專題突破練12　求數(shù)列的通項(xiàng)及前n項(xiàng)和1.(2021·湖南長郡中學(xué)月考)已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),且2a1+3a2=1,=9a2a6.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=log3a1+log3a2+…+log3an,求數(shù)列的前n項(xiàng)和Tn.2.(2021·河北張家口高三三模)已知數(shù)列{an}的前n項(xiàng)和為An,數(shù)列{bn}的前n項(xiàng)和為Bn,且An-Bn=2n+1-2.(1)求數(shù)列{an-bn}的通項(xiàng)公式;(2)若an+bn=,求數(shù)列{an·bn}的前n項(xiàng)和Tn.3.(2021·湖南郴州高三質(zhì)監(jiān))已知數(shù)列{an}的首項(xiàng)a1=1,前n項(xiàng)和為Sn,且數(shù)列是公差為1 的等差數(shù)列.(1)求數(shù)列的前n項(xiàng)和Tn;(2)設(shè)等比數(shù)列{cn}的首項(xiàng)為2,公比為q(q>0),其前n項(xiàng)和為Pn,若存在正整數(shù)m,使得是Sm與P3的等比中項(xiàng),求q的值.4.(2021·廣東珠海高三一模)在①6a1=a2+a3,②a4=2a1+a2+a3,③2(a3+2)=a2+a4這三個(gè)條件中任選一個(gè),補(bǔ)充在下面問題中,并作答.問題:各項(xiàng)均為正數(shù)的等比數(shù)列{an}的公比為q,滿足an0成立,求m的取值范圍.5.(2021·廣東揭陽檢測(cè))已知等差數(shù)列{an}與正項(xiàng)等比數(shù)列{bn}滿足a1=b1=3,且b3-a3,20,a5+b2既是等差數(shù)列,又是等比數(shù)列.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;(2)在①cn=+(-1)nbn,②cn=an·bn,③cn=這三個(gè)條件中任選一個(gè),補(bǔ)充在下面問題中,并完成求解.若　　　　　,求數(shù)列{cn}的前n項(xiàng)和Sn.? 6.(2021·山東菏澤一模)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且an+1=2Sn+2,數(shù)列{bn}滿足b1=2,(n+2)bn=nbn+1.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;(2)在an與an+1之間插入n個(gè)數(shù),使這n+2個(gè)數(shù)組成一個(gè)公差為cn的等差數(shù)列,求數(shù)列{bncn}的前n項(xiàng)和Tn.7.(2021·廣東廣州檢測(cè))已知數(shù)列{an}滿足a1=1,an+1=3an+3n+1.(1)求證:數(shù)列是等差數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式;(3)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,求證:.專題突破練12　求數(shù)列的通項(xiàng)及前n項(xiàng)和1.解(1)設(shè)等比數(shù)列{an}的公比為q(q>0),由=9a2a6,得=9,所以q2=,所以q=由2a1+3a2=1,得2a1+3a1=1,所以a1=故數(shù)列{an}的通項(xiàng)公式為an=(2)因?yàn)閎n=log3a1+log3a2+…+log3an=-(1+2+…+n)=-,所以=-=-2 所以Tn=+…+=-2+…+=-所以數(shù)列的前n項(xiàng)和Tn=-2.解(1)記數(shù)列{an-bn}的前n項(xiàng)和為Sn,所以Sn=An-Bn=2n+1-2,所以當(dāng)n≥2時(shí),Sn-1=2n-2,兩式作差,得an-bn=2n(n≥2).因?yàn)楫?dāng)n=1時(shí),S1=A1-B1=2,也符合上式,所以{an-bn}的通項(xiàng)公式為an-bn=2n.(2)由(1)知an-bn=2n,因?yàn)閍n+bn=,所以an·bn=[(an+bn)2-(an-bn)2]=(n-4n),所以數(shù)列{an·bn}的前n項(xiàng)和Tn=(1+2+…+n)-(41+42+…+4n)=,所以數(shù)列{an·bn}的前n項(xiàng)和Tn=3.解(1)由題設(shè)可得+n-1=n,即Sn=n2,當(dāng)n≥2時(shí),an=Sn-Sn-1=n2-(n-1)2=2n-1,當(dāng)n=1時(shí),a1=1也適合上式,∴an=2n-1,,∴Tn=1-+…+=1-=(2)由(1)可知Sn=n2,由S3=Sm·P3得9=m2(2+2q+2q2),=2+2q+2q2,∵q>0,>2,∵m∈N*,∴m=1或m=2.當(dāng)m=1時(shí),2q2+2q+2=9,解得q=(舍負(fù)),當(dāng)m=2時(shí),2q2+2q+2=,解得q=(舍負(fù)),∴q=或q=4.解(1)因?yàn)閍n>0,an1.選①:由解得a1=q=2,∴an=a1qn-1=2n.選②:由解得a1=q=2,∴an=a1qn-1=2n.選③:由解得a1=q=2,∴an=a1qn-1=2n.(2)bn=-anlog2an=-n·2n,故Sn=-(1×21+2×22+3×23+…+n×2n),2Sn=-[1×22+2×23+3×24+…+(n-1)·2n+n·2n+1],相減得Sn=21+22+23+…+2n-n·2n+1=-n·2n+1=(1-n)·2n+1-2.Sn+(n+m)an+1=(1-n)·2n+1-2+n·2n+1+m·2n+1>0,即m>-1+對(duì)任意正整數(shù)n恒成立,當(dāng)n=1時(shí),-1+取得最大值-,所以m>-,因此,實(shí)數(shù)m的取值范圍是-,+∞.5.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(q>0),由已知得20=b3-a3=a5+b2,即20=3q2-(3+2d),20=(3+4d)+3q,解得d=2,q=3,所以an=2n+1,bn=3n. (2)若選擇①,則cn=+(-1)nbn=+(-3)n=+(-3)n,所以Sn=c1+c2+…+cn=+(-3)1++(-3)2+…++(-3)n=若選擇②,則cn=an·bn=(2n+1)3n,所以Sn=c1+c2+…+cn=3×3+5×32+…+(2n+1)3n,3Sn=3×32+5×33+…+(2n+1)3n+1,兩式相減得-2Sn=32+2×32+2×33+…+2×3n-(2n+1)3n+1=-2n·3n+1,所以Sn=n·3n+1.若選擇③,則cn=,所以Sn=c1+c2+…+cn=++…+6.解(1)設(shè)等比數(shù)列{an}的公比為q,由an+1=2Sn+2,可得an=2Sn-1+2(n≥2),兩式相減得an+1-an=2Sn-2Sn-1=2an,整理得an+1=3an,可知q=3.令n=1,則a2=2a1+2,即3a1=2a1+2,解得a1=2.故an=2·3n-1.由b1=2,(n+2)bn=nbn+1,得,則當(dāng)n≥2時(shí),bn=…b1=…2=n(n+1).又b1=2滿足上式,所以bn=n(n+1).(2)若在an與an+1之間插入n個(gè)數(shù),使這n+2個(gè)數(shù)組成一個(gè)公差為cn的等差數(shù)列,則an+1-an=(n+1)cn,即2·3n-2·3n-1=(n+1)cn,整理得cn=,所以bncn=4n·3n-1,所以Tn=b1c1+b2c2+b3c3+…+bn-1cn-1+bncn=4×1×30+4×2×31+4×3×32+…+4·(n-1)3n-2+4·n·3n-1=4[1×30+2×31+3×32+…+(n-1)3n-2+n·3n-1],3Tn=4[1×31+2×32+…+(n-1)3n-1+n·3n],兩式相減得-2Tn=4(30+31+32+…+3n-1-n·3n)=4=(2-4n)·3n-2,所以Tn=(2n-1)3n+1.7.(1)證明由an+1=3an+3n+1,得+1,即=1.又,所以數(shù)列是以為首項(xiàng),1為公差的等差數(shù)列.(2)解由(1)得+(n-1)×1=n-,所以an=3n.(3)證明由(2)得Sn=31+32+…+3n-1+3n,3Sn=32+33+…+(n-1)-×3n+3n+1,兩式相減得2Sn=3n+1--1=n-×3n+1+,故Sn=3n+1+,從而