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.若数列{an}满足:a1=,a2=2,3(an+1-2an+an-1)=2.(1)证明:数列{an+1-an}是等差数列;(2)求使+…+成立的最小的正整数n.(1)证明由3(an+1-2an+an-1)=2可得an+1-2an+an-1=,即(an+1-an)-(an-an-1)=,故数列{an+1-an}是以a2-a1=为首项,为公差的等差数列.(2)解由(1)知an+1-an=(n-1)=(n+1),于是累加求和得an=a1+(2+3+…+n)=n(n+1),∴=3.∴+…+=3-,∴n>5.∴最小的正整数n为6.2.(2017广东揭阳二模,理17)已知数列{an},a1=1,an+1=+n+1.(1)求证:数列是等比教列;(2)求数列{an}的前n项和Sn.(1)证明∵an+1=+n+1,∴=2×+1,∴+1=2×,∴数列是等比教列,公比为2,首项为2.(2)解由(1)可得+1=2n,可得an=n·2n-n.设数列{n·2n}的前n项和为Tn,则Tn=2+2×22+3×23+…+n·2n,2Tn=22+2×23+…+(n-1)·2n+n·2n+1,两式相减可得-Tn=2+22+…+2n-n·2n+1=-n·2n+1,∴Tn=(n-1)·2n+1+2.∴Sn=(n-1)·2n+1+2-.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=,求λ.解(1)由题意得a1=S1=1+λa1,故λ≠1,a1=,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.因此{an}是首项为,公比为的等比数列,于是an=.(2)由(1)得Sn=1-.由S5=得1-,即.解得λ=-1.4.(2017吉林白山二模,理17)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=,证明数列{bn}为等差数列;(2)求数列{an}的前n项和Sn.(1)证明由已知得an+1=2an+2n,∴bn+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==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.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:为等差数列,并求bn.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}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)=,∴当n≥2时,bn=f(bn-1)=,∴bnbn-1+3bn=3bn-1,∴.∴是以1为首项,为公差的等差数列,∴=1+.又=1也符合,∴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),求数列的前n项和Tn,并求证Tn<.(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-,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,∴.∴Tn=+…+.∴Tn<.导学号〚16804192〛7.(2017湖南长郡中学模拟,理17)在数列{an}中,Sn为其前n项和,an>0,且4Sn=+2an+1(n∈N*),数列{bn}为等比数列,公比q>1,b1=a1,且2b2,b4,3b3成等差数列.(1)求{an}与{bn}的通项公式;(2)令cn=,若{cn}的前n项和为Tn,求证:Tn<6.(1)解4Sn=+2an+1(n∈N*),①当n=1时,4a1=+2a1+1,解得a1=1.当n≥2时,4Sn-1=+2an-1+1,②①-②得4an=(an+1)2-(an-1+1)2,即(an+an-1)(an-an-1-2)=0.又an>0,∴an-an-1-2=0,即an-an-1=2,∴数列{an}是等差数列,公差为2.∴an=1+2(n-1)=2n-1.∵2b2,b4,3b3成等差数列,∴2b4=2b2+3b3.∴2b2q2=2b2+3b2q,即2q2-3q-2=0,解得q=2.又b1=a1=1,∴bn=2n-1.(2)证明cn=,则{cn}的前n项和为Tn=1++…+Tn=+…+,∴Tn=1+2=1+2×,∴Tn=6-<6.导学号〚16804193〛8.已知数列{an}的前n项和为Sn,a1=1,且对任意正整数n,点(an+1,Sn)都在直线2x+y-2=0上.(1)求数列{an}的通项公式;(2)是否存在实数λ,使得数列为等差数列?若存在,求出λ的值;若不存在,请说明理由.解(1)由题意,得2an+1+Sn-2=0.①当n≥2时,2an+Sn-1-2=0.②①-②,得2an+1-2an+an=0(n≥2),所以(n≥2).因为a1=1,2a2+a1=2,所以a2=.所以{an}是首项为1,公比为的等比数列.所以数列{an}的通项公式为an=.(2)由(1)知,Sn==2-.若为等差数列,则S1+λ+,S2+2λ+,S3+3λ+成等差数列,则2=S1++S3+,即2=1+,解得λ=2.又当λ=2时,Sn+2n+=2n+2,显然{2n+2}是等差数列.故存在实数λ=2,使得数列为等差数列.
