高考大题专项练3高考中的数列 4页

高考大题专项练3　高考中的数列　高考大题专项练第6页　 1.(2015大连一模)等差数列{an}的前n项和为Sn,等比数列{bn}的公比为12,满足S3=15,a1+2b1=3,a2+4b2=6.(1)求数列{an},{bn}的通项公式an,bn;(2)求数列{an·bn}的前n项和Tn.解:(1)设{an}的公差为d,所以3a1+3d=15,a1+2b1=3,a1+d+2b1=6,解得a1=2,d=3,b1=12,所以an=3n-1,bn=12n.(2)由(1)知Tn=212+5122+8123+…+(3n-4)·12n-1+(3n-1)12n,①①12得12Tn=2122+5123+…+(3n-4)12n+(3n-1)12n+1,②①-②得12Tn=212+3122+123+…+12n-(3n-1)·12n+1=1+3141-12n-11-12-(3n-1)·12n+1,整理得Tn=-(3n+5)12n+5.〚导学号92950935〛2.已知数列{an}的前n项和为Sn,且Sn=2an-1;数列{bn}满足bn-1-bn=bnbn-1(n≥2,n∈N+),b1=1.(1)求数列{an},{bn}的通项公式;(2)求数列anbn的前n项和Tn.解:(1)由Sn=2an-1,得S1=a1=2a1-1,故a1=1.又Sn=2an-1,Sn-1=2an-1-1(n≥2),两式相减,得Sn-Sn-1=2an-2an-1,即an=2an-2an-1.故an=2an-1,n≥2.所以数列{an}是首项为1,公比为2的等比数列.故an=1·2n-1=2n-1.由bn-1-bn=bnbn-1(n≥2,n∈N+),得1bn-1bn-1=1.又b1=1,∴数列1bn是首项为1,公差为1的等差数列.∴1bn=1+(n-1)·1=n.∴bn=1n.(2)由(1)得anbn=n·2n-1.∴Tn=1·20+2·21+…+n·2n-1,∴2Tn=1·21+2·22+…+n·2n.两式相减,得-Tn=1+21+…+2n-1-n·2n=1-2n1-2-n·2n=-1+2n-n·2n.∴Tn=(n-1)·2n+1.〚导学号92950936〛3.已知正项数列{an}的首项a1=1,前n项和Sn满足an=Sn+Sn-1(n≥2).(1)求证:{Sn}为等差数列,并求数列{an}的通项公式;(2)记数列1anan+1的前n项和为Tn,若对任意的n∈N+,不等式4Tn0,所以a3=5,a5=9,公差d=a5-a35-3=2.所以an=a5+(n-5)d=2n-1.当n=1时,b1=S1=1-b12,解得b1=13.当n≥2时,bn=Sn-Sn-1=12(bn-1-bn),所以bnbn-1=13(n≥2).所以数列{bn}是首项b1=13,公比q=13的等比数列,所以bn=b1qn-1=13n.(2)由(1),知cn=anbn=2n-13n,cn+1=2n+13n+1,所以cn+1-cn=2n+13n+1-2n-13n=4(1-n)3n+1≤0.所以cn+1≤cn.(3)由(2),知cn=anbn=2n-13n,\n则Tn=131+332+533+…+2n-13n,①13Tn=132+333+534+…+2n-33n+2n-13n+1,②①-②,得23Tn=13+232+233+…+23n-2n-13n+1=13+2132+133+…+13n-2n-13n+1=23-2n+23n+1,化简得Tn=1-n+13n.故数列{cn}的前n项和Tn=1-n+13n.〚导学号92950939〛6.(2015长沙二模)已知数列{an}是公差不为零的等差数列,a10=15,且a3,a4,a7成等比数列.(1)求数列{an}的通项公式;(2)设bn=an2n,数列{bn}的前n项和为Tn,求证:-74≤Tn<-1(n∈N+).解:(1)设数列{an}的公差为d(d≠0),由已知得a10=15,a42=a3a7,即a1+9d=15,(a1+3d)2=(a1+2d)(a1+6d),解得a1=-3,d=2.∴an=2n-5.(2)证明:∵bn=an2n=2n-52n,n∈N+,∴Tn=-32+-122+123+…+2n-52n,①12Tn=-322+-123+124+…+2n-72n+2n-52n+1,②①-②,得12Tn=-32+2122+123+…+12n-2n-52n+1=-12+1-2n2n+1,∴Tn=-1-2n-12n(n∈N+).∵2n-12n>0(n∈N+),∴Tn<-1.Tn+1-Tn=-1-2n+12n+1--1-2n-12n=2n-32n+1,∴TnT2,∴T2最小,即Tn≥T2=-74.综上所述,-74≤Tn<-1(n∈N+).〚导学号92950940〛7.(2015天津红桥模拟)已知数列{an}的前n项和为Sn,且Sn=12n2+112n(n∈N+).(1)求数列{an}的通项公式.(2)设cn=1(2an-11)(2an-9),数列{cn}的前n项和为Tn,求使不等式Tn>k2015对一切n∈N+都成立的最大正整数k的值.(3)设f(n)=an,n=2k-1,k∈N+,3an-13,n=2k,k∈N+,是否存在m∈N+,使得f(m+15)=5f(m)成立?若存在,求出m的值;若不存在,请说明理由.解:(1)当n=1时,a1=S1=6,当n≥2时,an=Sn-Sn-1=12n2+112n-12(n-1)2+112(n-1)=n+5.而当n=1时,a1=6适合上式,∴an=n+5.(2)cn=1(2an-11)(2an-9)=1(2n-1)(2n+1)=1212n-1-12n+1,∴Tn=c1+c2+…+cn=121-13+13-15+…+12n-1-12n+1=n2n+1.\n∵Tn+1-Tn=n+12n+3-n2n+1=1(2n+3)(2n+1)>0,∴Tn单调递增,故(Tn)min=T1=13.令13>k2015,得k<67123,所以k的最大整数值为671.(3)f(n)=n+5,n=2k-1,k∈N+,3n+2,n=2k,k∈N+,当m为奇数时,m+15为偶数,∴3m+47=5m+25,m=11.当m为偶数时,m+15为奇数,∴m+20=15m+10,m=57∉N+(舍去).综上,存在唯一正整数m=11,使得f(m+15)=5f(m)成立.〚导学号92950941〛