高考大题专项三高考中的数列1.(2018山西吕梁一模,17)已知{an}是首项为1的等比数列,数列{bn}满足b1=2,b2=5,且anbn+1=anbn+an+1.(1)求数列{an}的通项公式;(2)求数列{bn}的前n项和.2.(2018福建龙岩4月质检,17)已知正项等比数列{an}的前n项和为Sn,且Sn=2an-1(n∈N+).(1)求数列{an}的通项公式;(2)若bn=lgan,求数列{an+bn}的前n项和Tn.13.(2018北京海淀期末,15)已知等差数列{an}的前n项和Sn,且a2=5,S3=a7.(1)数列{an}的通项公式;(2)若bn=2an,求数列{an+bn}的前n项和.4.(2018河北唐山一模,17)已知数列{an}为单调递增数列,Sn为其前n项和,2Sn=an2+n.(1)求{an}的通项公式;(2)若bn=an+22n+1·an·an+1,Tn为数列{bn}的前n项和,证明:Tn<12.25.(2018湖南衡阳二模,17)等差数列{an}中,a3=1,a7=9,Sn为等比数列{bn}的前n项和,且b1=2,若4S1,3S2,2S3成等差数列.(1)求数列{an},{bn}的通项公式;(2)设cn=|an|·bn,求数列{cn}的前n项和Tn.36.已知数列{an}的前n项和为Sn,Sn=(m+1)-man对任意的n∈N+都成立,其中m为常数,且m<-1.(1)求证:数列{an}是等比数列;(2)记数列{an}的公比为q,设q=f(m),若数列{bn}满足b1=a1,bn=f(bn-1)(n≥2,n∈N+).求证:数列{1bn}是等差数列;(3)在(2)的条件下,设cn=bn·bn+1,数列{cn}的前n项和为Tn,求证:Tn<1.7.(2018宿州十三所中学期中,17)已知数列{an}的前n项和为Sn,并且满足a1=1,nan+1=Sn+n(n+1).(1)求数列{an}的通项公式;(2)若bn=an2n,数列{bn}的前n项和为Tn,求Tn;(3)在(2)的条件下,是否存在常数λ,使得数列{Tn+λan+2}为等比数列?若存在,试求出λ;若不存在,说明理由.4参考答案高考大题专项三高考中的数列1.解(1)把n=1代入已知等式得a1b2=a1b1+a2,∴a2=a1b2-a1b1=3a1.∴{an}是首项为1,公比为3的等比数列,即an=3n-1.(2)由已知得bn+1-bn=an+1an=3,∴{bn}是首项为2,公差为3的等差数列,其通项公式为bn=3n-1,∴Sn=n(b1+bn)2=n(2+3n-1)2=3n2+n2.2.解(1)由Sn=2an-1(n∈N+),可得S1=2a1-1,∴a1=2a1-1,∴a1=1.又S2=2a2-1,∴a1+a2=2a2-1,∴a2=2.∵数列{an}是等比数列,∴公比q=a2a1=2,∴数列{an}的通项公式为an=2n-1.(2)由(1)知,bn=lgan=(n-1)lg2,∴Tn=(b1+a1)+(b2+a2)+…+(bn+an)=(0+1)+(lg2+2)+…+[(n-1)lg2+2n-1]=[lg2+2lg2+…+(n-1)lg2]+(1+2+…+2n-1)=n(n-1)2lg2+2n-1.3.解(1)设等差数列{an}的首项为a1,公差为d,则{a1+d=5,3a1+3d=a1+6d,解得a1=3,d=2.由an=a1+(n-1)d,则an=2n+1.因此,通项公式为an=2n+1.(2)由(1)可知:an=2n+1,则bn=22n+1,bn+1bn=22(n+1)+122n+1=4.因为b1=23=8,所以{bn}是首项为8,公比为q=4的等比数列.记{an+bn}的前n项和为Tn,则Tn=(a1+b1)+(a2+b2)+…+(an+bn)=(a1+a2+…+an)+(b1+b2+…+bn)=n(a1+an)2+b1(1-qn)1-q=n2+2n+8(4n-1)3.54.(1)解当n=1时,2S1=2a1=a12+1.所以(a1-1)2=0,即a1=1.又{an}为单调递增数列,所以an≥1.由2Sn=an2+n得2Sn+1=an+12+n+1,所以2Sn+1-2Sn=an+12-an2+1,整理得2an+1=an+12-an2+1,即an2=(an+1-1)2,所以an=an+1-1,即an+1-an=1,所以{an}是以1为首项,1为公差的等差数列,所以an=n.(2)证明因为bn=an+22n+1×an×an+1=n+22n+1×n×(n+1)=12n×n-12n+1×(n+1),所以Tn=121×1-122×2+122×2-123×3+…+12n×n-12n+1×(n+1)=121×1-12n+1×(n+1)<12.5.解(1)在等差数列{an}中,设公差为d,则a7-a3=4d=9-1=8,故d=2,∴an=a3+(n-3)d=1+2(n-3)=2n-5.设等比数列{bn}的公比为q,依题意有:6S2=4S1+2S3,故q=2,∴bn=2n.(2)∵cn=|2n-5|·2n.当n=1时,T1=6,当n=2时,T2=10,当n≥3时,2n-5>0,Tn=10+1×23+3×24+…+(2n-7)2n-1+(2n-5)2n,①2Tn=20+1×24+3×25+…+(2n-7)2n+(2n-5)2n+1,②由①-②,得-Tn=-10+8+2(24+…+2n)-(2n-5)2n+1,∴Tn=34+(2n-7)2n+1.∴Tn={6,n=1,10,n=2,34+(2n-7)2n+1,n≥3.6.证明(1)当n=1时,a1=S1=1.∵Sn=(m+1)-man,①∴Sn-1=(m+1)-man-1(n≥2),②由①-②,得an=man-1-man(n≥2),即(m+1)an=man-1.∵a1≠0,m<-1,∴an-1≠0,m+1≠0.∴anan-1=mm+1(n≥2).∴数列{an}是首项为1,公比为mm+1的等比数列.6(2)∵f(m)=mm+1,b1=a1=1,bn=f(bn-1)=bn-1bn-1+1(n≥2),∴1bn=bn-1+1bn-1(n≥2),∴1bn-1bn-1=1(n≥2),∴数列{1bn}是首项为1,公差为1的等差数列.(3)由(2)得1bn=n,则bn=1n,故cn=bn·bn+1=1n(n+1),因此,Tn=11×2+12×3+…+1n(n+1)=11-12+12-13+13-14+…+1n-1n+1=1-1n+1<1.7.解(1)∵nan+1=Sn+n(n+1),①∴当n≥2时,(n-1)an=Sn-1+n(n-1),②由①-②可得an+1-an=2(n≥2),且a1=1,a2=S1+1×(1+1)=3,∴数列{an}是首项为1,公差为2的等差数列,即an=2n-1.(2)由(1)知数列an=2n-1,∴bn=2n-12n,则Tn=121+322+523+…+2n-32n-1+2n-12n,①∴12Tn=122+323+524+…+2n-32n+2n-12n+1,②由①-②得,12Tn=12+2122+123+…+12n-2n-12n+1=12+2×14(1-12n-1)1-12-2n-12n+1,∴Tn=3-2n+32n.(3)由(2)知Tn=3-2n+32n,∴Tn+λan+2=(3-2n+32n+λ)2n+3=(3+λ)2n+3-12n,∴要使数列{Tn+λan+2}为等比数列,当且仅当3+λ=0,即λ=-3.故存在λ=-3,使得数列{Tn+λan+2}为等比数列.78
