高三數(shù)學(xué)復(fù)習(xí)知識(shí)點(diǎn)-數(shù)列數(shù)學(xué)薃薃知薃點(diǎn)數(shù)列-一、列念數(shù)概數(shù)數(shù)稱數(shù)數(shù)個(gè)數(shù)稱數(shù)列的定薃,按照一定次序排列的一列薃列～列中的每薃薃薃列1.的薃.理解列定薃的四要點(diǎn),數(shù)個(gè)?列中的是按一定“次序”排列的～在薃里～只強(qiáng)薃有“次序”～而不強(qiáng)薃有數(shù)數(shù)“薃律”,因此～如果薃成列的相同而次序不同～那薃薃就是不同的列,兩個(gè)數(shù)數(shù)它數(shù)?在列中同一可以重薃出薃,數(shù)個(gè)數(shù)?薃與數(shù)薃薃是根本不同的念,兩個(gè)概ann?列可以看作一定薃域薃正整集數(shù)個(gè)數(shù)或的有限子集它的函自薃量小到大數(shù)當(dāng)從()依次取薃薃薃薃的一列函薃～但函不一定是列,數(shù)數(shù)數(shù)通薃公式,如果列薃數(shù)的第薃序之薃可以用一式子表示與號(hào)個(gè)那薃薃公式叫個(gè)2.an,n,做薃個(gè)數(shù)即列的通薃公式～an,f(n).數(shù)列的前薃和通薃的公式與?～3.nSn,a1,a2,,,an?an,,,S1(n,1).薃Sn,Sn,1(n,2)?已知列的前數(shù)薃和公式～求的方法nSnan第一步,求～a1=S1第二步,當(dāng)薃～求～n?2an=Sn-Sn-1第三步,薃薃當(dāng)是否適合當(dāng)薃得到的～若適合薃將用一式子表示～個(gè)a1n?2anan若不適合～將分段表示出來an?已知與的薃系式～求方法anSnan根據(jù)已薃出薃薃系式～令;或,～出一寫個(gè);或,與n=n+1n=n-1an+1an-1Sn+1;或,～Sn-1的薃系式～然后式相。消去將兩減得到與;或與,的薃系～從Snanan+1anan-1而定確是{an}等差列薃是等比列或者其列～然和求出其通薃公式。數(shù)數(shù)它數(shù)例,薃列數(shù)的前薃和～求答案,1{an}nSn=nan.(2an,2n-1)例,已知列數(shù)的通薃與前薃和之薃薃足薃系式求2{an}annSnSn=2-3anan答案,(a,n13,()2n-1)4薃推公式,如果已知列薃數(shù)的第一薃;或前薃,～且任何一薃幾與它的前一4.an,an薃那薃薃式子叫做列薃個(gè)數(shù)的薃推公式如列薃數(shù)中～～其中an,.an,a1,1,an,2an,1;或前薃,薃的薃系可以用一式子表示～幾個(gè)來即或an,1an,f(an,1)～是列薃數(shù)的薃推公式an,f(an,1,an,2)an,2an,1an,.利用列的薃推公式求列的通薃公式一般有以下三薃方法數(shù)數(shù)?累加法,如果已知列數(shù)的相薃薃兩與的差的一薃系式～我薃可依次個(gè){an}an+1an寫出前薃中所有相薃薃的差的薃系式～然后把薃兩個(gè)數(shù)式子相加～整理求出列nn-1的通薃公式例,已知列數(shù)薃足～,～求且求的最小薃.{an}a1=33an+1an=2nanana;答案,nnn,,min212?累薃法,如果已知列數(shù)的相薃薃兩與的商的一薃系式～我薃可以依次個(gè){an}an+1an寫出前薃中所有相薃薃的商的薃系式～然后把薃兩個(gè)數(shù)式子相乘～整理求出列nn-1的通薃公式.薃薃薃薃例,在列數(shù)中且有;～,～;～,～{an},an>0,a1=1a=n+1anb=nan+1a與共薃～求列的通薃公式;答案,數(shù),ban,1n?造法,根據(jù)所薃的列的薃推公式以及其有薃薃系式～薃行薃形整理～造出構(gòu)數(shù)它構(gòu)一新的等差或等比列～再利用等差或等比列的相薃知薃點(diǎn)求解個(gè)數(shù)數(shù)如,形如～an+1=pan+q可以令;,薃;,an+1+,=pan+,an+1=pan+p-1*,薃;,p-1*,=q薃薃=qp-1從而得到;,薃薃是等比列數(shù)an+1+,qqq,=pan+an,,p-1p-1p-1,,例,列的前數(shù)薃和薃～已知～n,a1=1?～求;答案,～,an,1,n,2n-2,(n,1),nN+anSnan2n數(shù)列的表示方法,解析法、薃像法、列薃法、薃推法5..數(shù)數(shù)數(shù)數(shù)減數(shù)數(shù)數(shù)數(shù)列的分薃,有薃列～無薃列～薃增列～薃列～薃薃列～常列～有6.界列～無界列數(shù)數(shù).?薃增列數(shù)薃于任何均有:n,N,,an,1,an.?薃列減數(shù)薃于任何均有:n,N,,an,1,an.?薃薃列數(shù)例如::,1,1,,1,1,,1,,.?常列數(shù)數(shù)例如::6,6,6,6,??.?有界列數(shù)存在正數(shù)使:Man,M,n,N,.?無界列數(shù)薃于任何正數(shù)薃有薃使得:M,anan,M.二、等差列數(shù)等差列的念數(shù)概1.如果一列第二薃起～每一薃前一薃的差等于同一常個(gè)數(shù)從與它個(gè)數(shù)～薃列叫個(gè)數(shù)d做等差列～常數(shù)數(shù)d稱數(shù)薃等差列的公差.通薃公式前與薃和公式2.n?通薃公式;,～薃首薃～薃公差an,a1,(n,1)dn?1a1d.?前薃和公式nSn,或或函的角度～薃于數(shù)的一元二次22n(a1,an)1Sn,na1,n(n,1)d.22s,dn2,(a-d)n(n函數(shù))n1等差中薃3.如果成等差列～那薃數(shù)叫做與的等差中薃即,是與的等差中薃a,A,bAab.Aab薃～～成等差列數(shù)2A,a,b,aAb.等差列的判定方法數(shù)4.?定薃法,若當(dāng)～?薃～有;薃常,或數(shù)當(dāng)～?薃～有n?2nN*an,an-1,ddn?1nN*;薃常,～薃列數(shù)數(shù)an,1,an,dd?等差中薃法,若列薃數(shù)薃足～薃列薃數(shù)是等差列數(shù)數(shù),～an,2an,1,an,an,2(n,N,)an,.薃列薃數(shù)是等差列。數(shù)an,等差列的常用性薃數(shù)5.?或者?函法,若薃于列薃數(shù)數(shù)有;、薃常,或數(shù)an,am,(n,m)dd,an,an,kn,bkb者有;、薃常是等差列。數(shù)Sn=An+BnAB2,an,-nmn-m?若～薃m,n,p,q(m,n,p,q,N,)am,an,ap,aq?若列薃數(shù);,～薃是等差列～薃列薃數(shù)數(shù)、薃、薃;、an,n,mb,an,p,pan,pan,qbn,p薃常、qn數(shù)數(shù)數(shù),等列都是等差列～?若等差列薃數(shù)的前薃和～薃薃an,nSn成的列也是等差列～公差薃數(shù)數(shù)～薃是等差列～數(shù)、、nk2Sn,,SkS2k-SkS3k-S2k構(gòu)薃......n,偶San,1～薃奇San偶Sn,1當(dāng)數(shù)薃薃薃～薃奇,偶薃2n,1(n,N,)SSan,,奇Sn?薃薃薃當(dāng)數(shù)～薃偶,奇薃2n(n,N,)SSnd,?在等差列數(shù)中～～,～薃{an}Sn=aSm=b(nm)Sm,n=.n,m,,(ab)n,m在等差列,數(shù),中有薃的最薃薃薃,當(dāng)薃～薃足薃6.an,Sn(1)a1>0,d<0薃am,0的薃數(shù)薃am,1,0薃am,0使得取最大薃當(dāng)薃～薃足薃的薃數(shù)使得取最小薃。在解msm.(2)a1<0,d>0msm含a,0,m,1薃薃薃的列最薃薃薃薃數(shù)注意薃化思想的薃用。;,利用～薃薃行配方～,3Sns,dn2,(a-d)nn122求取最薃薃的薃Snn例薃,例、在列薃數(shù)中1an,an,4n-5,2～～其中、薃a1,a2,,,,,an,an,bn,cn,N,ab2常數(shù)薃。;答案,,.ab+c=-1例、等差列薃數(shù)中～是其前薃和～2an,Snna,-11,10-81018;答案,,薃～求,-112S11,薃an,,,例、已知列薃數(shù)薃足薃推公式～且薃薃薃等差列～數(shù)3an,an,1,2an,2-1(n,N)薃2,n,薃薃的薃。;答案,,-1例、已知列薃數(shù)的各薃均薃正～前數(shù)薃和薃～且薃足;,求薃薃薃等3an,nSn2Sn,1an,差列;數(shù),求薃的通薃公式;答案,2an,a2n,n-4an薃,n,2例、已知等差列薃數(shù)薃足4an,;,求和1anSn;,令～薃前薃和薃2a,7,a,a357,26an,nSnbn,1a2求列薃數(shù)的前薃和n-1(n,N),bn,nTn,;答案,;,;,～,12Tn,an,2n,1,Sn,n(n,2)n4(n,1)三、等比列數(shù)等比列的念數(shù)概1.如果一列第二薃起～每一薃前一薃的比等于同一常個(gè)數(shù)從與它個(gè)數(shù)～薃個(gè)數(shù)q(q,0)列叫做等比數(shù)列～常數(shù)稱數(shù)薃等比列的公比q.通薃公式前與薃和公式2.n?通薃公式,～薃首薃～薃公比an,a1qn,1a1q.?前薃和公式,?當(dāng)薃～nq,1Sn,na1?當(dāng)薃～a1(1,qn)a1,anqq,1Sn,.,1,q1,q等比中薃3.如果成等比列～那薃數(shù)叫做與的等比中薃即,是與的等差中薃a,G,bGab.Gab薃～～成等差列薃數(shù)aAbG2,a,b.等比列的判定方法數(shù)4.?定薃法,若當(dāng)～?薃～有～薃常數(shù)或當(dāng)～?薃～有n?2nN*an,q(q,0q)n?1nN*an-1～薃常數(shù)～薃列薃數(shù)是等比列。數(shù)an,1,q(q,0q)an,an?等比中薃法,且是等比列數(shù)an,1,an,an,2(n,N,)an,0,,an,.2?函法,如列薃數(shù)數(shù)通薃公式其函特征薃常指函的乘薃數(shù)數(shù)與數(shù)數(shù)～an,an=f(n)=cq,前n薃和公式其特征是的系常薃的相反數(shù)與數(shù)數(shù)nSn=g(n)=k(1-q)(q,1),q.等比列的常用性薃數(shù)5.?公比的等比列數(shù)an,am,q?若列薃數(shù)an,n,mnn(n,m,N,或者)qn-m薃anam2?如果是等比列～公比薃數(shù)～那薃～～～～～?～是以?薃{an}qa1a3a5a2n,1q12,an,薃薃薃薃是等比列～薃列、薃薃數(shù)數(shù);、、、bpaa,bn,,,,p,,nnann薃薃薃an,薃bn,薃常,等也是等比列～數(shù)數(shù)?若～薃～m,n,p,q(m,n,p,q,N,)am,an,ap,aq?若等比列薃數(shù)的前薃和～薃、、、是等比列數(shù)an,nSnSkS2k,SkS3k,S2kS4k,S3k.?若是以薃公比的等比列～薃有數(shù)，例薃,SnqSn,m=SmqSn例、若等比列薃數(shù)的前薃和薃～且1an,nSn例、已知～～成等比列～數(shù)～～和～～分薃成等差列～薃兩個(gè)數(shù)2abcambbnc例、在等比列薃數(shù)中～3an,例、已知列薃數(shù)的首薃～前薃和薃～且;,薃明,列薃數(shù)是等比4an,a1,5nSn1an,1,數(shù)列;,求薃的通薃公式以及;答案,2an,Sns3薃求;答案,,2,s6,18,10,?335ac薃,.2mn～薃;答案,,an,2,am,2(m,n)am,n,.1Sn,1,2Sn,n,5,n,Nn,mn,an,6,2-1,Sn,6,2-n-6n-1例、已知等差列薃數(shù)的前薃和薃～前薃和薃5an,368-4;,求列薃數(shù)的通薃公式1an,;,薃～求列薃數(shù)前薃和,2bn,(4-an)q(q,0,n,N)bn,nn-1;答案,;,～;,n(n,1),(q,1),,n,12n1an,4-n2Sn,,n-(n,1)q,1q,(q,1)2,(q-1),四、列求和數(shù)數(shù)列的求和,熟薃掌握等差列等比列的求和公式～數(shù)與數(shù)1,能用倒序相加、薃位相、薃相消等重要的方法薃行求和算～運(yùn)減拆數(shù)學(xué)運(yùn)2,熟薃一些常用的列的和的公式,數(shù)3,特殊列求和的方法,數(shù)4一、利用常用求和公式求和;,等差列的求和公式,數(shù)1Sn,n(a1,an)n(n,1),na1,d22薃;,等比列的求和公式數(shù);切薃,公比含字母薃一定要na1(q,1),n2Sn,,a1(1,q)薃薃,(q,1),,1,q;,薃3k2,12,22,32,,,n2,k,1nn(n,1)(2n,1)6薃n(n,1),k,1,2,3,,,n,,,,,2,k,133333n2例已知～求的前薃和[1]log3x,,123nx,x,x,,,,,x,,,,n.log23解,由log3x,,11,log3x,,log32,x,log232,,由等比列求和公式得,數(shù),n11(1,n)x(1,x)11Sn,x,x2,x3,,,,,xn=12n1,x1,2例薃,～?求的最大薃解,由等差列數(shù)[2]Sn1+2+3+?+nnN,f(n),*Sn.(n,32)Sn,1求和公式得Sn,?～,,,?當(dāng)f(n),11n(n,1)Sn,(n,1)(n,2)2211n1Sn,2～即,薃～例已知是64850n,34n,64(n,32)Sn,1n,34,(n,)2,50nnn,18n8f(n)max,50[3]{an}各薃均薃正的等比例列～且數(shù)數(shù)a1,a2,2(～11111,)a3,a4,a5,64(,,)a1a2a3a4a5.求列數(shù)的前薃和?求的通薃公式～;?,薃12),{bn}nTn.an(){an}bn,(an,解,;?,薃公比薃～薃由已知有qan,a1qn,1.薃薃11,a,aq,2,,11,,,aaq,,11,,,aq2,aq3,aq4,64,1,1,1,.,21134,,1aqaqaq,111,,～化薃得薃2,,a1q,226,,a1q,64.又～故～所以,a1,0q,2,a1,1an,2n,1薃1,112n,1;?,由;?,知bn,,an,,,an,2,2,4,n,1,2an,an4,21n1,4,11,1n,1Tn,,1,4,...,4,,,1,,...,n,1,,2n,,,2n,,4n,41,n,,2n,14,4,11,3,44n1,二、薃位相法求和減薃薃方法是在推薃等比列的前數(shù)薃和公式薃所用的方法～薃薃方法主要用于求列數(shù)n的前薃和～其中、分薃是等差列和等比列數(shù)數(shù)例求和,{an?bn}n{an}{bn}.[4]Sn,1,3x,5x2,7x3,,,,,(2n,1)xn,1解,由薃可知～的通薃是等差列數(shù),的通薃等比列與數(shù){(2n,1)xn,1}{2n1}{xn,1的}通薃之薃,薃?;薃制薃位,xSn,1x,3x2,5x3,7x4,,,,,(2n,1)xn?又?Sn,1,3x,5x2,7x3,,,,,(2n,1)xn,1????????;薃位相,得,減-(1,x)Sn,1,2x,2x2,2x3,2x4,,,,,2xn,1,(2n,1)xn1,xn,1,。再利用等比列的求和公式得,數(shù)(2n,1)xn(1,x)Sn,1,2x,1,x(2n,1)xn,1,(2n,1)xn,(1,x)?Sn,2(1,x)2462n前薃的和,2,3,,,,,n,,,,n22222n1解,由薃可知～的通薃是等差列數(shù)的通薃等比列與數(shù)的通薃之薃.{n}{2n}{n}222462n薃?Sn,,2,3,,,,,n???????????“?222212462n?Sn,2,3,4,,,,,n,1????2222212n1222222n?,?得(1,)Sn,,2,3,4,,,,,n,n,1,2,n,1,n,1222222222n,2?Sn,4,n,12例求列三、倒序相加法求和數(shù)[5]薃是推薃等差列的前數(shù)薃和公式薃所用的方法～就是一列倒薃排列;反將個(gè)數(shù)來n序,～再把原列相加～就可以得到它與數(shù)個(gè)n(a1,an).例求的薃[6]sin1,sin2,sin3,,,,,sin88,sin89解,薃?S,sin1,sin2,sin3,,,,,sin88,sin89????將?式右薃反序得,?S,sin89,sin88,,,,,sin3,sin2,sin1??2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,又因薃??,sinx,cos(90,,x),sin2x,cos2x,1+,2S,(sin21,,cos21,),(sin22,,cos22,),,,,,(sin289,,cos289,)89?,S44.5四、分薃法求和有一薃列～不是等差列～也不是等比列～若薃薃列適薃～可分薃數(shù)既數(shù)數(shù)將數(shù)當(dāng)拆幾個(gè)等差、等比或常薃的列～然后分薃求和～再其合可數(shù)將并即.例求列的前數(shù)薃和,～?[7]n1,1,111,4,2,7,,,,,n,1,3n,2aaa解,薃111Sn,(1,1),(,4),(2,7),,,,,(n,1,3n,2)aaa將拆其每一薃薃再重新薃合得;分薃,111,2,,,,,n,1),(1,4,7,,,,,3n,2)aaa當(dāng),薃～,;分薃求和,(3n,1)n(3n,1)na1Sn,n,22當(dāng)薃～,,11,n(3n,1)na,a1,n(3n,1)n,a,1Sn,a,1221,aSn,(1,例求列數(shù)的前薃和[8]{n(n+1)(2n+1)}n.解,薃ak,k(k,1)(2k,1),2k3,3k2,k?,薃Sn,,k(k,1)(2k,1)(2kk,1k,1將拆其每一薃薃再重新薃合得,n2nnn3,3k2,k),,Sn2,k,1nk,3,k,,k2(13,23,,,,,n3),3(12,22,,,,,n2),(1,2,,,,,n)3k,1k,1,n2(n,1)2n(n,1)(2n,1)n(n,1)n(n,1)2(n,2),,=2222五、裂薃法求和薃是分解薃合思想在列求和中的具薃用與數(shù)體裂薃法的薃薃是列中的每薃;通薃,將數(shù).分解～然后重新薃合～使之能消去一些薃～最薃到求和的目的達(dá)通薃分解;裂薃,.如,sin1,薃薃;,;,薃1an,f(n,1),f(n)2tan(n,1),tann,,cosncos(n,1);,;,111(2n)2111,,3an,4an,,1,(,)n(n,1)nn,1(2n,1)(2n,1)22n,12n,1;,5an,1111,[,]n(n,1)(n,2)2n(n,1)(n,1)(n,2)薃an,n,212(n,1),n1111,n,,n,,,S,1,nn,1nnn(n,1)2n(n,1)2n,2(n,1)2(n,1)211,2,12,1n,n,1,1的前薃和例求列解,薃數(shù)～薃2,,,,,,1n,n,1,,,,n.[9]an,,n,1,n1n,n,1Sn,1,,1,2,,,,,(2,),(3,2),,,,,(n,1,n)n,1,1例在列數(shù)中～[10]{an}an,前薃的和～又～求列數(shù)的n.12n2,,,,,,bn,{bn}n,1n,1n,1an,an,112nn,,,,,,,n,1n,1n,12?211bn,,8(,)nn,1,22?列數(shù)的前薃和,1111111)]{bn}nSn,8[(1,),(,),(,),,,,,(,22334nn,1,1)8(1,n,1,解,?8nn,1an,111cos1,薃薃薃薃薃,,,例求薃,[11]cos0,cos1,cos1,cos2,cos88,cos89,sin21,解,薃S,111,,,,,,cos0,cos1,cos1,cos2,cos88,cos89,sin1,薃薃?薃tan(n,1),tann,,cosncos(n,1)?S,111,,,,,,cos0,cos1,cos1,cos2,cos88,cos89,,薃1{(tan1,,tan0,),(tan2,,tan1,),(tan3,,tan2,),[tan89,,tan88,]}sin111cos1,薃薃薃,薃,?原等式成立,(tan89,tan0)cot12,sin1,sin1,sin1薃算,六、合法求和并薃薃一些特殊的列～某些薃合在一起就具有某薃特殊的性薃～因此～在求列數(shù)將并數(shù)的和薃～可薃些薃放在一起先求和～然后再求將Sn.例求的薃[12]cos1?+cos2?+cos3?+???+cos178?+cos179?.解,薃,Sncos1?+cos2?+cos3?+???+cos178?+cos179??;特殊性薃薃,找cosn,,,cos(180,,n,)?,;,;,;,Sncos1?+cos179?+cos2?+cos178?+cos3?+cos177?;,,;合求和,并+???+cos89?+cos91?+cos90?0例數(shù)列,～求[13]{an}a1,1,a2,3,a3,2,an,2,an,1,anS2002.解,薃,S2002a1,a2,a3,,,,,a2002由可得a1,1,a2,3,a3,2,an,2,an,1,ana4,,1,a5,,3,a6,,2,a7,1,a8,3,a9,2,a10,,1,a11,,3,a12,,2,??a6k,1,1,a6k,2,3,a6k,3,2,a6k,4,,1,a6k,5,,3,a6k,6,,2?a6k,1,a6k,2,a6k,3,a6k,4,a6k,5,a6k,6,0?,S2002a1,a2,a3,,,,,a2002=(a1,a2,a3,,,,a6),(a7,a8,,,,a12),,,,,(a6k,1,a6k,2,,,,,a6k,6),,,,,(a1993,a1994,,,,,a,1998),a1999,a2000,a2001,a2002=a1999,a2000,a2001,a2002,a6k,1,a6k,2,a6k,3,a6k,45例在各薃均薃正的等比列中～若數(shù)數(shù)求的[14]a5a6,9,log3a1,log3a2,,,,,log3a10薃。