（通用版）2020版高考數(shù)學(xué)復(fù)習(xí)專題四數(shù)列4.2數(shù)列解答題練習(xí)文.docx

4.2數(shù)列解答題高考命題規(guī)律1.高考命題的完全考題,常與解三角形解答題交替在第17題呈現(xiàn).2.解答題,12分,中檔難度.3.全國高考有3種命題角度,分布如下表.2020年高考必備2015年2016年2017年2018年2019年卷卷卷卷卷卷卷卷卷卷卷卷卷卷命題角度1等差、等比數(shù)列的判定與證明1717171719命題角度2等差、等比數(shù)列的通項(xiàng)公式與前n項(xiàng)和公式的應(yīng)用1717171718命題角度3一般數(shù)列的通項(xiàng)公式與前n項(xiàng)和的求解17命題角度1等差、等比數(shù)列的判定與證明高考真題體驗(yàn)對(duì)方向1.(2019全國19)已知數(shù)列an和bn滿足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)證明:an+bn是等比數(shù)列,an-bn是等差數(shù)列;(2)求an和bn的通項(xiàng)公式.(1)證明由題設(shè)得4(an+1+bn+1)=2(an+bn),即an+1+bn+1=12(an+bn).又因?yàn)閍1+b1=1,所以an+bn是首項(xiàng)為1,公比為12的等比數(shù)列.由題設(shè)得4(an+1-bn+1)=4(an-bn)+8,即an+1-bn+1=an-bn+2.又因?yàn)閍1-b1=1,所以an-bn是首項(xiàng)為1,公差為2的等差數(shù)列.(2)解由(1)知,an+bn=12n-1,an-bn=2n-1.所以an=12(an+bn)+(an-bn)=12n+n-12,bn=12(an+bn)-(an-bn)=12n-n+12.2.(2018全國17)已知數(shù)列an滿足a1=1,nan+1=2(n+1)an.設(shè)bn=ann.(1)求b1,b2,b3;(2)判斷數(shù)列bn是否為等比數(shù)列,并說明理由;(3)求an的通項(xiàng)公式.解(1)由條件可得an+1=2(n+1)nan.將n=1代入得,a2=4a1,而a1=1,所以a2=4.將n=2代入得,a3=3a2,所以a3=12.從而b1=1,b2=2,b3=4.(2)bn是首項(xiàng)為1,公比為2的等比數(shù)列.由條件可得an+1n+1=2ann,即bn+1=2bn,又b1=1,所以bn是首項(xiàng)為1,公比為2的等比數(shù)列.(3)由(2)可得ann=2n-1,所以an=n2n-1.3.(2017全國17)設(shè)Sn為等比數(shù)列an的前n項(xiàng)和,已知S2=2,S3=-6.(1)求an的通項(xiàng)公式;(2)求Sn,并判斷Sn+1,Sn,Sn+2是否成等差數(shù)列.解(1)設(shè)an的公比為q.由題設(shè)可得a1(1+q)=2,a1(1+q+q2)=-6.解得q=-2,a1=-2.故an的通項(xiàng)公式為an=(-2)n.(2)由(1)可得Sn=a1(1-qn)1-q=-23+(-1)n2n+13.由于Sn+2+Sn+1=-43+(-1)n2n+3-2n+23=2-23+(-1)n2n+13=2Sn,故Sn+1,Sn,Sn+2成等差數(shù)列.典題演練提能刷高分1.(2019黑龍江哈爾濱第三中學(xué)高三上學(xué)期期中考試)已知數(shù)列an中,a1=32且an=12(an-1+n+1)(n2,nN*).(1)求a2,a3,并證明an-n是等比數(shù)列;(2)設(shè)bn=2nan,求數(shù)列bn的前n項(xiàng)和Sn.解(1)由題意,可知:a2=12(a1+2+1)=1232+2+1=94,a3=12(a2+3+1)=1294+3+1=258.當(dāng)n=1時(shí),a1-1=32-1=12,當(dāng)n2時(shí),an-n=12(an-1+n+1)-n=12an-1+12n+12-n=12an-1-12n+12=12(an-1-n+1)=12an-1-(n-1).數(shù)列an-n是以12為首項(xiàng),12為公比的等比數(shù)列.(2)由(1)可知,an-n=12n,an=n+12n,nN*.bn=2nan=2nn+12n=n2n+2n12n=n2n+1.Sn=b1+b2+b3+bn=(121+1)+(222+1)+(323+1)+(n2n+1),Sn=121+222+323+n2n+n,2Sn=122+223+(n-1)2n+n2n+1+2n,由-,可得:-Sn=121+122+123+12n-n2n+1+n-2n=2-2n+11-2-n2n+1-n=(1-n)2n+1-n-2,Sn=(n-1)2n+1+n+2.2.已知數(shù)列an的通項(xiàng)公式為an=2n-11.(1)求證:數(shù)列an是等差數(shù)列;(2)令bn=|an|,求數(shù)列bn的前10項(xiàng)和S10.(1)證明an=2n-11,an+1-an=2(n+1)-11-2n+11=2(nN*),數(shù)列an為等差數(shù)列.(2)解由(1)得bn=an=|2n-11|,當(dāng)n5時(shí),bn=|2n-11|=11-2n;當(dāng)n6時(shí),bn=|2n-11|=2n-11.S10=55-2(1+2+3+4+5)+2(6+7+8+9+10)-55=50.3.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=2an-1.(1)證明數(shù)列an是等比數(shù)列;(2)設(shè)bn=(2n-1)an,求數(shù)列bn的前n項(xiàng)和Tn.解(1)當(dāng)n=1時(shí),a1=S1=2a1-1,所以a1=1,當(dāng)n2時(shí),an=Sn-Sn-1=(2an-1)-(2an-1-1),所以an=2an-1,所以數(shù)列an是以a1=1為首項(xiàng),以2為公比的等比數(shù)列.(2)由(1)知,an=2n-1,所以bn=(2n-1)2n-1,所以Tn=1+32+522+(2n-3)2n-2+(2n-1)2n-1,2Tn=12+322+(2n-3)2n-1+(2n-1)2n,-得-Tn=1+2(21+22+2n-1)-(2n-1)2n=1+22-2n-121-2-(2n-1)2n=(3-2n)2n-3,所以Tn=(2n-3)2n+3.4.設(shè)a1=2,a2=4,數(shù)列bn滿足:bn+1=2bn+2且an+1-an=bn.(1)求證:數(shù)列bn+2是等比數(shù)列;(2)求數(shù)列an的通項(xiàng)公式.解(1)由題知bn+1+2bn+2=2bn+2+2bn+2=2,又b1=a2-a1=4-2=2,b1+2=4,bn+2是以4為首項(xiàng),以2為公比的等比數(shù)列.(2)由(1)可得bn+2=42n-1,故bn=2n+1-2.an+1-an=bn,a2-a1=b1,a3-a2=b2,a4-a3=b3,an-an-1=bn-1.累加得an-a1=b1+b2+b3+bn-1,an=2+(22-2)+(23-2)+(24-2)+(2n-2)=2+22(1-2n-1)1-2-2(n-1)=2n+1-2n,即an=2n+1-2n(n2).而a1=2=21+1-21,an=2n+1-2n(nN*).命題角度2等差、等比數(shù)列的通項(xiàng)公式與前n項(xiàng)和公式的應(yīng)用高考真題體驗(yàn)對(duì)方向1.(2019全國18)記Sn為等差數(shù)列an的前n項(xiàng)和.已知S9=-a5.(1)若a3=4,求an的通項(xiàng)公式;(2)若a10,求使得Snan的n的取值范圍.解(1)設(shè)an的公差為d.由S9=-a5得a1+4d=0.由a3=4得a1+2d=4.于是a1=8,d=-2.因此an的通項(xiàng)公式為an=10-2n.(2)由(1)得a1=-4d,故an=(n-5)d,Sn=n(n-9)d2.由a10知d0;當(dāng)n6時(shí),an0.所以,Sn的最小值為S6=-30.3.(2018全國17)記Sn為等差數(shù)列an的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求an的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解(1)設(shè)an的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.所以an的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.4.(2018全國17)等比數(shù)列an中,a1=1,a5=4a3.(1)求an的通項(xiàng)公式;(2)記Sn為an的前n項(xiàng)和,若Sm=63,求m.解(1)設(shè)an的公比為q,由題設(shè)得an=qn-1.由已知得q4=4q2,解得q=0(舍去),q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,則Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程沒有正整數(shù)解.若an=2n-1,則Sn=2n-1.由Sm=63得2m=64,解得m=6.綜上,m=6.5.(2016全國17)已知an是公差為3的等差數(shù)列,數(shù)列bn滿足b1=1,b2=13,anbn+1+bn+1=nbn.(1)求an的通項(xiàng)公式;(2)求bn的前n項(xiàng)和.解(1)由已知,a1b2+b2=b1,b1=1,b2=13,得a1=2.所以數(shù)列an是首項(xiàng)為2,公差為3的等差數(shù)列,通項(xiàng)公式為an=3n-1.(2)由(1)和anbn+1+bn+1=nbn得bn+1=bn3,因此bn是首項(xiàng)為1,公比為13的等比數(shù)列.記bn的前n項(xiàng)和為Sn,則Sn=1-13n1-13=32-123n-1.6.(2016全國17)等差數(shù)列an中,a3+a4=4,a5+a7=6.(1)求an的通項(xiàng)公式;(2)設(shè)bn=an,求數(shù)列bn的前10項(xiàng)和,其中x表示不超過x的最大整數(shù),如0.9=0,2.6=2.解(1)設(shè)數(shù)列an的公差為d,由題意有2a1+5d=4,a1+5d=3,解得a1=1,d=25.所以an的通項(xiàng)公式為an=2n+35.(2)由(1)知,bn=2n+35.當(dāng)n=1,2,3時(shí),12n+352,bn=1;當(dāng)n=4,5時(shí),22n+353,bn=2;當(dāng)n=6,7,8時(shí),32n+354,bn=3;當(dāng)n=9,10時(shí),42n+3535成立的n的最小值.解(1)設(shè)等差數(shù)列an的公差為d,d0.因?yàn)閍2,a3,a6成等比數(shù)列,所以a32=a2a6,即(1+d)2=1+4d,解得d=2或d=0(舍去).所以an的通項(xiàng)公式為an=a2+(n-2)d=2n-3.(2)因?yàn)閍n=2n-3,所以Sn=n(a1+an)2=n(a2+an-1)2=n2-2n.依題意有n2-2n35,解得n7.故使Sn35成立的n的最小值為8.2.(2019廣東佛山第一中學(xué)高三上學(xué)期期中考試)等差數(shù)列an中,a1=3,前n項(xiàng)和為Sn,等比數(shù)列bn各項(xiàng)均為正數(shù),b1=1,且b2+S2=12,bn的公比q=S2b2.(1)求an與bn;(2)求Tn=Sb1+Sb2+Sb3+Sb4+Sbn.解(1)由已知可得q+3+a2=12,q=3+a2q,解得q=3或q=-4(舍去負(fù)值),a2=6.an=3n,bn=3n-1.(2)Sn=3n(n+1)2,Sbn=32bn(bn+1)=32(bn2+bn),Tn=Sb1+Sb2+Sb3+Sb4+Sbn=32(b1+b2+bn)+(b12+b22+bn2)=32(30+31+3n-1)+(30+32+32n-2)=321-3n1-3+1-32n1-32=32n+116+3n+14-1516.3.(2019西南名校聯(lián)盟重慶第八中學(xué)高三5月高考適應(yīng)性月考)已知Sn為等差數(shù)列an的前n項(xiàng)和,且S7=28,a2=2.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=4an-1,求數(shù)列bn的前n項(xiàng)和Tn.解(1)設(shè)an的首項(xiàng)為a1,公差為d,由a2=a1+d=2,S7=7a1+21d=28,解得a1=1,d=1,所以an=n.(2)bn=4n-1,所以bn的前n項(xiàng)和Tn=1-4n1-4=4n-13.4.(2019寧夏石嘴山第三中學(xué)高三下學(xué)期三模)已知等差數(shù)列an是遞增數(shù)列,且a1+a4=0,a2a3=-1.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=3an+4,數(shù)列bn的前n項(xiàng)和為Tn,是否存在常數(shù),使得Tn-bn+1恒為定值?若存在,求出的值;若不存在,請(qǐng)說明理由.解(1)已知等差數(shù)列an是遞增數(shù)列,設(shè)公差為d且a1+a4=0,a2a3=-1.則2a1+3d=0,(a1+d)(a1+2d)=-1,解得a1=-3,d=2.所以an=a1+(n-1)d=2n-5.(2)由于bn=3an+4,所以bn=32n-1.數(shù)列bn是以3為首項(xiàng),9為公比的等比數(shù)列.則Tn=3(1-9n)1-9=38(9n-1).所以Tn-bn+1=38(9n-1)-332n=38-19n-38.當(dāng)8-1=0,即=8時(shí),Tn-bn+1恒為定值-3.5.(2019西藏山南地區(qū)第二高級(jí)中學(xué)高三上學(xué)期期中模擬)已知等差數(shù)列an的前n項(xiàng)和為Sn,且S9=90,S15=240.(1)求數(shù)列an的通項(xiàng)公式和前n項(xiàng)和Sn;(2)設(shè)bn-(-1)nan是等比數(shù)列,且b2=7,b5=71,求數(shù)列bn的前n項(xiàng)和Tn.解(1)設(shè)等差數(shù)列an的首項(xiàng)為a1,公差為d.則由S9=90,S15=240,得9a1+36d=90,15a1+105d=240,解得a1=2,d=2.所以an=2+(n-1)2=2n,即an=2n.Sn=2n+n(n-1)22=n(n+1),即Sn=n(n+1).(2)令cn=bn-(-1)nan,設(shè)cn的公比為q,b2=7,b5=71,an=2n,c2=b2-(-1)2a2=3,c5=b5-(-1)5a5=81,q3=c5c2=27,q=3,cn=c2qn-2=3n-1,從而bn=3n-1+(-1)n2n,Tn=b1+b2+bn=(30+31+3n-1)+-2+4-6+(-1)n2n,當(dāng)n為偶數(shù)時(shí),Tn=3n+2n-12;當(dāng)n為奇數(shù)時(shí),Tn=3n-2n-32.所以Tn=3n+2n-12,n為偶數(shù),3n-2n-32,n為奇數(shù).6.(2019貴州貴陽高三5月適應(yīng)性考試)等差數(shù)列an的前n項(xiàng)和為Sn,公差d0,已知S4=16,a1,a2,a5成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)記點(diǎn)A(n,Sn),B(n+1,Sn+1),C(n+2,Sn+2),求證:ABC的面積為1.(1)解由題意得4a1+432d=16,(a1+d)2=a1(a1+4d),由于d0,解得a1=1,d=2.an=1+(n-1)2=2n-1.(2)證明由(1)知Sn=n1+n(n-1)22=n2,ABC的面積S=12(Sn+Sn+2)2-12(Sn+Sn+1)1-12(Sn+1+Sn+2)1=12(Sn+Sn+2-2Sn+1)=12n2+(n+2)2-2(n+1)2=1.命題角度3一般數(shù)列的通項(xiàng)公式與前n項(xiàng)和的求解高考真題體驗(yàn)對(duì)方向1.(2019天津18)設(shè)an是等差數(shù)列,bn是等比數(shù)列,公比大于0.已知a1=b1=3,b2=a3,b3=4a2+3.(1)求an和bn的通項(xiàng)公式;(2)設(shè)數(shù)列cn滿足cn=1,n為奇數(shù),bn2,n為偶數(shù),求a1c1+a2c2+a2nc2n(nN*).解(1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q.依題意,得3q=3+2d,3q2=15+4d.解得d=3,q=3,故an=3+3(n-1)=3n,bn=33n-1=3n.所以,an的通項(xiàng)公式為an=3n,bn的通項(xiàng)公式為bn=3n.(2)a1c1+a2c2+a2nc2n=(a1+a3+a5+a2n-1)+(a2b1+a4b2+a6b3+a2nbn)=n3+n(n-1)26+(631+1232+1833+6n3n)=3n2+6(131+232+n3n).記Tn=131+232+n3n,則3Tn=132+233+n3n+1,-得,2Tn=-3-32-33-3n+n3n+1=-3(1-3n)1-3+n3n+1=(2n-1)3n+1+32.所以,a1c1+a2c2+a2nc2n=3n2+6Tn=3n2+3(2n-1)3n+1+32=(2n-1)3n+2+6n2+92(nN*).2.(2017全國17)設(shè)數(shù)列an滿足a1+3a2+(2n-1)an=2n.(1)求an的通項(xiàng)公式;(2)求數(shù)列an2n+1的前n項(xiàng)和.解(1)因?yàn)閍1+3a2+(2n-1)an=2n,故當(dāng)n2時(shí),a1+3a2+(2n-3)an-1=2(n-1).兩式相減得(2n-1)an=2.所以an=22n-1(n2).又由題設(shè)可得a1=2,從而an的通項(xiàng)公式為an=22n-1.(2)記an2n+1的前n項(xiàng)和為Sn.由(1)知an2n+1=2(2n+1)(2n-1)=12n-1-12n+1.則Sn=11-13+13-15+12n-1-12n+1=2n2n+1.3.(2017天津18)已知an為等差數(shù)列,前n項(xiàng)和為Sn(nN*),bn是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.(1)求an和bn的通項(xiàng)公式;(2)求數(shù)列a2nbn的前n項(xiàng)和(nN*).解(1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q.由已知b2+b3=12,得b1(q+q2)=12,而b1=2,所以q2+q-6=0.又因?yàn)閝0,解得q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.由S11=11b4,可得a1+5d=16,聯(lián)立,解得a1=1,d=3,由此可得an=3n-2.所以,an的通項(xiàng)公式為an=3n-2,bn的通項(xiàng)公式為bn=2n.(2)設(shè)數(shù)列a2nbn的前n項(xiàng)和為Tn,由a2n=6n-2,有Tn=42+1022+1623+(6n-2)2n,2Tn=422+1023+1624+(6n-8)2n+(6n-2)2n+1,上述兩式相減,得-Tn=42+622+623+62n-(6n-2)2n+1=12(1-2n)1-2-4-(6n-2)2n+1=-(3n-4)2n+2-16.得Tn=(3n-4)2n+2+16.所以,數(shù)列a2nbn的前n項(xiàng)和為(3n-4)2n+2+16.典題演練提能刷高分1.已知等比數(shù)列an的前n項(xiàng)和為Sn,且滿足2Sn=2n+1+m(mR).(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足bn=1(2n+1)log2(anan+1),求數(shù)列bn的前n項(xiàng)和Tn.解(1)由2Sn=2n+1+m(mR)得2Sn-1=2n+m(mR),當(dāng)n2時(shí),2an=2Sn-2Sn-1=2n,即an=2n-1(n2),又a1=S1=2+m2,當(dāng)m=-2時(shí)符合上式,所以通項(xiàng)公式為an=2n-1.(2)由(1)可得log2(anan+1)=log2(2n-12n)=2n-1,bn=1(2n+1)(2n-1)=1212n-1-12n+1,Tn=b1+b2+bn=121-13+13-15+12n-1-12n+1=n2n+1.2.(2019北京豐臺(tái)區(qū)高三第二學(xué)期綜合練習(xí)二)已知數(shù)列an滿足a1=1,an+1=ean(e是自然對(duì)數(shù)的底數(shù),nN*).(1)求an的通項(xiàng)公式;(2)設(shè)數(shù)列l(wèi)n an的前n項(xiàng)和為Tn,求證:當(dāng)n2時(shí),1T2+1T3+1Tn0,所以1-1n1.所以21-1n2.即1T2+1T3+1Tn2.3.已知等比數(shù)列an的前n項(xiàng)和為Sn,滿足S4=2a4-1,S3=2a3-1.(1)求an的通項(xiàng)公式;(2)記bn=log2(anan+1),數(shù)列bn的前n項(xiàng)和為Tn,求證:1T1+1T2+1Tn2.解(1)設(shè)an的公比為q,由S4-S3=a4得2a4-2a3=a4,所以a4a3=2,所以q=2.又因?yàn)镾3=2a3-1,所以a1+2a1+4a1=8a1-1,所以a1=1.所以an=2n-1.(2)由(1)知bn=log2(an+1an)=log2(2n2n-1)=2n-1,所以Tn=1+(2n-1)2n=n2,所以1T1+1T2+1Tn=112+122+1n21+112+123+1(n-1)n=1+1-12+12-13+1n-1-1n=2-1n2.4.(2019河北保定高三第二次模擬考試)已知函數(shù)f(x)=log3(ax+b)的圖象經(jīng)過點(diǎn)A(2,1)和B(5,2),an=an+b,nN*.(1)求an;(2)設(shè)數(shù)列an的前n項(xiàng)和為Sn,bn=2n+2Sn,求bn的前n項(xiàng)和Tn.解(1)由函數(shù)f(x)=log3(ax+b)的圖象經(jīng)過點(diǎn)A(2,1)和B(5,2),得log3(2a+b)=1,log3(5a+b)=2,解得a=2,b=-1.所以an=2n-1,nN*.(2)由(1)知數(shù)列an為以1為首項(xiàng),2為公差的等差數(shù)列,所以Sn=n+n(n-1)22=n2,