bn=2n-=anbn=(2n-1)2n.数列cn的前n项和Mn=12+322+523+(2n-3)2n-1+(2n-1)2n,2Mn=122+323+524+(2n-3)2n+(2n-1)2n+1,由-,得-Mn=2+222+223+224+22n-(2n-1)2n+1=2+222-2n+11-2-(2n-1)2n+1,即Mn=6+(2n-3)2n+1.