2、數(shù)列J｝的前n項(xiàng)宛S,滿足n^NAnn1'+且&、a?5>a3成等差數(shù)列.(I)求a的值；1(U)求數(shù)列Jn〉的通項(xiàng)公式；(3)(附加)明：證3a1n」+—+(U+—<—aa24、設(shè)數(shù)列』也前n項(xiàng)S,已a(bǔ)i==+=(Na28,Sn14Sn+i5Sirn2,和為n{知}n2,T是數(shù)列Iog{2aj的前n項(xiàng)和.n(1)求數(shù)列a的通項(xiàng)公式;n(2)求T；(3)(附加)求滿足c一2高三第二輪復(fù)習(xí)1、設(shè)｛a｝是公比大于+1的等比數(shù)列,+3,(2)3^2，83+Ina,‘3n+111丄、T丿1010的最大正整數(shù)n的值.2013數(shù)列專題201
3、3-3-26S為數(shù)列｛aj的前n項(xiàng)和.己知S37,4構(gòu)成等差數(shù)列.(1)求數(shù)列｛an｝的通項(xiàng)公式;-III,求數(shù)列｛bn｝的前n項(xiàng)和T?nn1,2,+=-I2^解：(1)由已知得：1(3)a1"3~4)3a?2解得a2=2.2設(shè)數(shù)列｛a｝的公比為q,由a?2,可得ia?aq?2+q=可知2q7,坯解得q12,q21_?由題意得q22.｝的通項(xiàng)為+n（2）由于bn1,2,3n9+3n+IH+8312nIn2In2又bn1bn3ln2｛b｝是等差數(shù)列.n2n(3ln23ln2)2+In2.3n(n23n(n1)ln222、設(shè)數(shù)列｛&
4、｝為前n項(xiàng)狗S,數(shù)列｛bn｝滿足：bn=nan,且數(shù)列｛bn｝的前*(neN)?(1)求ai,a2的值；⑵求證：數(shù)列｛So+2｝是等比數(shù)列；⑶求數(shù)列解：(1)由題意得：ai+2a2+3as+…+nan=(rM)Sn+2n；當(dāng)時(shí)，則有ai=(1-1)S+2,解得：ai=2；當(dāng)n二2時(shí)，則有ai+2a2=(2-1)S+4,即2+2a2=(2+a2)+4,解得：(2)由ai+2a2+3a3+--+nan=(n-1)S+2n,①得ai+2a2+3a3+??-+nan+(n+1)an+i=nSn+i+2(n+1),②｛｝n項(xiàng)稲(n-1)S
5、+2na的通項(xiàng)公式na2=4.②■①得:(n+1)an+i=nSn+i-(n-1)S+2,(4分)即(n+1)(S)+i-Sn)=nSn+i-(n-1)S+2,得S+i=2S+2；/.S+i+2=2(S+2),S++2―n-1=由S+2二ai+2=4H0知+2S2數(shù)列｛Sn+2｝是以4為首項(xiàng)，2為公比的等比數(shù)列。(1)方法壬=_=+___=當(dāng)n2時(shí),1(202)(2n2)也滿足上式，2+n2.+=+法2：由②x①得:(n=1)an+—1nSn1(n1)Sn22,⑤-④得:+2?2,a1nai￡?2n2a2S24,2a「++_二數(shù)
6、列｛a］是以@2為首項(xiàng),2為公比的等比數(shù)列3、＜2012年高考(廣東理))設(shè)數(shù)列a的前n項(xiàng)和為S,滿足2s｛｝n幾a、a25、^3成等差數(shù)列.1(I)求a的值;1+一+((
7、+—<-(U)求數(shù)列a'的通項(xiàng)公式;n:(+)=(3)證明:1au1+.3212n=+一++=—+2aa312解析:(I)由2aaa7,解得ai1.1232a25a1a3(U)山2snn1na24可得2Sa21(n2),兩式相減，n1n1n可得2a=a+_a一即a+=3a^2,nn1nnn即aV2n+=3(a+2°),(宀nn由羽=a-3可得，a2=5,所以
8、a2+2"=3(ai+2)aya++1nan所以數(shù)列+a2是一全以J3為首珈,3為公比的等比數(shù)列——<(III)因?yàn)閚_i
9、
10、(+23<3、)231a<-inn所以a2(71^2),n于是14、設(shè)數(shù)列an的前n{}+2-項(xiàng)宛已知822,,Sn14Sn15SnH是數(shù)列Ipg2an白術(shù)nI-一IiI*111?I八丿I求數(shù)列a的通項(xiàng)公式；（2）求T；(3)求滿足1010u2的最大正整數(shù)門(mén)的值(1)解：???當(dāng)n+=2時(shí),T3sn4STn5S,n20131=?n丁ai分8,824ai3???數(shù)列是以&2為首項(xiàng)，公比為4的等比數(shù)列.2n1
11、(2)解:由（1）得:log2=log25分2n_12=2一1,(3)解:log+anlog2+(III1-K1八12n?川-II1III?IIPloga/2n)=III?22122232?3.2412.4?2n12■III(n-)(?III-2n[[分令n12
