廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練31 數(shù)列求和 文.docx

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考點(diǎn)規(guī)范練31數(shù)列求和一、基礎(chǔ)鞏固1.數(shù)列112,314,518,7116,(2n-1)+12n,的前n項(xiàng)和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案A解析該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.已知數(shù)列an滿足a1=1,且對(duì)任意的nN*都有an+1=a1+an+n,則1an的前100項(xiàng)和為()A.100101B.99100C.101100D.200101答案D解析an+1=a1+an+n,a1=1,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項(xiàng)和為21-12+12-13+1100-1101=21-1101=200101.故選D.3.已知數(shù)列an滿足an+1-an=2,a1=-5,則|a1|+|a2|+|a6|=()A.9B.15C.18D.30答案C解析an+1-an=2,a1=-5,數(shù)列an是首項(xiàng)為-5,公差為2的等差數(shù)列.an=-5+2(n-1)=2n-7.數(shù)列an的前n項(xiàng)和Sn=n(-5+2n-7)2=n2-6n.令an=2n-70,解得n72.當(dāng)n3時(shí),|an|=-an;當(dāng)n4時(shí),|an|=an.|a1|+|a2|+|a6|=-a1-a2-a3+a4+a5+a6=S6-2S3=62-66-2(32-63)=18.4.已知函數(shù)f(x)=xa的圖象過點(diǎn)(4,2),令an=1f(n+1)+f(n),nN*.記數(shù)列an的前n項(xiàng)和為Sn,則S2 018等于()A.2018-1B.2018+1C.2019-1D.2019+1答案C解析由f(4)=2,可得4a=2,解得a=12,則f(x)=x12.an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2 018=a1+a2+a3+a2 018=(2-1)+(3-2)+(4-3)+(2019-2018)=2019-1.5.已知數(shù)列an滿足an+1+(-1)nan=2n-1,則an的前60項(xiàng)和為()A.3 690B.3 660C.1 845D.1 830答案D解析an+1+(-1)nan=2n-1,當(dāng)n=2k(kN*)時(shí),a2k+1+a2k=4k-1,當(dāng)n=2k+1(kN*)時(shí),a2k+2-a2k+1=4k+1,+得:a2k+a2k+2=8k.則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.已知等差數(shù)列an,a5=2.若函數(shù)f(x)=sin 2x+1,記yn=f(an),則數(shù)列yn的前9項(xiàng)和為.答案9解析由題意,得yn=sin2an+1,所以數(shù)列yn的前9項(xiàng)和為sin2a1+sin2a2+sin2a3+sin2a8+sin2a9+9.由a5=2,得sin2a5=0.a1+a9=2a5=,2a1+2a9=4a5=2,2a1=2-2a9,sin2a1=sin2-2a9=-sin2a9.由倒序相加可得12(sin2a1+sin2a2+sin2a3+sin2a8+sin2a9+sin2a1+sin2a2+sin2a3+sin2a8+sin2a9)=0,y1+y2+y3+y8+y9=9.7.已知數(shù)列an滿足:a3=15,an-an+1=2anan+1,則數(shù)列anan+1前10項(xiàng)的和為.答案1021解析an-an+1=2anan+1,an-an+1anan+1=2,即1an+1-1an=2.數(shù)列1an是以2為公差的等差數(shù)列.1a3=5,1an=5+2(n-3)=2n-1.an=12n-1.anan+1=1(2n-1)(2n+1)=1212n-1-12n+1.數(shù)列anan+1前10項(xiàng)的和為121-13+13-15+1210-1-1210+1=121-121=122021=1021.8.(2018云南昆明第二次統(tǒng)考)在數(shù)列an中,a1=3,an的前n項(xiàng)和Sn滿足Sn+1=an+n2.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列bn滿足bn=(-1)n+2an,求數(shù)列bn的前n項(xiàng)和Tn.解(1)由Sn+1=an+n2,得Sn+1+1=an+1+(n+1)2,-,得an=2n+1.a1=3滿足上式,所以數(shù)列an的通項(xiàng)公式為an=2n+1.(2)由(1)得bn=(-1)n+22n+1,所以Tn=b1+b2+bn=(-1)+(-1)2+(-1)n+(23+25+22n+1)=(-1)1-(-1)n1-(-1)+23(1-4n)1-4=(-1)n-12+83(4n-1).9.設(shè)等差數(shù)列an的公差為d,前n項(xiàng)和為Sn,等比數(shù)列bn的公比為q,已知b1=a1,b2=2,q=d,S10=100.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)當(dāng)d1時(shí),記cn=anbn,求數(shù)列cn的前n項(xiàng)和Tn.解(1)由題意,得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an=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.已知Sn為數(shù)列an的前n項(xiàng)和,an0,an2+2an=4Sn+3.(1)求an的通項(xiàng)公式;(2)設(shè)bn=1anan+1,求數(shù)列bn的前n項(xiàng)和.解(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.兩式相減可得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,可得an+1-an=2.又a12+2a1=4a1+3,解得a1=-1(舍去),a1=3.所以an是首項(xiàng)為3,公差為2的等差數(shù)列,故an的通項(xiàng)公式為an=2n+1.(2)由an=2n+1可知bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.設(shè)數(shù)列bn的前n項(xiàng)和為Tn,則Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).11.已知各項(xiàng)均為正數(shù)的數(shù)列an的前n項(xiàng)和為Sn,滿足an+12=2Sn+n+4,a2-1,a3,a7恰為等比數(shù)列bn的前3項(xiàng).(1)求數(shù)列an,bn的通項(xiàng)公式;(2)若cn=(-1)nlog2bn-1anan+1,求數(shù)列cn的前n項(xiàng)和Tn.解(1)因?yàn)閍n+12=2Sn+n+4,所以an2=2Sn-1+n-1+4(n2).兩式相減,得an+12-an2=2an+1,所以an+12=an2+2an+1=(an+1)2.因?yàn)閍n是各項(xiàng)均為正數(shù)的數(shù)列,所以an+1=an+1,即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為首項(xiàng),1為公差的等差數(shù)列,所以an=n+1.由題意知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).設(shè)Fn=-1+2-3+(-1)nn,則當(dāng)n為偶數(shù)時(shí),Fn=(-1+2)+(-3+4)+-(n-1)+n=n2;當(dāng)n為奇數(shù)時(shí),Fn=Fn-1+(-n)=n-12-n=-(n+1)2.設(shè)Gn=123+134+1(n+1)(n+2),則Gn=12-13+13-14+1n+1-1n+2=12-1n+2.所以Tn=n-12+1n+2,n為偶數(shù),-n+22+1n+2,n為奇數(shù).二、能力提升12.在數(shù)列an中,a1=1,且an+1=an2an+1.若bn=anan+1,則數(shù)列bn的前n項(xiàng)和Sn為()A.2n2n+1B.n2n+1C.2n2n-1D.2n-12n+1答案B解析由an+1=an2an+1,得1an+1=1an+2,數(shù)列1an是以1為首項(xiàng),2為公差的等差數(shù)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,故選B.13.(2018福建寧德期末)今要在一個(gè)圓周上標(biāo)出一些數(shù),第一次先把圓周二等分,在這兩個(gè)分點(diǎn)處分別標(biāo)上1,如圖(1)所示;第二次把兩段半圓弧二等分,在這兩個(gè)分點(diǎn)處分別標(biāo)上2,如圖(2)所示;第三次把四段圓弧二等分,并在這4個(gè)分點(diǎn)處分別標(biāo)上3,如圖(3)所示.如此繼續(xù)下去,當(dāng)?shù)趎次標(biāo)完數(shù)以后,這個(gè)圓周上所有已標(biāo)出的數(shù)的總和是.答案(n-1)2n+2解析由題意可得,第n次標(biāo)完后,圓周上所有已標(biāo)出的數(shù)的總和為Tn=1+1+22+322+n2n-1.設(shè)S=1+22+322+n2n-1,則2S=2+222+(n-1)2n-1+n2n,兩式相減可得-S=1+2+22+2n-1-n2n=(1-n)2n-1,則S=(n-1)2n+1,故Tn=(n-1)2n+2.14.已知首項(xiàng)為32的等比數(shù)列an不是遞減數(shù)列,其前n項(xiàng)和為Sn(nN*),且S3+a3,S5+a5,S4+a4成等差數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=(-1)n+1n(nN*),求數(shù)列anbn的前n項(xiàng)和Tn.解(1)設(shè)等比數(shù)列an的公比為q.由S3+a3,S5+a5,S4+a4成等差數(shù)列,可得2(S5+a5)=S3+a3+S4+a4,即2(S3+a4+2a5)=2S3+a3+2a4,即4a5=a3,則q2=a5a3=14,解得q=12.由等比數(shù)列an不是遞減數(shù)列,可得q=-12,故an=32-12n-1=(-1)n-132n.(2)由bn=(-1)n+1n,可得anbn=(-1)n-132n(-1)n+1n=3n12n.故前n項(xiàng)和Tn=3112+2122+n12n,則12Tn=31122+2123+n12n+1,兩式相減可得,12Tn=312+122+12n-n12n+1=3121-12n1-12-n12n+1,化簡可得Tn=61-n+22n+1.15.(2018湖南長沙雅禮中學(xué)模擬)若數(shù)列an的前n項(xiàng)和Sn滿足Sn=2an-(0,nN*).(1)證明:數(shù)列an為等比數(shù)列,并求an;(2)若=4,bn=an,n是奇數(shù),log2an,n是偶數(shù)(nN*),求數(shù)列bn的前2n項(xiàng)和T2n.(1)證明Sn=2an-,當(dāng)n=1時(shí),得a1=,當(dāng)n2時(shí),Sn-1=2an-1-,則Sn-Sn-1=2an-2an-1,即an=2an-2an-1,an=2an-1,數(shù)列an是以為首項(xiàng),2為公比的等比數(shù)列,an=2n-1.(2)解=4,an=42n-1=2n+1,bn=2n+1,n是奇數(shù),n+1,n是偶數(shù).T2n=22+3+24+5+26+7+22n+2n+1=(22+24+26+22n)+(3+5+2n+1)=4-22n41-4+n(3+2n+1)2=4n+1-43+n(n+2),T2n=4n+13+n2+2n-43.三、高考預(yù)測16.已知數(shù)列an的前n項(xiàng)和為Sn,且a1=2,Sn=2an+k,等差數(shù)列bn的前n項(xiàng)和為Tn,且Tn=n2.(1)求k和Sn;(2)若cn=anbn,求數(shù)列cn的前n項(xiàng)和Mn.解(1)Sn=2an+k,當(dāng)n=1時(shí),S1=2a1+k.a1=-k=2,即k=-2.Sn=2an-2.當(dāng)n2時(shí),Sn-1=2an-1-2.an=Sn-Sn-1=2an-2an-1.an=2an-1.數(shù)列an是以2為首項(xiàng),2為公比的等比數(shù)列.即an=2n.Sn=2n+1-2.(2)等差數(shù)列bn的前n項(xiàng)和為Tn,且Tn=n2,當(dāng)n2時(shí),bn=Tn-Tn-1=2n-1.又b1=T1=1符合bn=2n-1,bn=2n-1.cn=anbn=(2n-1)2n.數(shù)列cn的前n項(xiàng)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.