（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 大題專項(xiàng)練（二）數(shù)列 理

大題專項(xiàng)練(二)數(shù)列A組基礎(chǔ)通關(guān)1.已知等差數(shù)列an滿足a3-a2=3,a2+a4=14.(1)求an的通項(xiàng)公式;(2)設(shè)Sn是等比數(shù)列bn的前n項(xiàng)和,若b2=a2,b4=a6,求S7.解(1)設(shè)等差數(shù)列an的公差為d,a3-a2=3,a2+a4=14.d=3,2a1+4d=14,解得a1=1,d=3,an=1+3(n-1)=3n-2.(2)設(shè)等比數(shù)列bn的公比為q,b2=a2=4=b1q,b4=a6=16=b1q3,聯(lián)立解得b1=2,q=2,或b1=-2,q=-2.S7=2×(27-1)2-1=254,或S7=-2×1-(-2)71-(-2)=-86.2.已知數(shù)列an的前n項(xiàng)和為Sn,滿足a2=15,Sn+1=Sn+3an+6.(1)證明:an+3是等比數(shù)列;(2)求數(shù)列an的通項(xiàng)公式以及前n項(xiàng)和Sn.(1)證明在Sn+1=Sn+3an+6中,令n=1,得S2=S1+3a1+6,得a1+a2=a1+3a1+6,即a1+15=4a1+6,解得a1=3.因?yàn)镾n+1=Sn+3an+6,所以an+1=3an+6.所以an+1+3an+3=3an+9an+3=3.所以an+3是以6為首項(xiàng),3為公比的等比數(shù)列.(2)解由(1)得an+3=6×3n-1=2×3n,所以an=2×3n-3.Sn=2×(3+32+33+3n)-3n=2×3×(1-3n)1-3-3n=3n+1-3-3n.3.設(shè)數(shù)列an的前n項(xiàng)和為Sn,Sn=1-an(nN*).(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=log2an,求數(shù)列1bnbn+1的前n項(xiàng)和Tn.解(1)因?yàn)镾n=1-an(nN*),所以Sn-1=1-an-1(nN*,且n2),則Sn-Sn-1=(1-an)-(1-an-1)(nN*,且n2).即an=12an-1(nN*,且n2).因?yàn)镾n=1-an(nN*),所以S1=1-a1=a1,即a1=12.所以an是以12為首項(xiàng),12為公比的等比數(shù)列.故an=12n(nN*).(2)bn=log2an,所以bn=log212n=-n.所以1bnbn+1=1n(n+1)=1n-1n+1,故Tn=1-12+12-13+1n-1n+1=1-1n+1=nn+1.4.設(shè)等差數(shù)列an的公差為d,d為整數(shù),前n項(xiàng)和為Sn,等比數(shù)列bn的公比為q,已知a1=b1,b2=2,d=q,S10=100,nN*.(1)求數(shù)列an與bn的通項(xiàng)公式;(2)設(shè)cn=anbn,求數(shù)列cn的前n項(xiàng)和Tn.解(1)由題意可得10a1+45d=100,a1d=2,解得a1=9,d=29(舍去)或a1=1,d=2,所以an=2n-1,bn=2n-1.(2)cn=anbn,cn=2n-12n-1,Tn=1+32+522+723+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-1.5.已知正項(xiàng)數(shù)列an的前n項(xiàng)和為Sn,滿足2Sn+1=2an2+an(nN*).(1)求數(shù)列an的通項(xiàng)公式;(2)已知對于nN*,不等式1S1+1S2+1S3+1Sn<M恒成立,求實(shí)數(shù)M的最小值.解(1)n=1時,2a1+1=2a12+a1,又an>0,所以a1=1,當(dāng)n2時,2Sn+1=2an2+an(nN*),2Sn-1+1=2an-12+an-1(nN*),作差整理,得an+an-1=2(an+an-1)(an-an-1),因?yàn)閍n>0,故an+an-1>0,所以an-an-1=12,故數(shù)列an為等差數(shù)列,所以an=n+12.(2)由(1)知Sn=n(n+3)4,所以1Sn=4n(n+3)=431n-1n+3,從而1S1+1S2+1S3+1Sn=431-14+12-15+13-16+1n-2-1n+1+1n-1-1n+2+1n-1n+3=431+12+13-1n+1-1n+2-1n+3=43116-1n+1-1n+2-1n+3<229.所以M229,故M的最小值為229.6.已知數(shù)列an是公比為q的正項(xiàng)等比數(shù)列,bn是公差d為負(fù)數(shù)的等差數(shù)列,滿足1a2-1a3=da1,b1+b2+b3=21,b1b2b3=315.(1)求數(shù)列an的公比q與數(shù)列bn的通項(xiàng)公式;(2)求數(shù)列|bn|的前10項(xiàng)和S10.解(1)由已知,b1+b2+b3=3b2=21,得b2=7,又b1b2b3=(b2-d)·b2·(b2+d)=(7-d)·7·(7+d)=343-7d2=315,得d=-2或2(舍),b1=7+2=9,bn=-2n+11.于是1a2-1a3=-2a1,又an是公比為q的等比數(shù)列,故1a1q-1a1q2=-2a1,所以,2q2+q-1=0,q=-1(舍)或12.綜上,q=12,d=-2,bn=11-2n.(2)設(shè)bn的前n項(xiàng)和為Tn;令bn0,11-2n0,得n5,于是,S5=T5=5(b1+b5)2=25.易知,n>6時,bn<0,|b6|+|b7|+|b10|=-b6-b7-b10=-(b6+b7+b10)=-(T10-T5)=-(0-25)=25,所以,S10=50.B組能力提升7.已知數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)(n,Sn)(nN*)在函數(shù)f(x)=12x2+12x的圖象上.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列1anan+2的前n項(xiàng)和為Tn,不等式Tn>13loga(1-a)對任意正整數(shù)n恒成立,求實(shí)數(shù)a的取值范圍.解(1)點(diǎn)(n,Sn)在函數(shù)f(x)=12x2+12x的圖象上,Sn=12n2+12n.當(dāng)n2時,Sn-1=12(n-1)2+12(n-1),-,得an=n.當(dāng)n=1時,a1=S1=1,符合上式.an=n(nN*).(2)由(1),得1anan+2=1n(n+2)=121n-1n+2,Tn=1a1a3+1a2a4+1anan+2=121-13+12-14+1n-1n+2=34-121n+1+1n+2.Tn+1-Tn=1(n+1)(n+3)>0,數(shù)列Tn單調(diào)遞增,Tn中的最小項(xiàng)為T1=13.要使不等式Tn>13loga(1-a)對任意正整數(shù)n恒成立,只要13>13loga(1-a),即loga(1-a)<logaa.解得0<a<12,即實(shí)數(shù)a的取值范圍為0,12.8.設(shè)an是各項(xiàng)均不相等的數(shù)列,Sn為它的前n項(xiàng)和,滿足nan+1=Sn+1(nN*,R).(1)若a1=1,且a1,a2,a3成等差數(shù)列,求的值;(2)若an的各項(xiàng)均不為零,問當(dāng)且僅當(dāng)為何值時,a2,a3,an,成等差數(shù)列?試說明理由.解(1)令n=1,2,得a2=a1+1=2,2a3=S2+1=a1+a2+1,又由a1,a2,a3成等差數(shù)列,所以2a2=a1+a3=1+a3,解得=3±52.(2)當(dāng)且僅當(dāng)=12時,a2,a3,an,成等差數(shù)列,證明如下:由已知nan+1=Sn+1,當(dāng)n2時,(n-1)an=Sn-1+1,兩式相減得nan+1-nan+an=an,即n(an+1-an)=(1-)an,由于an的各項(xiàng)均不相等,所以n1-=anan+1-an(n2),當(dāng)n3時,有(n-1)1-=an-1an-an-1,兩式相減可得1-=anan+1-an-an-1an-an-1,當(dāng)=12,得anan+1-an=an-1an-an-1+1=anan-an-1,由于an0,所以an+1-an=an-an-1,即2an=an+1+an-1(n3),故a2,a3,an,成等差數(shù)列.再證當(dāng)a2,a3,an,成等差數(shù)列時,=12,因?yàn)閍2,a3,an,成等差數(shù)列,所以an+1-an=an-an-1(n3),可得anan+1-an-an-1an-an-1=anan-an-1-an-1an-an-1=1=1-,所以=12,所以當(dāng)且僅當(dāng)=12時,a2,a3,an,成等差數(shù)列.8