數(shù)列證明題型總結(jié)教師版附答案

1、、解答題1在數(shù)列an中，ai=1,an+i=2an+2n.(I)設(shè)bn，證明：數(shù)列%是等差數(shù)列；(n)求數(shù)列an的前n項(xiàng)的和Sn.【答案】an+ianan+i2an2n(I)因?yàn)閎n+1bn=-2n21=2=2n=1所以數(shù)列bn為等差數(shù)列(n)因?yàn)閎n=b1+(n1)¥=n所以an=n2n1所以Sn=1>20+2X21+nx2n12sl=1><2+2X22+nX2n兩式相減得Sn=(n-1)2n+11112在數(shù)列an中，a1=2，an+1=2an+2n+1.(I)設(shè)bn=2nan,證明：數(shù)列bn是等差數(shù)列;(n)求數(shù)列an的前n項(xiàng)和Sn.1.，(I)由an+1=2a

2、n+2n+v得2n1an+1=2nan+1bn+1=bn+1,則bn是首項(xiàng)b1=1,公差為1的等差數(shù)歹U.故bn=n,an=2n.(n)Sn=1名+2.+34+-+(n-1)>2；1T+n>；Sn=1+24+3+(n1)斗+n>2兩式相減，得：11,1,1,1n2Sn=2+22+23+2n2n+11212n+12n2n11/18Sn=2一2n2n3.數(shù)列an的各項(xiàng)均為正數(shù)，前n項(xiàng)和為Si,且滿足4Sn=(an+1)2?anan12an2an1=0?(an+an1)(anan1)2=0(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=an+2an

3、(nN*),求數(shù)列bn的前n項(xiàng)和Tn.【答案】(I)n=1時(shí)，4ai=(ai+1)2?a22ai+1=0,即ai=1nn2時(shí),4an=4Sn4Sn1=(an+1)2(an-1+1)2=anan1+2an2an1?a2a212an2an-1=0(an+ant)(anan-1)-2=0,an>0.anan1=2故數(shù)列an是首項(xiàng)為a1=1,公差為d=2的等差數(shù)列，且an=2n1(nCN*)(n)由(I)知bn=an+2an=(2n-1)+22n1Tn=b1+b2+bn=(1+21)+(3+2,an>0"an一an-1=2故數(shù)列an是首項(xiàng)為a1=1,公差為d=2的等差數(shù)列，且an

4、=2n1(nCN*)+(2n1)+22n1=1+3+(2n1)+(2+23+22n1)=n2+2(122n)1-422n+1+n222n+1+3n2234.數(shù)列an的各項(xiàng)均為正數(shù)，前n項(xiàng)和為Sn,且滿足2<&=an+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；(nbn=an2n(nCN*),求數(shù)列bn的前n項(xiàng)和Tn.【答案】(I)由2VS1=an+1(nCN*)可以彳導(dǎo)到4Sn=(an+1)2(nCN*)n=1時(shí)，4a1=(a1+1)2?a22aI+1=0,即a=1n>2時(shí)，4an=4Sn4Sn1=(an+1)2(an-1+1)2222/18ana

5、ni+2an2an-1(n)由(I)知bn=an2=(2n1)2-Tn=(121)+(322)+(2n3)2n1+(2n1)2n則2Tn=(122)+(323)+(2n3)2n+(2n-1)2n+1兩式相減得：-Tn=(121)+(222)+(22n)-(2n-1)2n+12(1-2)cn+f=22-(2n-1)2n1=(32n)2n+161-Tn=(2n-3)2n+1+6(或Tn=(4n6)2n+6)3o7*5.已知數(shù)列an,其前n項(xiàng)和為Sn=2n2+2n(nN).(I)求a1,a2;(n)求數(shù)列an的通項(xiàng)公式，并證明數(shù)列an是等差數(shù)列；(出)如果數(shù)列bn滿足an=log2bn,請(qǐng)證明數(shù)列b

6、n是等比數(shù)列，并求其前n項(xiàng)和Tn.【答案】(I)a1=S1=5,a13一27-2+解得a2=8.(n)當(dāng)n>2時(shí)，an=SnSn137=2【n2(n1)2+沖(n1)=2(2n-1)+7=3n+2.又ai=5滿足an=3n+2,an=3n+2(nN*).,an一an-1=3n+23(n-1)+2=3(n"nCN*),，數(shù)歹Uan是以5為首項(xiàng)，3為公差的等差數(shù)列.(出)由已知得bn=2an(nN*),h-onn+1.1=2-=2an+1an=23=8(nN*),bn23/18又bi=2a1=32,，數(shù)列bn是以32為首項(xiàng)，8為公比的等比數(shù)列.丁32(18n)32n八Tn=一二8一

7、U(8t)2x_46.已知函數(shù)f(x)=,數(shù)列an?兩足：a1=3，an+1=f(an).x十231（i）求證：數(shù)列a7為等差數(shù)列，并求數(shù)列an的通項(xiàng)公式;(n)t己Sn=aia2+a2a3+anan+i,求證:8SnCZ.3=一+5,an+1an2an證明：(I)an+1=f(an)=a_1_2，一.1則短成等差數(shù)列,an=2n+1所以=+(n-1)R=3+(n1)>1=2n1ana12424=8T7一二二2n+12n+344(n)anan+1=c'22n+12n+3Sn=a1a2+a2a3+-+anan+1=835+5-尹+2-2T37.已知數(shù)列an的前三項(xiàng)依次為2,8,24

8、,且an2an-1是等比數(shù)列.(I)證明,是等差數(shù)列；(n)試求數(shù)列an的前n項(xiàng)和S的公式.【答案】(I).1a22a1=4,a32a2=8,an2an-1是以2為公比的等比數(shù)列.an2an-1=4浸"2=2n.等式兩邊同除以2n,得齊齊=1,an四是等差數(shù)列.(n)根據(jù)(I)可知a1埸+(n1)M=n,an=n2n.Sn=1X2+2X22+3X23+-+n2n,電2Sn=1>22+2X23+(n-1)2n+n2n+J一得：4/18-Sn=2+22+23+2nn2n+12(1-20)n+1n+1n+1-n2n1=2n12n2n1,1-2Sn=(n-1)2n+1+2.8.已知數(shù)列

9、an的各項(xiàng)為正數(shù)，前n項(xiàng)和為Si,且滿足：Sn=jan+(nCN).2an(I)證明：數(shù)列S2是等差數(shù)列；12121212,(n)設(shè)Tn=,s1+落2+2s3+2ns2，求Tn.【答案】1.1一*(I)證明：當(dāng)n=1時(shí)，a1=S1,又Sn=2an+an(nCN),一11一一-,S1=7Si+m,斛仔S1=1.2S1當(dāng)n>2時(shí)，an=SnSnT,.c1cc,1&=%St+q0nd，2SnSn1一1即Sn+SnT：,化間付SnSn-1=1,SnSn1S2是以S2=1為首項(xiàng)，1為公差的等差數(shù)列.(n)由(I)知島=n,12212Tn=2S2+咨+11212n11一2Ln2n+1=1一4

10、一n2n2n+1=1-？n+15.Tn+22一2n即Tn=11+2+(n-1)$1+nj.小1小11,11,得Tn=1加+1)27+n酒.1111一信Tn=2+級(jí)+2Kn22T19.數(shù)列an滿足a=1,an+14=1(nCN*),記Sn=a2+a2+a2.(I)證明：是等差數(shù)列；5/18(n)對(duì)任意的nCN*,如果S2n+1Snqmj恒成立，求正整數(shù)m的最小值.11,c11,.c1.C(I)證明：有一瞽=4?#就(ni)>4?ar4n3,1即-12是等差數(shù)列.an(n)令g(n)=S2n+1Sn=4n+14n+51卜+_!8n+1g(n+1)-g(n)<0,g(n)在nCN*上單調(diào)

11、遞減,g(n)max=g(1)=14.普老恒成立？m片4545303又mCN,.正整數(shù)m的最小值為10.11.10.已知數(shù)列an是首項(xiàng)a1=,公比為的等比數(shù)列，設(shè)bn+1510g3an=t,常數(shù)tCN.33(I)求證：bn為等差數(shù)歹U；(11)設(shè)數(shù)列5滿足Cn=anbn,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；若不存在，請(qǐng)說(shuō)明理由.【答案】(I)證明：an=3n,bn+1bn=15log3a=5,3an，bn是首項(xiàng)為b1=t+5,公差為5的等差數(shù)歹U.(n)cn=(5n+t)3-n,令5n+t=x,則Cn=x-n33n+1n+2Cn+1=(x+5)3-,C

12、n+2=(x+10)3,33若C2=Cn+1Cn+2,則(x3n)2=(x+5)3n(x+10)335化簡(jiǎn)得2x215x50=0,解得x=10或一5(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，a1=1,an+1=2an+2n+1.(I)設(shè)bn=an+1an+2,(nCN*),證明：數(shù)列bn是等比數(shù)歹U；(n)求數(shù)列an的通項(xiàng)an.6/18【答案】(I)由已知an+i=2an+2n+1得an+2=2an+1+2n+3，得an+2an+i=2an+i2an+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+i

13、+2=2(an+ian+2)由bn=an+1an+2,得bn+1=2bn,由ai=1,a2=2ai+3=5,得bi=a2ai+2=6,故數(shù)列bn是首項(xiàng)為6,公比為2的等比數(shù)列(11)由(1)知，bn=6>2n1=3>2n則an+ian=bn2=3>2n_2,所以an=ai+(a2ai)+(asa2)+(anan1)=1+(3>21-2)+(3>22-2)+-+(3>2n-1-2)=1+3(2+22+23+-+2n1)-2(n-1)an=32n2n3,當(dāng)n=1時(shí)，ai=3X21-2X1-3=6-5=1,故也滿足上式故數(shù)列an的通項(xiàng)為an=3>2n-2n-

14、3(nN*).12.在數(shù)列an中，ai=,an=、1+；4(nCN*且n>2)1(I)證明：an+或是等比數(shù)列；(n)求數(shù)列an的通項(xiàng)公式；1(ID)設(shè)$為數(shù)列an的前n項(xiàng)和，求證【答案】1111,_JLan+1+n+12an23n1+3產(chǎn)111(I)由已知，得an+印是等比數(shù)列.an+Tnan+TnoJ11111(n)設(shè)An=an+則Al=ai+1=+且Q=OOOdZN1則An=(jn,7/18an+zni=Tn,可得an=”-Tn322311、，11、，11、(m)Sn=(21-31)+(22-32)+/一我1-31111_123n2n12”+23K226n<213 .已知數(shù)列

15、an滿足a1=2,an+1=2ann+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)數(shù)列bn滿足：bn=cnc(neN*),求數(shù)列bn的前n項(xiàng)和Sn.2an2n(I)nE法*:由an+1=2ann+1可得an+1(n+1)=2(ann),又a1=2,則a11=1,，數(shù)列ann是以ai-1=1為首項(xiàng)，且公比為2的等比數(shù)歹U,則ann=1X2n一1，an=2n一1+n.、丁丹_.an+1(n+1)2ann+1(n+1)證法一：-=二annann2an2n=2,ann又a1=2,則a11=1,，數(shù)列ann是以ai-1=1為首項(xiàng)，且公比為2的等比數(shù)歹U,則ann=1

16、X2n一1，an=2n一1+n.n.nn(n).bn=c，-bn=cCcn'22an2n2an2n2Sn=b1+b2+bn=2+2(2)2+n(1)n.,%口=(2)2+2(；)3+(n1)(2)n+ng)1,1、n111cle11由一，佝Sn=2+(萬(wàn))2+(萬(wàn))3+(2)nn(3)n+1/7"J(n+2n+11n-Sn=2-(n+2)(2)n.14 .在數(shù)歹Uan中，a=1,2nan+1=(n+1)an,nCN*.8/18(I)設(shè)bn=7，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】bn+ian+in1(I)因?yàn)楣?bnn+1an21所以bn是首項(xiàng)為

17、1,公比為2的等比數(shù)列.(n)由(I)可知胃=27，即an=2,_2,3,4,n4=1+2+'22+23+24-1，一,一，1，1上式兩邊乘以2,得11,2,3,n-1,n2&=2+了+-2?十葉，11111n兩式相減，得2Sn=1+2+才一了,12+n25=2-2n)所以&=4記115 .設(shè)數(shù)列an的前n項(xiàng)和為Sn,且Sn=(1+»入n,其中入d1,0.(I)證明：數(shù)列an是等比數(shù)列；1(n)設(shè)數(shù)列an的公比q=f(?),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.【答案】(I)由Sn=(1+入)-入anSn-1=(1

18、+入)-入an1(n>2)相減得：an=一入anb入an-1,an=(n>2)an-11+八-1bn1+1,數(shù)列an是等比數(shù)列(n)f(MLl-bn=bn一?J'111+入1+bn1bn，：是首項(xiàng)為，=2,公差為1的等差數(shù)歹u；11;-=2+(n1)=n+1,.bn=bn''n+1.16 .在等差數(shù)列an中，a10=30,a20=50.9/18(I)求數(shù)列an的通項(xiàng)an；(n)令bn=2an-10,證明：數(shù)列bn為等比數(shù)列;(出)求數(shù)列nbn的前n項(xiàng)和Tn.【答案】(1)由2門(mén)=21+51)5ai0=30,a20=50,ai+9d=30得方程組,解得a1=i

19、2,d=2.ai+19d=50-an=12+(n-1)2=2n+10.一一一bn+14n+1(n)由()得bn=2an10=22n1010=22n=4n,=-n-=4bn4，bn是首項(xiàng)是4,公比q=4的等比數(shù)列.(出)由nbn=nW得：Tn=1>4+2X42+nX4n4Tn=1>42+(n1)>4n+nX4n+1相減可得:c+“4(14n)+“-3Tn=4+42+-+4n-nX4n1=4-nX4n1一3(3n1)Mn+410/18(ID)求數(shù)列an的前n項(xiàng)和Si.【答案】(I)-ai=3,an=2an1+n2(n>2,且nN),-82=2ai+22=6,as=2a2+3

20、2=13.an+n(2an-1-|-n-2)+n(n)證明：=丁丁=1：an-1+(n1)an1+n12ani+2n2_=-=2,ani+n1，數(shù)列an+n是首項(xiàng)為ai+1=4,公比為2的等比數(shù)列.an+n=42n-1=2n+1,即an=2n+l-n,an的通項(xiàng)公式為an=2n+l-n(nN*).(ID),an的通項(xiàng)公式為an=2n+l-n(nN*),.Sn=(22+23+24+2n+l)-(1+2+3+-+n)2?x(12")_nx(n+1)1-22_9n+2-+n+8=2-219 .已知數(shù)列an滿足ai=2,an+1=3an+2(nN).(I)求證:數(shù)列an+1是等比數(shù)列;(n)

21、求數(shù)列an的通項(xiàng)公式.【答案】(I)證明：由an+i=3an+2得an+1=3(an+1),7Han+i+1從而-=3,an+1即數(shù)列an+1是首項(xiàng)為3,公比為3的等比數(shù)列.(n)由(I)知，an+1=33n1=3n?an=3n-1-20 .已知數(shù)列an滿足ai=2,an+i=4an+2nS)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)列，并求數(shù)列an的通項(xiàng)公式;(n)設(shè)Tn=g,n=1,2,3,，證明：Ti<2.11/18【答案】(I)因?yàn)閎n+1=an+i+2n1=(4an+2n1)+2n1=4(an+2n)=4bn,且bi=ai+2=4,所以bn是以4為首項(xiàng)，以

22、q=4為公比的等比數(shù)列.所以bn=b1qn1=4n,所以an=4n2n.(n)Sn=a+a2+an=(4+42+4n)-(2+22+2n)=|(4n-1)-2(2n-1)=1(2n+1)2-32n+1+2=3(2n+1-1)(2n+1-2)=|(2n+1-1)(2n-1),2n3所以Tn=-=-XSn22n312n一1)(2n1)=2X2n112n+11，n3n11因此Ti=22門(mén)_1_2口+11i=12i=1223_J1322112n+1-1<221.已知數(shù)列an的前n項(xiàng)和為Sn,且S=4an3(nCN*).(I)證明：數(shù)列an是等比數(shù)列；(n)若數(shù)列bn滿足bn+1=an+bn(nC

23、N*),且b1=2,求數(shù)列bn的通項(xiàng)公式.【答案】(I)證明：由Sn=4an-3,n=1時(shí)，a1=4a13,解得a1=1.4n>2時(shí)，SiT=4anT3,所以當(dāng)n>2時(shí)，an=SnSn1=4an4an1,得an=qan1.3Q所以an是首項(xiàng)為1,公比為4的等比數(shù)歹U.3(n)因?yàn)閍n=3，由bn+1=an+bn(nCN),得bn+1bn=|.4n-1可得bn=b1+(b2b1)+(b3b2)+(bn-bn1)=2+34=33-1(n>2)1_13當(dāng)n=1時(shí)也滿足，所以數(shù)列bn的通項(xiàng)公式為bn=34-1.3222.在各項(xiàng)均為負(fù)數(shù)的數(shù)列an中，已知點(diǎn)(an,an+1)(nCN)在

24、函數(shù)y=3x的圖象上，且a2a527.(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S1.【答案】12/18(I)證明：因?yàn)辄c(diǎn)(an,an+i)(nCN*)在函數(shù)丫=2*的圖象上,3所以an+i=2an,即史上=2,故數(shù)列an是公比q=2的等比數(shù)歹U.3an33一.8.8一c2523因?yàn)閍2a5=,則aiqaiq4=,即a23=3,272733由于數(shù)列an的各項(xiàng)均為負(fù)數(shù)，則ai=-2 n一2(n)由(I)知，an=3,bn=所以s=3，|n-1+nj±n-9.3 23,所以an2n223n2+n,23.已知數(shù)列an的前n項(xiàng)和為Sn

25、,且Si=32ni_2,bn=an+i.(I)求數(shù)列an的通項(xiàng)公式；(n)證明：數(shù)列bn是等比數(shù)列，并求其前n項(xiàng)和Tn.【答案】(I).Sn=32nJ2,當(dāng)n>2時(shí)，an=Sn-Sni=32n1一23n2+2=32n2,當(dāng)n=1時(shí)，ai=i不滿足上式.i(n=i),an=o32n2(n>2).(n)bn=an+i=3Mni,bi=a2=3,nCN.皿=bni3X23X2nCN*,.數(shù)列bn是首項(xiàng)為3,公比為2的等比數(shù)歹U;由等比數(shù)列前n項(xiàng)和公式得Tn=3-ffnL=3或3.24.設(shè)數(shù)列an的前n項(xiàng)和為Sn,已知ai=5,an+i=S+3n(nCN*).(I)令bn=Sn-3n,求證

26、：bn是等比數(shù)列;(II)令Cn=：-,設(shè)Tn是數(shù)列Cn的前n項(xiàng)和，求滿足不等式Tn>2的n的最''log2bn+ilog2bn+2',4026小值.【答案】(I)證明：bi=Si3=2WQSn+iSn=Sn+3：即Sn+i=2Sn+3n,bn+i_Sn+i3n+1_203n+3nnn2AqbnSn3Sn3所以bn是等比數(shù)列.13/18則Cn=(n)由(I)知bn=2n,10g2bn+1log2bn+2(n+1)-(n+2)n+1n+2'11Tn=2-nZ2，112011Tn=2-n+2,4026'n>2011,即nmm=2012.25.已知

27、數(shù)列an滿足：a1=1,an+1=a+2(nCN).(I)求證：數(shù)列1是等比數(shù)列；(n)若為三=+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】(I)證明：=1+-,+1=2-+1,an+1anan+1an111 +1=2WO,所以數(shù)列1+1是等比數(shù)列.a1an1 ,1(n)an+1=2,而=2口_-b:+1=2n,bn+1=2n(n卜n入an八bn=2n1(n-1-?)(n>2)b1=入適合，所以bn=2n1(n-1-4(nCN*),由bn+1>bn得2n+1(n+1g2n(n5，Kn+2)<(n+2)min=3,，入的取值范圍為彳3.26.已知數(shù)列a

28、n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；(n)記bn=2(a：一立,數(shù)列bn的前n項(xiàng)和為Sn,求使0>2010的n的最小值.an【答案】(I)an+1=3an-2an1(n>2)(an+1一an)=2(anan1)(n>2.),a1=2,a2=4,a2a1=2wqanan-wq14/18(n)由(I)知bn=2an1)an=21Aan=2-X2n1故數(shù)列an+1an是首項(xiàng)為2,公比為2的等比數(shù)列,an+ian=(a2ai)2n1=2n,an=(anan-1)+(an-1an-2

29、)+(an-2an-3)+(a2ai)+ai=2n-2=2n(n>2)又a1=2滿足上式，an=2n(nCN).+2n、+2n飛+21+22(12L1.c.1,1,1-Sn=2n-1+”+22+m1=2n1孑2n21>"2+12n-1.由Sn>2010得：一一11.一2n2+2n1>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)和為滿足$+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=log2(an+2),求數(shù)列-的前n項(xiàng)和Tn.bn【答案】(

30、I)證明：由Sn+2n=2an,得Sn=2an-2n,當(dāng)nCN*時(shí)，Sn=2an-2n,當(dāng)n=1時(shí)，S1=2a1-2,貝Ua1=2,當(dāng)n>2時(shí)，Sn-1=2an-12(n1),-，得an=2an-2an1-2,即an=2an1+2,15/18-3n+2=2(an-1+2),an+2=2an1+2，an+2是以ai+2為首項(xiàng),以2為公比的等比數(shù)列.an+2=4*2n1,an=2n+12.(n)解：an=2n+12,bn=n(n+1),11_11bnnn+1nn+11Tn=+b11一+1b21'+一bn=1-1+1-1+1223nn+11=1一n+1=dn+1【解析】考點(diǎn)：數(shù)列的求和

31、；等比數(shù)列的通項(xiàng)公式.專題：綜合題.分析：(I)由Sn+2n=2an,得Sn=2an-2n,由此利用構(gòu)造法能夠證明數(shù)列an+2是等比數(shù)列，并求出數(shù)列an的通項(xiàng)公式an.11111(n)由an=2n+12,得,二=1,由此利用錯(cuò)位相減法能夠求出數(shù)列,bnnn+1nn+1bn的前n項(xiàng)和Tn.28.數(shù)列an中，a1=1,當(dāng)n>2時(shí)，其前n項(xiàng)的和&滿足Sn=an(Sn-1).1.(I)證明：數(shù)列是等差數(shù)列；Sn(n)設(shè)bn=log2；S-,數(shù)列bn的前n項(xiàng)和為T(mén)n,求滿足Tn>6的最小正整數(shù)n.16/18(I).Si=an(Sh-1),-S2=(Sn-Sni)(Sn-1)(n>：2)11.SnSn1=Sn1-Sn,即