（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練22 數(shù)列的概念與表示（含解析）新人教A版

《（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練22 數(shù)列的概念與表示（含解析）新人教A版》由會(huì)員分享，可在線閱讀，更多相關(guān)《（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練22 數(shù)列的概念與表示（含解析）新人教A版（6頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、考點(diǎn)規(guī)范練22數(shù)列的概念與表示一、基礎(chǔ)鞏固1.數(shù)列1,23,35,47,59,的一個(gè)通項(xiàng)公式an=()A.n2n+1B.n2n-1C.n2n-3D.n2n+32.若Sn為數(shù)列an的前n項(xiàng)和,且Sn=nn+1,則1a5等于()A.56B.65C.130D.303.已知數(shù)列an滿足an+1+an=n,若a1=2,則a4-a2=()A.4B.3C.2D.14.若數(shù)列an滿足1an+1-1an=d(nN*,d為常數(shù)),則稱數(shù)列an為調(diào)和數(shù)列.已知數(shù)列1xn為調(diào)和數(shù)列,且x1+x2+x20=200,則x5+x16=()A.10B.20C.30D.405.已知數(shù)列an滿足:m,nN*,都有anam=an+

2、m,且a1=12,則a5=()A.132B.116C.14D.126.已知數(shù)列an的前4項(xiàng)分別是32,1,710,917,則這個(gè)數(shù)列的一個(gè)通項(xiàng)公式是an=.7.已知數(shù)列an滿足:a1+3a2+5a3+(2n-1)an=(n-1)3n+1+3(nN*),則數(shù)列an的通項(xiàng)公式an=.8.已知數(shù)列an的通項(xiàng)公式為an=(n+2)78n,則當(dāng)an取得最大值時(shí),n=.9.若數(shù)列an的通項(xiàng)為an=(-1)n(2n+1)sinn2+1,前n項(xiàng)和為Sn,則S100=.10.已知數(shù)列an的前n項(xiàng)和為Sn.(1)若Sn=(-1)n+1n,求a5+a6及an;(2)若Sn=3n+2n+1,求an.二、能力提升11.

3、已知數(shù)列an的通項(xiàng)an=nn2+90,則an的最大值是()A.310B.19C.119D.106012.已知數(shù)列an滿足an+1=2an,0an12,2an-1,12anm(其中m,nN*),Sn-Sm的最大值是.考點(diǎn)規(guī)范練22數(shù)列的概念與表示1.B2.D解析當(dāng)n2時(shí),an=Sn-Sn-1=nn+1-n-1n=1n(n+1),1a5=5(5+1)=30.3.D解析由an+1+an=n,得an+2+an+1=n+1,兩式相減得an+2-an=1,令n=2,得a4-a2=1.4.B解析數(shù)列1xn為調(diào)和數(shù)列,11xn+1-11xn=xn+1-xn=d.xn是等差數(shù)列.又x1+x2+x20=200=2

4、0(x1+x20)2,x1+x20=20.又x1+x20=x5+x16,x5+x16=20.5.A解析數(shù)列an滿足:m,nN*,都有anam=an+m,且a1=12,a2=a1a1=14,a3=a1a2=18,a5=a3a2=132.6.2n+1n2+1解析數(shù)列an的前4項(xiàng)可分別變形為21+112+1,22+122+1,23+132+1,24+142+1,故an=2n+1n2+1.7.3n解析a1+3a2+5a3+(2n-3)an-1+(2n-1)an=(n-1)3n+1+3,把n換成n-1,得a1+3a2+5a3+(2n-3)an-1=(n-2)3n+3,兩式相減得an=3n.8.5或6解析

5、由題意令anan-1,anan+1,(n+2)78n(n+1)78n-1,(n+2)78n(n+3)78n+1,解得n6,n5.n=5或n=6.9.200解析當(dāng)n為偶數(shù)時(shí),則sinn2=0,即an=(2n+1)sinn2+1=1(n為偶數(shù)).當(dāng)n為奇數(shù)時(shí),若n=4k+1,kZ,則sinn2=sin2k+2=1,即an=-2n;若n=4k+3,kZ,則sinn2=sin2k+32=-1,即an=2n+2.故a4k+1+a4k+2+a4k+3+a4k+4=-2(4k+1)+1+2+2(4k+3)+1=8,因此S100=10048=200.10.解(1)因?yàn)镾n=(-1)n+1n,所以a5+a6=S

6、6-S4=(-6)-(-4)=-2.當(dāng)n=1時(shí),a1=S1=1;當(dāng)n2時(shí),an=Sn-Sn-1=(-1)n+1n-(-1)n(n-1)=(-1)n+1n+(n-1)=(-1)n+1(2n-1).又a1也適合于此式,所以an=(-1)n+1(2n-1).(2)當(dāng)n=1時(shí),a1=S1=6;當(dāng)n2時(shí),an=Sn-Sn-1=(3n+2n+1)-3n-1+2(n-1)+1=23n-1+2.因?yàn)閍1不適合式,所以an=6,n=1,23n-1+2,n2.11.C解析令f(x)=x+90x(x0),運(yùn)用基本不等式得f(x)290,當(dāng)且僅當(dāng)x=310時(shí)等號(hào)成立.因?yàn)閍n=1n+90n,所以1n+90n1290,

7、由于nN*,不難發(fā)現(xiàn)當(dāng)n=9或n=10時(shí),an取得最大值,故an=119最大.12.15解析由已知可得,a2=235-1=15,a3=215=25,a4=225=45,a5=245-1=35,an為周期數(shù)列且T=4,a2018=a5044+2=a2=15.13.66解析由題得,這個(gè)數(shù)列各項(xiàng)的值分別為1,1,3,1,5,3,7,1,9,5,11,3,a64+a65=a32+65=a16+65=a8+65=a4+65=1+65=66.14.2n-1解析當(dāng)n2時(shí),an=Sn-Sn-1=2an-n-2an-1+(n-1),即an=2an-1+1,an+1=2(an-1+1).又S1=2a1-1,a1=

8、1.數(shù)列an+1是以首項(xiàng)為a1+1=2,公比為2的等比數(shù)列,an+1=22n-1=2n,an=2n-1.15.解(1)因?yàn)閍n+1=Sn+3n,所以Sn+1-Sn=an+1=Sn+3n,即Sn+1=2Sn+3n,由此得Sn+1-3n+1=2(Sn-3n),即bn+1=2bn.又b1=S1-3=a-3,故bn的通項(xiàng)公式為bn=(a-3)2n-1.(2)由題意可知,a2a1對(duì)任意的a都成立.由(1)知Sn=3n+(a-3)2n-1.于是,當(dāng)n2時(shí),an=Sn-Sn-1=3n+(a-3)2n-1-3n-1-(a-3)2n-2=23n-1+(a-3)2n-2,故an+1-an=43n-1+(a-3)2n-2=2n-21232n-2+a-3.當(dāng)n2時(shí),由an+1an,可知1232n-2+a-30,即a-9.又a3,故所求的a的取值范圍是-9,3)(3,+).16.10解析由an=-n2+12n-32=-(n-4)(n-8)0得4n8,即在數(shù)列an中,前三項(xiàng)以及從第9項(xiàng)起后的各項(xiàng)均為負(fù)且a4=a8=0,因此Sn-Sm的最大值是a5+a6+a7=3+4+3=10.6