數(shù)列證明題型總結(jié)附答案

1、、解答題1.在數(shù)列an中，a1=1,an+1=2an+2n.(I)設(shè)酬=品，證明：數(shù)列bn是等差數(shù)列;(n)求數(shù)列an的前n項(xiàng)的和S.c-cn，一E,.an41anan+1-2an2,(I)因?yàn)閎n+1bn=2n2=2n=m=1所以數(shù)列bn為等差數(shù)列(n)因?yàn)閎n=bi+(n1)x1=n所以an=n2n-所以S=1X2°+2X21+nX2nT2s=1X212><22+nx2n兩式相減得S=(n1)2n+11 1.12 .在數(shù)歹Uan中，a1=2，an+1=an+2+1.3 I)設(shè)bn=2Zn,證明：數(shù)列bn是等差數(shù)列;4 n)求數(shù)列an的前n項(xiàng)和S【答案】1 .1(I)由

2、an+1=an+2Tl,得2an+1=2an+1bn+1=bn+1,則bn是首項(xiàng)bl=1,公差為1的等差數(shù)列.n故bn=n,an=2-n"(n)S=1x-+2xm+3*2+(n1)x2門-1+nx2K.1-11112s=1''22+2'2'3+3*了+(n-1)X2n+nX2rm兩式相減，得:;(1-J)2 2_2112n+1-12n2n+11-123,數(shù)列an的各項(xiàng)均為正數(shù)，前n項(xiàng)和為Sn,且滿足4s=(an+1)2(neN*).(I)證明：數(shù)列an是等差數(shù)列，并求出其通項(xiàng)公式*(n)設(shè)bn=an+2an(nCN),求數(shù)列bn的刖n項(xiàng)和Tn.(I)n

3、=1時(shí)，4a1=(d+1)2?a22a1+1=0,即a=1n>2時(shí)，4an=4Sn4Sn1=(an+1)(an-1+1)=anan-1+2an2an-1_22?anan-12an2an1=0?(an+an1)(an-an1)-2=0-an>0anan-1=2一一、*.一、.*故數(shù)列an是首項(xiàng)為ai=1,公差為d=2的等差數(shù)列，且an=2n1(nN)2n1(n)由(I)知bn=an+2an=(2n-1)+2Tn=bi+b2+bn=(1+21)+(3+23)+-+(2n-1)+22n1=1+3+(2n-1)+(21+23+22nJ=n2+-/_2nx_2n+12(12)21-4卜n2I

4、，"n+1+3n2-24,數(shù)列an的各項(xiàng)均為正數(shù)，前n項(xiàng)和為$,且滿足2yq=a+1(nCN).(I)證明：數(shù)列an是等差數(shù)列，并求出其通項(xiàng)公式an；(n)設(shè)bn=an2n(neN),求數(shù)歹Ubn的前n項(xiàng)和Tn.【答案】(l)由2痣=an+1(nCN*)可以得到43=(an+1)2(nCN)n=1時(shí)，4a1=(a1+1)2?a22a1+1=0,即a=1n>2時(shí))4an=434S-1=(an+1)2(an-1+1)222=an-an-1+2an2an-122?an一an-1一2an2an-1=0(an+anT)(anan-1)2=0an>0.anan-1=2故數(shù)列an是首項(xiàng)

5、為a1=1,公差為d=2的等差數(shù)列，且*.an=2n-1(nCN)(n)由(I)知bn=an-2n=(2n-1)2n.Tn=(121)+(322)+(2n3)2n1+(2n-1)-2n則2Tn=(1-22)+(323)+-+(2n-3)-2n+(2n-1)-2n+1兩式相減得：-Tn=(121)+(222)+-+(22n)-(2n-1)-2n+1=22(；-；)-2-(2n-1)-2n+112=(3-2n)-2n+1-6.Tn=(2n-3)-2-1+6(或Tn=(4n6)2n+6)327*5,已知數(shù)列an,其刖n項(xiàng)和為S=n+n(nCN).(I)求a1,a2；(n)求數(shù)列an的通項(xiàng)公式，并證明

6、數(shù)列an是等差數(shù)列；n項(xiàng)和Tn.(出)如果數(shù)列bn滿足an=log2bn,請(qǐng)證明數(shù)列bn是等比數(shù)列，并求其前【答案】(I)a1=S=5,21+22=82=5x22+5x2=13,解得a2=8.(n)當(dāng)n>2時(shí),3n=SnSn-1=2n?-(ni)2+n-(n-i)=|(2n-1)+|=3n+2.又ai=5滿足an=3n+2,*、an=3n+2(nCN).anan-1=3n+23(n1)+2*.=3(n>2,nN),數(shù)列an是以5為首項(xiàng)，3為公差的等差數(shù)列.(出)由已知得bn=2an(nCN*),h9nn+ia=2an+ian=23=8(nN),bn2又bi=2ai=32,.數(shù)列bn

7、是以32為首項(xiàng)，8為公比的等比數(shù)列.32(i8n)32n,Tn=o="y(8-i),I876.已知函數(shù)f(x)=2xx+2,數(shù)列an滿足:4一、ai=,an+i=f(an).3an的通項(xiàng)公式;i(I)求證：數(shù)列一為等差數(shù)列，并求數(shù)列an(n)記Sn=aia2+a2a3+anan+i,求證:【答案】2an111111證明：(I)an+1=f(an)=,=F-,即=,&+2an+1an2an+1an2,1則一成等差數(shù)列，an11,、13,、12n+14所以字=1十(n1)2=4+(n1)2=T，則an=2TT44-11(n)anan+1=-=8,'22n+12n+32n+

8、12n+3'Sn=a1a2+a2a3+1111,1anan+1=83-5+5-7+12n+3118<-32n3<3.7,已知數(shù)列an的前三項(xiàng)依次為2,8,24,且an2an-1是等比數(shù)歹U.(I)證明，是等差數(shù)列；(n)試求數(shù)列an的前n項(xiàng)和Sn的公式.【答案】(I)-a22a1=4,a32&=8,an2an-1是以2為公比的等比數(shù)列."an一2an-1=4X2=2.等式兩邊同除以2n，得ina=1,an1n是等差數(shù)列.(n)根據(jù)(I)可知1n=a1+(n1)X1=n,.an=n2n.S=1X2+2X22+3X23+n2n,'2&=1X22

9、+2X23+(n-1)-2n+n-2n+1.'一得:S=2+22+23+2nn2n+12(1"-n-2n+1=2n+1-2-n-2n+1,1-2，&=(n1),2+2.8,已知數(shù)列an的各項(xiàng)為正數(shù)，前n項(xiàng)和為S,且滿足：S=：an+1(nCN*).2an(I)證明：數(shù)列S2是等差數(shù)列；1212121.(n)設(shè)Tn=2S1+11+23S3+2nSh,求Tn.【答案】(I)證明：當(dāng)n=1時(shí)，a1=S,又S=：an+(nCN),2an'11-.S=2S+s,斛得S=1.當(dāng)n>2時(shí)，an=SnSi-1,-11.Sn=SSiS1-1+QQ2ChSi-11即S+S1

10、=，化簡(jiǎn)得S2St1=1,SS-1$是以s1=1為首項(xiàng)，1為公差的等差數(shù)列.(n)由(I)知S2=n,Ti=；5+2'nSn,1.1.11一即Tn=1金+21+(n-1)歹+n開(kāi)1m1_.1.11一x2行2T=1,了+(n1)2Tl+n2n+1.一得2'=2+22+2nn2inn11-2nn+1=1-nn221.n+2TnTT=1-n+1,22'.Tn=2-2n9.數(shù)列an滿足31=1,an+11*、222-2+4=1(nCN),記Sn=a1+a2+an.3n(I)證明：3是等差數(shù)列;an(n)對(duì)任意的nCN,如果&n+1日恒成立，求正整數(shù)m的最小值.3U【答案

11、】11_11,、，c1,C(I)證明:孑-一,=4?左=孑+(n1)x4?2=4n3,1即；2是等差數(shù)列.an111(n)令g(n)=S2n+1Si="+,+4n+14n+58n+1.g(n+1)g(n)<u,.g(n)在nCN上單倜遞減，1414m28g(n)max=g(1)=45.30恒成乂？m>y,又mN,.正整數(shù)m的最小值為1U.10.已知數(shù)列an是首項(xiàng)31=,公比為一的等比數(shù)列，設(shè)bn+1510g33n=t,常數(shù)tCN*.(I)求證：bn為等差數(shù)列;(II)設(shè)數(shù)列Cn滿足Cn=3nbn,是否存在正整數(shù)k,使Ck+1,Ck,Ck+2成等比數(shù)列？若存在,求k,t的值

12、；若不存在，請(qǐng)說(shuō)明理由.(I)證明：3n=3-,bn+1bn=-1510g3=5,3日bn是首項(xiàng)為bi=t+5,公差為5的等差數(shù)列.nn(n)Cn=(5n+t)-3-令5n+t=x,貝U6=x一33n+1n+2Cn+1=(x+5)3-,Cn+2=(x+10)3-,332n2育Ck=Cn+1Ci+2)貝U(x,3-3)=(x+5),3n+1n+2丁(x+10)3332.5人化間得2x-15x-50=0,解得x=10或一2(舍)，進(jìn)而求得n=1,t=5,綜上，存在n=1,t=5適合題意.11 .在數(shù)列an中,31=1,3n+1=2an+2n+1.(I)設(shè)bn=3n+1an+2,(nCN),證明：數(shù)

13、列bn是等比數(shù)列;(n)求數(shù)列an的通項(xiàng)an.(I)由已知3n+1=2an+2n+1得an+2=2an+i+2n+3一，得an+2an+1=2a+12an+2設(shè)an+2an+1+c=2(an+1-an+c).展開(kāi)與上式對(duì)比，得c=2因此，有an+2an+1+2=2(an+1an+2)由bn=an+1-an+2,得bn+1=2bn,由a1=1,a2=2a1+3=5,得b=a2a+2=6,故數(shù)列bn是首項(xiàng)為6,公比為2的等比數(shù)列n-1n(n)由(I)知)bn=6X2=3X2則an+1an=bn2=3X22)所以an=a+(aa)+(a3si2)+，+(anan-1)=1+(3X212)+(3X22

14、2)+(3X2n1-2)=1+3(2+22+23+2n1)-2(n-1)an=3X2n-2n-3,當(dāng)n=1時(shí)，a=3X212X13=65=1,故a1也滿足上式故數(shù)列an的通項(xiàng)為an=3X2n-2n-3(nN).12 .在數(shù)列an中，a1=1,an=！an1+!*工(nCN*且nR2).6223、一1一(I)證明：an+了是等比數(shù)列；(n)求數(shù)列an的通項(xiàng)公式；1(出)設(shè)S為數(shù)列an的前n項(xiàng)和，求證$</.【答案】(I)由已知，得1an+了111an+1+3Tr（2an+31、1nT!）+3門+111-an+-n是等比數(shù)列.23.1（n）設(shè)An=an+不,3i.1111則A=a+1=友+鼻

15、=m，且q=-6322則An=（2）n,11an+屋=，可得c11an2n-3n11,11,11（出）s=（初一耍）+（落?。?（藝3n）111112-3n2n1=2一簿29=2一丁丁<2*13.已知數(shù)列an滿足a=2,an+1=2an+1(nCN).(I)證明：數(shù)列ann是等比數(shù)列，并求出數(shù)列an的通項(xiàng)公式；n*(n)數(shù)列bn滿足：bn=-(nN),求數(shù)列bn的前n項(xiàng)和S.2an2n【答案】(I)證法一：由an+i=2an-n+1可得an+i(n+1)=2(an-n),又ai=2,則ai-1=1,.數(shù)列ann是以a1-1=1為首項(xiàng)，且公比為2的等比數(shù)列,則ann=1X2nT,1,an=

16、2nn.ch+1(n+1)2ann+1(n+1)2an2n證法二：ann=0=二n=2'.數(shù)列ann是以a1-1=1為首項(xiàng)，且公比為2的等比數(shù)列,則ann=1X2nT,1,an=2nn.n.nn(n)-bn=-,.-bn=-=-n'/2an2n2an2n2Sn=b1+b2+bn=2+2(/)2+n,(2)n1 12131n1n+11-2$=()+2()+(n1)(2)+n.()11n21E1n+1由，得2&=2+(；)2+(；)3+弓)nnJ)n+11-n-(-)=1(n.21、n+1+2)(2),.Sn=2-(n+2)(；)n.14 .在數(shù)歹Uan中，a=1,2nan

17、+1=(n+1)an,nN.(I)設(shè)bn=胃，證明：數(shù)列bn是等比數(shù)列；(n)求數(shù)列an的前n項(xiàng)和Sn.【答案】,、巾且b+1an+1n1(I)因?yàn)榫?干*=2,1所以bn是首項(xiàng)為1,公比為2的等比數(shù)歹U.,一an1n(n)由(I)可知n=2n1,即a=2n1,2.34.n&十萬(wàn)+5+5+*，一，一，1，r上式兩邊乘以2,得1123n-1n尹=2+尹?+n+產(chǎn)11111n兩式相減，得2s=1+2+了+23+2,所以Sn=4-2nF15 .設(shè)數(shù)列an的前n項(xiàng)和為Sn,且&=(1+入)一入an，其中入w1,0.(I)證明：數(shù)歹Uan是等比數(shù)歹U；(n)設(shè)數(shù)列an的公比q=f(入)，

18、數(shù)列bn滿足b1=2，bn=f(bn1)(nN,n>2),求數(shù)列bn的通項(xiàng)公式.(I)由Sn=(1+入)一入an?Sn-1=(1+入)一入an-1(n>2),相減得：an=一入an+入an1,1an=(n>2),an-11+入'八數(shù)列an是等比數(shù)列(n)f(入)=bn1+入，-bn-1+bn1?bn-bn1+1，的等差數(shù)列;1n+T1一、一，1、，bn1首項(xiàng)為b1=2,公差為11c,，,、，,，=2+(n1)=n+1,.bn=bn16 .在等差數(shù)列d中，310=30,320=50.(I)求數(shù)列3n的通項(xiàng)3n；(n)令bn=2an-10,證明：數(shù)列bn為等比數(shù)列;(出)

19、求數(shù)列nbn的前n項(xiàng)和Tn.【答案】(I)由3n=31+(n1)d,310=30,320=50,得方程組31+9d=3031+19d=50解得31=12,d=2.3n=12+(n1)2=2n+10.(II)由(I)得bn=23n10=22n+10T°=22n=4n,n+1bn+14-bn=4n.bn是首項(xiàng)是4,公比q=4的等比數(shù)歹U.(出)由nbn=nx4n得：1=1X4+2X42+nx4n4Tn=1X42+(n1)X4n+nX4n+1相減可得:3Tn=4+42+4nnx4n+14(14n)-3n+1nX4Tn=(3n1)X4+4917.已知an是等差數(shù)列，其前n項(xiàng)和為&,已

20、知a3=11,4=153,(I)求數(shù)列Jan的通項(xiàng)公式;(n)設(shè)an=log2bn,證明bn是等比數(shù)列，并求其前n項(xiàng)和Tn.(I)9X8解得：d=3,a1=5,，an=3n+29a1+-2d=153an+1(n)bn=2an,bn+122an+1-an之&bn2bn是公比為8的等比數(shù)列，a1_32(18n)32n又b1=2=32,Tn=18=y(81).18.在數(shù)列an中，a1=3,an=2an1+n-2(n>2,且nCN).(I)求a2,as的值；(n)證明：數(shù)列an+n是等比數(shù)列，并求an的通項(xiàng)公式；(出)求數(shù)列an的前n項(xiàng)和&.【答案】(I)-.-a1=3,*an=

21、2an1+n-2(n>2,且nCN),-32=23i+22=6)83=232H-32=13.(n)證明:3n+n(2an-1n2)+nan1+(n1)3n-1H-n12an-12n23n-1+n1，數(shù)列an+n是首項(xiàng)為ad1=4,公比為2的等比數(shù)列.,.八n1_n+1口口_n+13n+n=4,2=2,IP3n=2n,an的通項(xiàng)公式為an=2n+1-n(nN).(m)an)的通項(xiàng)公式為an=2n+1-n(nN),c,2_3_4_n+1x.S=(2+2+2+2)-(1+2+3+n)2n2x(1-2)nx(n+1)=1-222.m+2n+n+819.已知數(shù)列an滿足a=2,an+i=3a+2(

22、nN).(I)求證：數(shù)列an+1是等比數(shù)列;(n)求數(shù)列an的通項(xiàng)公式.(I)證明：由2卜1=32-2得2”-1=3(排+1),即數(shù)列an+1是首項(xiàng)為3,公比為3的等比數(shù)歹U.(n)由(I)知，an+1=3-3nT=3n?an=3n-1.20.已知數(shù)列an滿足a=2,an+1=4a+2n+1,S為an的前n項(xiàng)和.(I)設(shè)bn=an+2n,證明數(shù)列bn是等比數(shù)歹U,并求數(shù)列%的通項(xiàng)公式;2ni=1(n)設(shè)Tn=,n=1,2,3,，證明:Sn(I)因?yàn)閎n+1=an+1+2n+1=(4an+2n+1)+2“+1=4(an+2n)=4bn,且氏=21+2=4,所以bn是以4為首項(xiàng)，以q=4為公比的等

23、比數(shù)歹U.所以bn=biqi=4n,所以an=4n2n.(n)Si=a+a2+an=(4+4、+4”)一(2+2?+2”)=3(4n-1)-2(2n-1)=3(2n+1)2-3-2n+1+23(2n+1-1)(2n+1-2)=|(2n+1-1)(2n-1),2n3所以Tn=XSi22n(2n+1-1)(2n1)3=-x22n-<2n+1-1，因此t=32i=1J=12n1-2n+1-12211-2n+13<-1221.已知數(shù)列an的前n項(xiàng)和為*Sn,且S=4an3(nN).(I)證明：數(shù)歹Uan是等比數(shù)歹u；(n)右數(shù)列bn滿足bn+1=an+bn(nN),且b1=2,求數(shù)列bn的

24、通項(xiàng)公式.4an=an-1.3n-11(n>2),(I)證明：由Sn=4an-3,n=1時(shí)，ai=4ai-3,解得ai=1.n>2時(shí)，Sn1=4an-13,所以當(dāng)n>2時(shí)，an=SSn1=4an4an1,得,一、，,，4,，又a1=1w0,所以an是首項(xiàng)為1,公比為a的等比數(shù)列.34n1*4n-1(n)因?yàn)閍n=-,由bn+1=an+bn(nCN),得bn+1bn=-334n11一,3.4可信bn=b+(b2b1)+(b3b2)+(bnbn1)=2+4=3§13當(dāng)n=1時(shí)也滿足，所以數(shù)列bn的通項(xiàng)公式為bn=34n1一1.322.在各項(xiàng)均為負(fù)數(shù)的數(shù)列an中，已知點(diǎn)(

25、an,an+1)(nCN)在函數(shù)丫=宗的圖象上，且a2as38=27(I)求證：數(shù)列an是等比數(shù)列，并求出其通項(xiàng)；(n)若數(shù)列bn的前n項(xiàng)和為Sn,且bn=an+n,求S.【答案】(I)證明：因?yàn)辄c(diǎn)(an,an+1)(neN)在函數(shù)y=；x的圖象上，3所以an+1=ian,即.=2,故數(shù)列an是公比q=2的等比數(shù)列.3an33即a235=|3,由于數(shù)列an的各項(xiàng)均為負(fù)數(shù)，則ai=2，所以an=32n2(n)由(i)知)an=三,bn=32n2鼻+n,3所以Sn=3.1nT+n3223.已知數(shù)列an的前n項(xiàng)和為且S=3-2nT2,bn=an+i.(I)求數(shù)列Jan的通項(xiàng)公式;(n)證明：數(shù)列bn

26、是等比數(shù)列，并求其前n項(xiàng)和Tn.(I)-.-Sn=3.2nT_2,當(dāng)n>2時(shí))an=$1=3，223,n2+2=3-2n2,當(dāng)n=1時(shí)，ai=1不滿足上式.n=(n=1),(n>2).848因?yàn)閍2a5=否，則aiq,aiq=萬(wàn)，(n)bn=an+i=3X2nbi=a2=3,nN.bn3X21-=-2=2,nCN,.數(shù)列bn是首項(xiàng)為3,公比為2的等比數(shù)歹U;bni3X23112bn由等比數(shù)列前n項(xiàng)和公式得Tn=:/=3X2n3.1224.設(shè)數(shù)列an的前n項(xiàng)和為已知a=5,an+i=$+3n(nCN).(I)令bn=S3n,求證：bn是等比數(shù)列;“12011,(n)令cn='

27、一-,設(shè)Tn是數(shù)列cn的前n項(xiàng)和，求滿足不等式1>的n的log2bn+1log2bn+24026最小值.【答案】(I)證明：bi=Si-3=20,S+1Sn=S+3n,即S+i=2Sn+3n,bn+1Sn+13n+12S3n+1+3nSn-3n=Sn3n=2*0所以bn是等比數(shù)列.(n)由(i)知bn=2n,1log2bn+1,log2bn+21(n+1),(n+2)J1=-.n+1n+2',11Tn=_,2n+2n>2011,即nmin=2012._12011T1-2-n+2>4026an*25.已知數(shù)列an滿足：a=1,an+1=a+？(nCN).(I)求證：數(shù)列

28、7+1是等比數(shù)列；an,bn+11(n)若一-=-+1,且數(shù)列bn是單調(diào)遞增數(shù)列，求實(shí)數(shù)入的取值范圍.n一人an、”【答案】3n+11一+1=2L，一12(I)證明：-=1+t,an+1an1_,.1-，曾仁2”所以數(shù)列盤1是等比數(shù)列、1n1(n)an+1=2,an=21'-b21+1=2n,bn+1=2n(n-入),n-aan、'bn=2(n1入)(n>2),bi=入適合,所以bn=2(n1入)(neN),由bn+i>bn得2(n+1入)>2(n入)X<n+2,入<(n+2)min=3,入的取值范圍為入|入3.>乙一、r>>*_

29、r_一_._-*.26.已知數(shù)列an中，a=2,a2=4,an+1=3an2an1(n>2,nCN).(I)證明：數(shù)列an+1an是等比數(shù)列，并求出數(shù)列an的通項(xiàng)公式；2(an-1)一、，一一.一.(n)記bn=-,數(shù)列Jbn的前n項(xiàng)和為3b求使S>2010的n的最小值.an【答案】(I)an+1=3an2an-1(n>2),.(an+1an)=2(anSh1)(n>2).a1=2)a2=4,.a2a1=2w0)ana-1才0)故數(shù)列an+1an是首項(xiàng)為2,公比為2的等比數(shù)列，an+1an=(a2a1)2nT=21-3n=(3nan-1)+(3n-13n2)+(3n-2

30、3n3)+(a2-ai)+ai=2-+*2+213+21+2-/-n1、2(12)；-十212=2n(n>2).又d=2滿足上式，an=2n(neN).,.2(an1)(n)由(I)知bn=-an=21-=2an.11一2n一一.111.S=2n一1十大爐十+2=2n1=2n212n=2n2+?n-1.1-2由S>2010得:一一11.2n2+?n1>2010,即n+2n>1006,因?yàn)閚為正整數(shù)，所以n的最小值為1006.27.已知數(shù)列an的前n項(xiàng)和為Sn,滿足S+2n=2an.(I)證明：數(shù)列an+2是等比數(shù)列，并求數(shù)列an的通項(xiàng)公式an；(n)若數(shù)列bn滿足bn=

31、log2(an+2),求數(shù)列2的前n項(xiàng)和Tn.bn【答案】(I)證明：由$+2n=2an,得S=2an-2n,當(dāng)nCN時(shí)，S=2a2n,當(dāng)n=1時(shí))Si=2ai-2,則ai=2,當(dāng)n>2時(shí)，S-i=2an-12(n1),-)得an=2an-2an-1-2,即an=2an-1+2,.an+2=2(an-1+2),an+2=2,an1+2.伯n+2是以21+2為首項(xiàng)，以2為公比的等比數(shù)列.an+2=4c2n1,an=2n+12.(n)解：an=2n+12,bn=n(n+1),1_1_11=b1bnnn+1nn+1bn=1一1+1221+1X3nn+1=1n+1=JL.n+1【解析】考點(diǎn)：數(shù)列的求和；等比數(shù)列的通項(xiàng)公式.專題：綜合題.分析：(I)由$+2n=2an,得$=2an-2n,由此利用構(gòu)造法能夠證明數(shù)列an