2010届高考专题复习：19数列的应用(1)ppt课件

典型例题解:设第二个数为a,则第三个数为12-a.∵前三个数成等差数列,∴第一个数为3a-12.从而第四个数为16-(3a-12)=28-3a.依题意得:(12-a)2=a(28-3a).化简整理得a2-13a+36=0.解得a=4或9.∴这四个数分别为0,4,8,16或15,9,3,1.1.有四个数,前三个数成等差数列,后三个数成等比数列,并且第一个数与第四个数的和是16,第二个数与第三个数的和是12,求这四个数.∴a2=1从而a1=1-d,a3=1+d.整理得4(2d)2-17(2d)+4=0.故an=2n-3或an=-2n+5.2.设{an}是等差数列,bn=()an,已知b1+b2+b3=,b1b2b3=,求等差数列an.1282118解:设{an}的公差为d.∵b1b3=()a1()a3=()a1+a3=()2a2=b22,12121212∴由b1b2b3=得b23=.1818∴b2=.12又由b1+b2+b3=得()1-d++()1+d=.821121212821解得2d=22或2-2.∴d=2或-2.当d=2时,an=a2+(n-2)d=1+2n-4=2n-3;当d=-2时,an=a2+(n-2)d=1-2n+4=-2n+5.∴f(x)=2-104x.(2)由已知an=log2f(n)=log2(2-104n)=2n-10.3.已知函数f(x)=a∙bx的图象过点A(4,)和B(5,1).(1)求函数f(x)的解析式;(2)记an=log2f(n),n为正整数,Sn是数列{an}的前n项和,解关于n的不等式anSn≤0;(3)对于(2)中的an与Sn,整数104是否为数列{anSn}中的项?若是,则求出相应的项数;若不是,则说明理由.14解:(1)由已知a∙b4=,a∙b5=1,14解得b=4,a=2-10.∴Sn=n(n-9).∴anSn=2n(n-5)(n-9).∵nN*,∴由anSn≤0得(n-5)(n-9)≤0.解得5≤n≤9,nN*.∴n=5,6,7,8,9.(3)a1S1=64,a2S2=84,a3S3=72,a4S4=40;当5≤n≤9时,anSn≤0;当10≤n≤22时,anSn≤a22S22=9724104;当n≥23时,anSn≥a23S23=11529104;故整数104不是数列{anSn}中的项.解:(1)由已知数列{an+1-an}是首项为-2,公差为1的等差数列.∴an+1-an=(a2-a1)+(n-1)1=n-3.∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)4.设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3,且数列{an+1-an}(nN*)是等差数列,数列{bn-2}(nN*)是等比数列.(1)求数列{an}和{bn}的通项公式;(2)是否存在kN*,使ak-bk(0,)?若存在,求出k,若不存在,说明理由.12∴an-an-1=n-4(n≥2).=6+(-2)+(-1)+0+1+2+…+(n-4)=(n2-7n+18)(n≥2).12而a1=2亦适合上式,=(n2-7n+18)(nN*).12∴an又数列{bn-2}是首项为b1-2=4,公比为的等比数列,12∴bn-2=4()n-1=()n-3.1212∴bn=()n-3+2.12故数列{an}和{bn}的通项公式分别为:an=(n2-7n+18),12bn=()n-3+2.12解:(2)显然当k=1,2,3时,ak-bk=0,不适合题意;∴数列{ak}是递增数列,{bk}是递减数列.∴不存在kN*,使ak-bk(0,).12当k≥4时,∵ak=(k2-7k+18),bk=()k-3+2,1212∴数列{ak-bk}是递增数列.∴ak-bk≥a4-b4=3-(+2)=1212(0,).124.设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3,且数列{an+1-an}(nN*)是等差数列,数列{bn-2}(nN*)是等比数列.(1)求数列{an}和{bn}的通项公式;(2)是否存在kN*,使ak-bk(0,)?若存在,求出k,若不存在,说明理由.125.已知等比数列{an}的各项均为正数,公比q1,数列{bn}满足b1=20,b7=5,且(bn+1-bn+2)logma1+(bn+2-bn)logma3+(bn-bn+1)logma5=0.(1)求数列{bn}的通项公式;(2)设Sn=|b1|+|b2|+…+|bn|,求Sn.解:(1)将logma3=logma1+2logmq,logma5=logma1+4logmq代入已知等式整理得:2(bn-2bn+1+bn+2)logmq=0.∴bn-2bn+1+bn+2=0.∵q1,∴logmq0.即bn+bn+2=2bn+1.∴数列{bn}是等差数列.设其公差为d,52-.∴bn=20+(n-1)(-).52即bn=-n+.52245则由b7=b1+6d可得d=解:(2)令bn=0,得n=9.当n≤9时,bn≥0.则Sn=b1+b2+…+bn=20n+(-)n(n-1)252=-n2+n.54485当n9时,bn0,有:Sn=b1+b2+…+b9-b10-b11-…-bn=2(b1+b2+…+b9)-(b1+b2+…+bn)=180-(-n2+n)54485=n2-n+180.54485n2-n+180,n9.54485-n2+n,n≤9,54485∴Sn=5.已知等比数列{an}的各项均为正数,公比q1,数列{bn}满足b1=20,b7=5,且(bn+1-bn+2)logma1+(bn+2-bn)logma3+(bn-bn+1)logma5=0.(1)求数列{bn}的通项公式;(2)设Sn=|b1|+|b2|+…+|bn|,求Sn.6.设a0为常数,且an=3n-1-2an-1(nN*).(1)证明:对任意n≥1,an=[3n+(-1)n-12n]+(-1)n2na0;(2)假设对于任意n≥1,anan-1,求a0的取值范围.15(1)证:由an=3n-1-2an-1知:3+12令=-得=-.15则{an-3n}是以a0-为首项,公比为-2的等比数列.1515∴an-3n=(a0-)(-2)n.1515即an=[3n+(-1)n-12n]+(-1)n2na0.15(2)解:由anan-1及an=3n-1-2an-1知:an-an-1=3n-1-3an-10.∴an-13n-2.∵an+1-an=3n-3an=3n-3(3n-1-2an-1)=6an-10.∴0an-13n-2.13∴a0(0,).an+3n=3n-1-2an-1+3n=-2(an-1-3n-1).3+12∵(Sk)2=Sk2,∵kN*,∴k=4.∴满足(Sk)2=Sk2的正整数k的值是4.解:(1)由已知易得Sn=n2+n.127.设无穷等差数列{an}的前n项和为Sn.(1)若首项a1=,公差d=1,求满足(Sk)2=Sk2的正整数k;(2)求所有的无穷等差数列{an},使得对于一切正整数k都有(Sk)2=Sk2成立.3212∴(k2+k)2=(k2)2+k2.12整理得k3(k-4)=0.解:(2)设{an}的公差为d,则在(Sk)2=Sk2中分别取k=1,2得:(S1)2=S1且(S2)2=S4.即(a1)2=a1且(2a1+d)2=4a1+6d.解得a1=0,d=0或a1=0,d=6或a1=1,d=0或a1=1,d=2.若a1=0,d=0,则an=0,Sn=0,(Sk)2=Sk2成立;若a1=0,d=6,则S3=18,S9=216,(S3)2S9;若a1=1,d=0,则an=1,Sn=n,(Sk)2=Sk2成立;若a1=1,d=2,则an=2n-1,Sn=n2,(Sk)2=Sk2成立;综上所述,共有三个满足条件的等差数列:①{an}:an=0,即0,0,0,…;②{an}:an=1,即1,1,1,…;③{an}:an=2n-1,即1,3,5,….7.设无穷等差数列{an}的前n项和为Sn.(1)若首项a1=,公差d=1,求满足(Sk)2=Sk2的正整数k;(2)求所有的无穷等差数列{an},使得对于一切正整数k都有(Sk)2=Sk2成立.328.已知递增的等比数列{an}满足a2+a3+a4=28,且a3+2是a2,a4的等差中项.(1)求{an}的通项公式an;(2)若bn=anlogan,Sn=b1+b2+…+bn,求使Sn+n2n+130成立的n的最小值.12解:(1)由已知可设等比数列{an}的公比为q,依题意得:a1q+a1q2+a1q3=28,a1q+a1q3=2(a1q2+2),解得或(舍去)a1=2,q=2,a1=32,q=,12∴an=22n-1=2n.即{an}的通项公式为an=2n.(2)bn=anlogan=-n2n,12∴-Sn=12+222+323+…+n2n.∴-2Sn=122+223+324+…+n2n+1.∴Sn=2+22+23+…+2n-n2n+1=2n+1-2-n2n+1.为使Sn+n2n+130成立,应有2n+132.∴n4.∴使Sn+n2n+130成立的n的最小值为5.∴Sn=-(12+222+323+…+n2n).9.以数列{an}的任意相邻两项为坐标的点Pn(an,an+1)(nN*)均在一次函数y=2x+k的图象上,数列{bn}满足条件:bn=an+1-an(nN*,b10).(1)求证:数列{bn}是等比数列;(2)设数列{an},{bn}的前n项和分别为Sn,Tn,若S6=T4,S5=-9,求k的值.(1)证:∵Pn(an,an+1)(nN*)均在一次函数y=2x+k的图象上,∴an+1=2an+k,即:an+1+k=2(an+k).又bn=an+1-an=an+k,则bn+1=an+1+k,bn+1bn∴==2.an+1+kan+k∴数列{bn}是等比数列.解得:k=8.(2)解:b1=a1+k,bn=(a1+k)·2n-1,an=bn-k=(a1+k)·2n-1-k,S6=T6-6k=(a1+k)(26-1)-6k=63a1+5k,T4=(a1+k)(25-1)=15(a1+k),S5=31a1+26k=-9,S6=T4a1=-k,7810.(1)已知数列{cn},其中cn=2n+3n,且数列{cn+1-pcn}为等比数列,求常数p;(2)设{an},{bn}是公比不相等的两个等比数列,cn=an+bn,证明:数列{cn}不是等比数列.(1)解:∵数列{cn+1-pcn}为等比数列,∴(cn+1-pcn)2=(cn+2-pcn+1)(cn-pcn-1).又cn=2n+3n,∴[2n+1+3n+1-p(2n+3n)]2=[2n+2+3n+2-p(2n+1+3n+1)][2n+3n-p(2n-1+3n-1)].即[(2-p)2n+(3-p)3n]2=[(2-p)2n+1+(3-p)3n+1][(2-p)2n-1+(3-p)3n-1].整理得(2-p)(3-p)2n3n＝0.16解得p=2或3.(2)证:设{an},{bn}的公比分别为p,q,pq.为证{cn}不是等比数列,只须证c22c1c3.事实上,c22=(a1p+b1q)2=a12p2+b12q2+2a1b1pq,c1c3=(a1+b1)(a1p2+b1q2)=a12p2+b12q2+a1b1(p2+q2).∵pq,∴p2+q22pq.又a1,b1不为零,∴c22c1c3.故{cn}不是等比数列.11.设等比数列{an}的各项为实数,前n项的和为Sn,公比为q.(1)若S5,S15,S10成等差数列,求证:2S5,S10,S20-S10成等比数列;(2)若2S5,S10,S20-S10成等比数列,试问若S5,S15,S1