2020版高考数学大二轮培优理科通用版大题专项练（二）　数列 WORD版含解析.docx

1、大题专项练(二)数列A组基础通关1.已知等差数列an满足a3-a2=3,a2+a4=14.(1)求an的通项公式;(2)设Sn是等比数列bn的前n项和,若b2=a2,b4=a6,求S7.解(1)设等差数列an的公差为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)设等比数列bn的公比为q,b2=a2=4=b1q,b4=a6=16=b1q3,联立解得b1=2,q=2,或b1=-2,q=-2.S7=2(27-1)2-1=254,或S7=-21-(-2)71-(-2)=-86.2.已知数列an的前n项和为Sn,满足a2=

2、15,Sn+1=Sn+3an+6.(1)证明:an+3是等比数列;(2)求数列an的通项公式以及前n项和Sn.(1)证明在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.因为Sn+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为首项,3为公比的等比数列.(2)解由(1)得an+3=63n-1=23n,所以an=23n-3.Sn=2(3+32+33+3n)-3n=23(1-3n)1-3-3n=3n+1-3-3n.3.设数列an的前n项和为

3、Sn,Sn=1-an(nN*).(1)求数列an的通项公式;(2)设bn=log2an,求数列1bnbn+1的前n项和Tn.解(1)因为Sn=1-an(nN*),所以Sn-1=1-an-1(nN*,且n2),则Sn-Sn-1=(1-an)-(1-an-1)(nN*,且n2).即an=12an-1(nN*,且n2).因为Sn=1-an(nN*),所以S1=1-a1=a1,即a1=12.所以an是以12为首项,12为公比的等比数列.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+1

4、n-1n+1=1-1n+1=nn+1.4.设等差数列an的公差为d,d为整数,前n项和为Sn,等比数列bn的公比为q,已知a1=b1,b2=2,d=q,S10=100,nN*.(1)求数列an与bn的通项公式;(2)设cn=anbn,求数列cn的前n项和Tn.解(1)由题意可得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

5、-2-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.5.已知正项数列an的前n项和为Sn,满足2Sn+1=2an2+an(nN*).(1)求数列an的通项公式;(2)已知对于nN*,不等式1S1+1S2+1S3+1Sn0,所以a1=1,当n2时,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),因为an0,故an+an-10,所以an-an-1=12,故数列an为等差数列,所以an=n+12.(2)由(1)知Sn=n(n+3)4,所以1Sn=4n(n+3)=431n-1n+

6、3,从而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+36时,bn13loga(1-a)对任意正整数n恒成立,求实数a的取值范围.解(1)点(n,Sn)在函数f(x)=12x2+12x的图象上,Sn=12n2+12n.当n2时,Sn-1=12(n-1)2+12(n-1),-,得an=n.当n=1时,a1=S1=1,符合上式.an=n(nN*).(2)由(1),得1anan+2=1n(n+2)=121n-1n+2,Tn=1a1a3

7、+1a2a4+1anan+2=121-13+12-14+1n-1n+2=34-121n+1+1n+2.Tn+1-Tn=1(n+1)(n+3)0,数列Tn单调递增,Tn中的最小项为T1=13.要使不等式Tn13loga(1-a)对任意正整数n恒成立,只要1313loga(1-a),即loga(1-a)logaa.解得0a12,即实数a的取值范围为0,12.8.设an是各项均不相等的数列,Sn为它的前n项和,满足nan+1=Sn+1(nN*,R).(1)若a1=1,且a1,a2,a3成等差数列,求的值;(2)若an的各项均不为零,问当且仅当为何值时,a2,a3,an,成等差数列?试说明理由.解(1

8、)令n=1,2,得a2=a1+1=2,2a3=S2+1=a1+a2+1,又由a1,a2,a3成等差数列,所以2a2=a1+a3=1+a3,解得=352.(2)当且仅当=12时,a2,a3,an,成等差数列,证明如下:由已知nan+1=Sn+1,当n2时,(n-1)an=Sn-1+1,两式相减得nan+1-nan+an=an,即n(an+1-an)=(1-)an,由于an的各项均不相等,所以n1-=anan+1-an(n2),当n3时,有(n-1)1-=an-1an-an-1,两式相减可得1-=anan+1-an-an-1an-an-1,当=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,成等差数列.再证当a2,a3,an,成等差数列时,=12,因为a2,a3,an,成等差数列,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,所以当且仅当=12时,a2,a3,an,成等差数列.