高三数学课件18等比数列高三数学课件

一、概念与公式1.定义2.通项公式3.前n项和公式二、等比数列的性质1.首尾项性质:有穷等比数列中,与首末两项距离相等的两项积相等,即:特别地,若项数为奇数,还等于中间项的平方,即:a1an=a2an-1=a3an-2=….若数列{an}满足:=q(常数),则称{an}为等比数列.an+1anan=a1qn-1=amqn-m.na1(q=1);Sn=a1-anq1-q=(q≠1).a1(1-qn)1-qa1an=a2an-1=a3an-2=…=a中2.特别地,若m+n=2p,则aman=ap2.2.若p+q=r+s(p、q、r、s∈N*),则apaq=aras.3.等比中项如果在两个数a、b中间插入一个数G,使a、G、b成等比数列,则G叫做a与b的等比中项.5.顺次n项和性质4.若数列{an}是等比数列,m,p,n成等差数列,则am,ap,an成等比数列.6.若数列{an},{bn}是等比数列,则数列{anbn},{}也是等比数列.anbnG=ab.若{an}是公比为q的等比数列,则ak,ak,ak也成等比数列,且公比为qn.k=2n+13nk=1nk=n+12n7.单调性8.若数列{an}是等差数列,则{ban}是等比数列;若数列{an}是正项等比数列,则{logban}是等差数列.三、判断、证明方法1.定义法;2.通项公式法;3.等比中项法.a10,q1,a10,0q1,或{an}是递增数列;或a10,0q1,a10,q1,{an}是递减数列;q=1{an}是常数列;q0{an}是摆动数列.典型例题1.设数列{an}的前n项和为Sn,若S1=1,S2=3,且Sn+1-3Sn+2Sn-1=0(n≥2),试判断{an}是不是等比数列.2.设等比数列{an}的前n项和为Sn,若S3+S6=2S9,求数列的公比q.3.三个数成等比数列,若将第三项减去32,则成等差数列,再将此等差数列的第二项减去4,又成等比数列,求原来的三个数.4.已知数列{an}的各项均为正数,且前n和Sn满足:6Sn=an2+3an+2.若a2,a4,a9成等比数列,求数列的通项公式.a1=1,a2=2,Sn+1-Sn=2(Sn-Sn-1),an=2n-1,{an}是等比数列.设三数为a,b,c,得b=2+4a,c=7a+36.2,10,50或,,.933892629an+1-an=3,a1=1,an=3n-2.12-436.已知{an}是首项为a1,公比为q的等比数列.(1)求和:a1C2-a2C2+a3C2,a1C3-a2C3+a3C3-a4C3;(2)由(1)的结果归纳概括出关于正整数n的一个结论,并加以证明;(3)设q≠1,Sn是{an}的前n项和,求S1Cn-S2Cn+S3Cn-S4Cn+…+(-1)nSn+1Cn.00011122233n(1)a1(1-q)2,a1(1-q)3;(2)a1Cn-a2Cn+a3Cn-a4Cn+…+(-1)nan+1Cn=a1(1-q)n(nN*);0123n(3)-a1q(1-q)n-1.(2)bn=3qn-1.5.数列{an}中,a1=1,a2=2.数列{anan+1}是公比为q(q0)的等比数列.(1)求使anan+1+an+1an+2an+2an+3(nN*)成立的q的取值范围;(2)若bn=a2n-1+a2n(nN*),求{bn}的通项公式.(1)0q;1+52∴(n+2)Sn=n(Sn+1-Sn).证:(1)∵an+1=Sn+1-Sn,又an+1=Sn,n+2n整理得nSn+1=2(n+1)Sn.n+2n∴Sn+1-Sn=Sn,SnnSn+1n+1∴=2.7.数列{an}的前n项和记为Sn,已知a1=1,an+1=Sn(n=1,2,3,…),证明:(1)数列{}是等比数列;(2)Sn+1=4an.n+2nSnnSnn∴{}是以1为首项,2为公比的等比数列.(2)由(1)知=4(n≥2),Sn+1n+1Sn-1n-1于是Sn+1=4(n+1)=4an(n≥2),Sn-1n-1又a2=3S1=3a1=3,故S2=a1+a2=4=4a1.因此对于任意正整数n,都有Sn+1=4an.(2)证法2:由(1)知=2n-1.Snn∴Sn=n2n-1.∴Sn+1=(n+1)2n.∵an=Sn-Sn-1=n2n-1-(n-1)2n-2=(n+1)2n-2(n≥2).而a1=1也适合上式,∴an=(n+1)2n-2(nN*).∴4an=(n+1)2n=Sn+1.即Sn+1=4an.7.数列{an}的前n项和记为Sn,已知a1=1,an+1=Sn(n=1,2,3,…),证明:(1)数列{}是等比数列;(2)Sn+1=4an.n+2nSnn8.数列{an}中,a1=1,Sn+1=4an+2,(1)设bn=an+1-2an,求证:{bn}是等比数列,并求其通项;(2)设cn=,求证:数列{cn}是等差数列;(3)求Sn=a1+a2+…+an.an2n(1)证:由已知an+1=Sn+1-Sn=4an+2-4an-1-2,∴an+1=4an-4an-1(n≥2).∴bn=an+1-2an=4an-4an-1-2an=2(an-2an-1)=2bn-1.∴=2(n≥2).bn-1bn∴{bn}是以3为首项,2为公比的等比数列.又由a1=1,a1+a2=S2=4a1+2得a2=5,∴b1=a2-2a1=3.∴bn=32n-1.∴数列{cn}是等差数列.∴Sn=4an-1+2=4(3n-4)2n-3+2=(3n-4)2n-1+2.∴an=2ncn=(3n-1)2n-2.(2)证:∵cn+1-cn=-an2nan+12n+1an+1-2an2n+1=bn2n+1==32n-12n+1=(常数),34(3)解:由已知c1==,12a12∴由(2)得cn=+(n-1)=(3n-1).123414∴an-1=(3n-4)2n-3(n≥2).而S1=a1=1亦适合上式,∴Sn=(3n-4)2n-1+2(nN*).8.数列{an}中,a1=1,Sn+1=4an+2,(1)设bn=an+1-2an,求证:{bn}是等比数列,并求其通项;(2)设cn=,求证:数列{cn}是等差数列;(3)求Sn=a1+a2+…+an.an2n1.四个正数,前三个数成等差数列,其和为48,后三个数成等比数列,其最后一个数是25,求此四数.解:由已知可设前三个数为a-d,a,a+d(d为公差)且a+d0.∵后三数成等比数列,其最后一个数是25,解得:a=16,d=4.故所求四数分别为12,16,20,25.∴a-d+a+a+d=48,且(a+d)2=25a.∴a-d=12,a+d=20.课后练习题2.在等比数列{an}中,a1+a6=33,a3a4=32,an+1an.(1)求an;(2)若Tn=lga1+lga2+…+lgan,求Tn.解:(1)∵{an}是等比数列,∴a1a6=a3a4=32.又∵a1+a6=33,∴a1,a6是方程x2-33x+32=0的两实根.∵an+1an,∴a6a1,∴a1=32,a6=1.∴32q5=1∴an=26-n.q=.12(2)由已知lgan=(6-n)lg2.∴Tn=lga1+lga2+…+lgan=[6n-(1+2+…+n)]lg2=[6n-]lg2n(n+1)2=[(6-1)+(6-2)+…+(6-n)]lg2=(-n2+n)lg2.122113.已知数列{an}为等比数列,a2=6,a5=162,(1)求数列{an}的通项公式;(2)设Sn是数列{an}的前n项和,证明:≤1.SnSn+2Sn+12(1)解:设等比数列{an}的公比为q,依题意得:a1q=6且a1q4=162.解得:a1=2,q=3.∴数列{an}的通项公式为an=23n-1.(2)证:Sn==3n-1,2(1-3n)1-3SnSn+2Sn+12(3n-1)(3n+2-1)(3n+1-1)2=32n+2-(3n+3n+2)+132n+2-23n+1+1=32n+2-23n3n+2+132n+2-23n+1+1≤=1.故有≤1成立.SnSn+2Sn+124.设{an}为等比数列,Tn=na1+(n-1)a2+…+2an-1+an,已知T1=1,T2=4.(1)求数列{an}的首项和公比;(2)求数列{Tn}的通项公式.解:(1)设等比数列{an}的公比为q,则:T1=a1,T2=2a1+a2.又T1=1,T2=4,∴a1=1,2a1+a2=4a2=2.∴q=2.∴数列{an}的首项为1,公比为2.(2)解法1由(1)知:a1=1,q=2,∴an=2n-1.∴Tn=n1+(n-1)2+(n-2)22+…+22n-2+12n-1.∴2Tn=n2+(n-1)22+(n-2)23+…+22n-1+12n.∴Tn=-n+2+22+…+2n-1+2n=2n+1-n-2.解法2设Sn=a1+a2+…+an.∵an=2n-1,∴Sn=2n-1.∴Tn=na1+(n-1)a2+…+2an-1+an=a1+(a1+a2)+(a1+a2+a3)+…+(a1+a2+…+an)=S1+S2+…+Sn=2+22+…+2n-n=2n+1-n-2.5.在公差为d(d0)的等差数列{an}和公比为q的等比数列{bn}中,已知a1=b1=1,a2=b2,a8=b3,(1)求d,q的值;(2)是否存在常数a,b,使得对于一切正整数n,都有an=lgabn+b成立?若存在,求出a和b,若不存在,说明理由.解:(1)∵a1=b1=1,a2=b2,a8=b3,∵d0,∴解得d=5,q=6.故d,q的值分别为5,6.∴1+d=q且1+7d=q2.(2)由(1)及已知得an=5n-4,bn=6n-1.假设存在常数a,b,使得对于一切正整数n,都有an=lgabn+b成立,则5n-4=loga6n-1+b对一切正整数n都成立.即5n-4=nloga6+b-loga6对一切正整数n都成立.∴loga6＝5,b-loga6=-4.∴a=6,b=1.5故存在常数a,b,它们的值分别为6,1,使得对于一切正整数n,都有an=lgabn+b成立.56.设Sn为数列{an}的前n项和,且满足2Sn=3(an-1),(1)证明数列{an}是等比数列并求Sn;(2)若bn=4n+5,将数列{an}和{bn}的公共项按它们在原数列中的顺序排成一个新的数列{dn},证明{dn}是等比数列并求其通项公式.证:(1)由已知a1=3,当n≥2时,an=Sn-Sn-1.∴2an=2(Sn-Sn-1)=2Sn-2Sn-1=3(an-an-1)an=3an-1.故数列{an}是首项与公比均为3的等比数列.从而an=3n,Sn=(3n+1-3).12(2)易知d1=a2=b1=9.设dn是{an}中的第k项,又是{bn}中的第m项,即dn=3k=4m+5.∵ak+1=3k+1=3(4m+5)=4(3m+3)+3不是数列{bn}中的项,而ak+2=3k+2=9(4m+5)=4(9m+10)+5是{bn}的第(9m+10)项,∴dn+1=ak+2=3k+2.dn+1dn由=9知:{dn}是首项与公比均为9的等比数列,故dn=9n.7.{an},{bn}都是各项为正的数列,对于任意的自然数n,都有an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列.(1)求证{bn}是等差数列;(2)如果a1=1,b1=2,Sn=++…+,求Sn.2221a11a21an∵an0,bn0,∴由②式得an+1=bnbn+1.(1)证:依题意有:2bn=an+an+1,①an+1=bnbn+1.②2222从而当n≥2时,an=bn-1bn,代入①得2bn=bn-1bn+bnbn+1.2∴2bn=bn-1+bn+1(n≥2).∴{bn}是等差数列.(2)解:由a1=1,b1=2及①②两式