2019届高考数学二轮复习专题对点练14数列与数列的证明及数列中的存在性问题.docx

专题对点练14数列与数列不等式的证明及数列中的存在性问题1.已知等比数列an,a1=13,公比q=13.(1)Sn为an的前n项和,证明:Sn=1-an2;(2)设bn=log3a1+log3a2+log3an,求数列bn的通项公式.2.已知数列an满足a1=3,an+1=3an-1an+1.(1)证明:数列1an-1是等差数列,并求an的通项公式;(2)令bn=a1a2an,求数列1bn的前n项和Sn.3.已知数列an的前n项和Sn=1+an,其中0.(1)证明an是等比数列,并求其通项公式;(2)若S5=3132,求的值.4.在数列an中,设f(n)=an,且f(n)满足f(n+1)-2f(n)=2n(nN*),且a1=1.(1)设bn=an2n-1,证明数列bn为等差数列;(2)求数列an的前n项和Sn.5.设数列an的前n项和为Sn,且(3-m)Sn+2man=m+3(nN*),其中m为常数,且m-3.(1)求证:an是等比数列;(2)若数列an的公比q=f(m),数列bn满足b1=a1,bn=32f(bn-1)(nN*,n2),求证:1bn为等差数列,并求bn.6.已知数列an的前n项和为Sn,a1=-2,且满足Sn=12an+1+n+1(nN*).(1)求数列an的通项公式;(2)若bn=log3(-an+1),求数列1bnbn+2的前n项和Tn,并求证Tn34.7.(2018天津模拟)已知正项数列an,a1=1,a2=2,前n项和为Sn,且满足Sn+1Sn-1+Sn-1Sn+1=4Sn2Sn+1Sn-1-2(n2,nN*).(1)求数列an的通项公式;(2)记cn=1SnSn+1,数列cn的前n项和为Tn,求证:13Tn12.8.已知数列an的前n项和为Sn,a1=2,2Sn=(n+1)2an-n2an+1,数列bn满足b1=1,bnbn+1=2an.(1)求数列an的通项公式;(2)是否存在正实数,使得bn为等比数列?并说明理由.专题对点练14答案1.(1)证明 因为an=1313n-1=13n,Sn=131-13n1-13=1-13n2,所以Sn=1-an2.(2)解 bn=log3a1+log3a2+log3an=-(1+2+n)=-n(n+1)2.所以bn的通项公式为bn=-n(n+1)2.2.解 (1)an+1=3an-1an+1,an+1-1=3an-1an+1-1=2(an-1)an+1,1an+1-1=an+12(an-1)=1an-1+12,1an+1-1-1an-1=12.a1=3,1a1-1=12,数列1an-1是以12为首项,12为公差的等差数列,1an-1=12+12(n-1)=12n,an=n+2n.(2)bn=a1a2an,bn=314253nn-2n+1n-1n+2n=(n+1)(n+2)2,1bn=2(n+1)(n+2)=21n+1-1n+2,Sn=212-13+13-14+1n+1-1n+2=212-1n+2=nn+2.3.解 (1)由题意得a1=S1=1+a1,故1,a1=11-,a10.由Sn=1+an,Sn+1=1+an+1得an+1=an+1-an,即an+1(-1)=an.由a10,0得an0,所以an+1an=-1.因此an是首项为11-,公比为-1的等比数列,于是an=11-1n-1.(2)由(1)得Sn=1-1n.由S5=3132得1-15=3132,即-15=132.解得=-1.4.(1)证明 由已知得an+1=2an+2n,bn+1=an+12n=2an+2n2n=an2n-1+1=bn+1,bn+1-bn=1.又a1=1,b1=1,bn是首项为1,公差为1的等差数列.(2)解 由(1)知,bn=an2n-1=n,an=n2n-1.Sn=1+221+322+n2n-1,2Sn=121+222+(n-1)2n-1+n2n,两式相减得-Sn=1+21+22+2n-1-n2n=2n-1-n2n=(1-n)2n-1,Sn=(n-1)2n+1.5.证明 (1)由(3-m)Sn+2man=m+3,得(3-m)Sn+1+2man+1=m+3,两式相减,得(3+m)an+1=2man.m-3,an+1an=2mm+3,an是等比数列.(2)由(3-m)Sn+2man=m+3,得(3-m)S1+2ma1=m+3,即a1=1,b1=1.数列an的公比q=f(m)=2mm+3,当n2时,bn=32f(bn-1)=322bn-1bn-1+3,bnbn-1+3bn=3bn-1,1bn-1bn-1=13.1bn是以1为首项,13为公差的等差数列,1bn=1+n-13=n+23.又1b1=1也符合,bn=3n+2.6.(1)解 Sn=12an+1+n+1(nN*),当n=1时,-2=12a2+2,解得a2=-8.当n2时,an=Sn-Sn-1=12an+1+n+1-12an+n,即an+1=3an-2,可得an+1-1=3(an-1).当n=1时,a2-1=3(a1-1)=-9,数列an-1是等比数列,首项为-3,公比为3.an-1=-3n,即an=1-3n.(2)证明 bn=log3(-an+1)=n,1bnbn+2=121n-1n+2.Tn=121-13+12-14+13-15+1n-1-1n+1+1n-1n+2=121+12-1n+1-1n+234.Tn34.7.(1)解 由Sn+1Sn-1+Sn-1Sn+1=4Sn2Sn+1Sn-1-2(n2,nN*),得Sn+12+2Sn+1Sn-1+Sn-12=4Sn2,即(Sn+1+Sn-1)2=(2Sn)2.由数列an的各项均为正数,得Sn+1+Sn-1=2Sn,所以数列Sn为等差数列.由a1=1,a2=2,得S1=a1=1,S2=a1+a2=3,则数列Sn的公差为d=S2-S1=2,所以Sn=1+(n-1)2=2n-1.当n2时,an=Sn-Sn-1=(2n-1)-(2n-3)=2,而a1=1不适合上式,所以数列an的通项公式为an=1,n=1,2,n2.(2)证明 由(1)得cn=1SnSn+1=1(2n-1)(2n+1)=1212n-1-12n+1,则Tn=c1+c2+c3+cn=121-13+13-15+15-17+12n-1-12n+1=121-12n+112.又Tn=121-12n+1是关于n的增函数,则TnT1=13,因此,13Tn12.8.解 (1)由2Sn=(n+1)2an-n2an+1,得2Sn-1=n2an-1-(n-1)2an,2an=(n+1)2an-n2an+1-n2an-1+(n-1)2an,2an=an+1+an-1,数列an为等差数列.2S1=(1+1)2a1-a2,4=8-a2.a2=4.d=a2-a1=4-2=2.