2022版高考数学二轮复习专题四数列专题对点练14数列与数列不等式的证明及数列中的存在性问题文

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=,公比q=.(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=a1a2…an,求数列1bn的前n项和Sn.3.已知数列{an}的前n项和Sn=1+λan,其中λ≠0.(1)证明{an}是等比数列,并求其通项公式;(2)若S5=3132,求λ的值.4.在数列{an}中,设f(n)=an,且f(n)满足f(n+1)-2f(n)=2n(n∈N*),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.4\n5.设数列{an}的前n项和为Sn,且(3-m)Sn+2man=m+3(n∈N*),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=f(bn-1)(n∈N*,n≥2),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=an+1+n+1(n∈N*).(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Tn<.7.(2022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(n≥2,n∈N*).(1)求数列{an}的通项公式;(2)记cn=1Sn·Sn+1,数列{cn}的前n项和为Tn,求证:≤Tn<.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=λ·2an.(1)求数列{an}的通项公式;(2)是否存在正实数λ,使得{bn}为等比数列?并说明理由.4\n专题对点练14答案1.(1)证明因为an=13×13n-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是以为首项,为公差的等差数列,∴1an-1=12+12(n-1)=n,∴an=n+2n.(2)∵bn=a1a2…an,∴bn=31×42×53×…×nn-2×n+1n-1×n+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+λa1,故λ≠1,a1=11-λ,a1≠0.由Sn=1+λan,Sn+1=1+λan+1得an+1=λan+1-λan,即an+1(λ-1)=λan.由a1≠0,λ≠0得an≠0,所以an+1an=λλ-1.因此{an}是首项为11-λ,公比为λλ-1的等比数列,于是an=11-λλλ-1n-1.(2)由(1)得Sn=1-λλ-1n.由S5=3132得1-λλ-15=3132,即λλ-15=132.解得λ=-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=n·2n-1.∴Sn=1+2×21+3×22+…+n·2n-1,2Sn=1×21+2×22+…+(n-1)·2n-1+n·2n,两式相减得-Sn=1+21+22+…+2n-1-n·2n=2n-1-n·2n=(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.4\n∵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n≥2时,bn=f(bn-1)=32·2bn-1bn-1+3,∴bnbn-1+3bn=3bn-1,∴1bn-1bn-1=13.∴1bn是以1为首项,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=an+1+n+1(n∈N*),∴当n=1时,-2=a2+2,解得a2=-8.当n≥2时,an=Sn-Sn-1=an+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+2<34.∴Tn<.7.(1)解由Sn+1Sn-1+Sn-1Sn+1=4Sn2Sn+1Sn-1-2(n≥2,n∈N*),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n≥2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,n≥2.(2)证明由(1)得cn=1Sn·Sn+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=12　　1-12n+1<12.又Tn=121-12n+1是关于n的增函数,则Tn≥T1=,因此,≤Tn<.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.∴an=2+2(n-1)=2n.(2)∵bnbn+1=λ·2an=λ·4n,b1=1,∴b2b1=4λ,∴b2=4λ,∴bn+1bn+2=λ·4n+1,∴bn+1bn+2bnbn+1=4,∴bn+2=4bn,∴b3=4b1=4.若{bn}为等比数列,则b22=b3·b1,∴16λ2=4×1,∴λ=.故存在正实数λ=,使得{bn}为等比数列.4