2010年湖北黄冈中学高三数学《专题四 等差、等比数列的综合运用》

2010年湖北黄冈中学第一课时：等差数列、等比数列第一课时：等差数列、等比数列[课前导引]第一课时：等差数列、等比数列._______________,3432*,}{}{.1483759bbabbannTSNnTSnbannnnnn则都有：对任意的、项和分别为的前、若两个等差数列[课前导引][解析].41191121122221111111111666396369483759TSbbaababaabababbabba[解析].41191121122221111111111666396369483759TSbbaababaabababbabba[答案]4119.}{,12}{.2nnnnnnaanaSSna的通项求数列满足项和的前已知数列[解]242,422,22:,,1)1(2,1211111nnnnnnnnnnaaaaaanaSnaS即得两式相减.}{,12}{.2nnnnnnaanaSSna的通项求数列满足项和的前已知数列.)21(2,)21()21(212,21,21}2{.212,23,3,2122111111nnnnnnnnaaaaaaSaa从而为公比的等比数列为首项以是即：数列故而[链接高考][链接高考].}{}{)2()1().1(211).1(0521681}{4321111nnnnnnnnnnnSnbabbbbbnabnaaaaaa项和的前的通项公式及数列求数列的值；、、、求记且满足数列[例1].320,2013;421431,43;3821871,87;22111,1(1)44332211babababa故故故故[解析]).1(81625),2(2.2,32}34{:)34()34)(34(,)34()34(,)34(3832)34)(34((2)12231222231naaaaaqbbbbbbbnnnnnn故导致矛盾代入递推公式会否则将等比数列的公比是首项为故猜想034,3436162038212)34(2,361620343681634211341111bbaaabaaaaabnnnnnnnnnnn.,121:211),1(34231,23134,3234.2}34{22111nnnnnnnnnnnnnbababaSbbaabnbbbqb故得由故的等比数列确是公比为故).152(313521)21(31)(2121nnnbbbnnn.320,4,38,21,342,0364,052168,211:211(1)4321111111bbbbabbbbbbaaaabaabnnnnnnnnnnnnnn有由即整理得代入递推关系得由[法二]).1(34231,23134,2,32}34{,03234),34(234,342(2)111nbbqbbbbbbnnnnnnnnn即故的等比数列公比是首项为由).152(313521)21(31)(21,121211212211nnnbbbbababaSbbaabnnnnnnnnnnn故得由.(1)同解法一[法三]).1(81625,2,231,2,32}{,)34(3832,38,34,32(2)1112342312naaaabbqbbbbbbbbnnnnnnnnn故又的等比数列公比是首项为猜想;36810366368161222181625121121111nnnnnnnnnnnnaaaaaaaaaabb因此).(23616203681636243636816368162112111111212nnnnnnnnnnnnnnnnbbaaaaaaaaaaaabb).1(342312)22(312)222(31)()()(,231,2}{,0321211122111112nbbbbbbbbbbqbbbbnnnnnnnnnnnnnn从而的等比数列是公比).152(313521)21(31)(21,121211212211nnnbbbbababaSbbaabnnnnnnnnnnn故得由.}{)2()1().,4,3(21}{211nnnnnnSnnacnaaaaac项和的前求数列的值；求且满足的首项的等比数列若公比为[例2].}{)2()1().,4,3(21}{211nnnnnnSnnacnaaaaac项和的前求数列的值；求且满足的首项的等比数列若公比为[例2],212,,,3,(1)2212122nnnnnnnnacaaacaaacan时当由题设[解析].211.012,21,0222ccccccan或解得即因此由题设条件可得.211.012,21,0222ccccccan或解得即因此由题设条件可得.2)1(321}{,*).(1,}{,1)1((2)nnnSnnaNnaacnnnn项和的前数列这时即是一个常数列数列时当知要分两种情况讨论：由:,2111)21()21(3)21(21}{,*).()21(,21}{,21121得式两边同乘项和的前数列这时即的等比数列是一个公比为数列时当nnnnnnnSnnaNnaacnnnnnnnSnnS)21()21()21()21(1)211(:,212)21()21)(1()21(221211212得式式减去*).(]223)1(4[91)21(211)21(11NnnSnnnnnn[例3].}{,,,,,,.,0,}{2131412nnkkknkkaaaaaaaadan通项的求数列成等比数列已知数列的等比中项与是公差中在等差数列.}{,,,,,,.,0,}{2131412nnkkknkkaaaaaaaadan通项的求数列成等比数列已知数列的等比中项与是公差中在等差数列[例3])3()(.,)1(112141221daadaaaadnaan依题设得[解析],313,1,,,,,,3,1,0.,,,,,3,,,,0,:2121112qkkkddkdkdkddndaadddadnnnn公比为首项为等比数列也是数列由是等比数列由已知得得整理得.3}{),,3,2,1(39,3,9}{.9:11111nnnnnnnkknqkqkkk的通项即得数列公比的首项等比数列由此得.,,2,,,2}{)2()1(.,,,}{231并说明理由的大小与较比时当项和为其前等差数列为公差的为首项是以设的值；求成等差数列且的等比数列是公比为已知nnnnnbSnSnqbqaaaqa[例4].211.012,0,2,2(1)211121213或即由题设qqqaqaaqaaaa[解析].211.012,0,2,2(1)211121213或即由题设qqqaqaaqaaaa[解析],2312)1(2,1(2)2nnnnnSqn则若.49)21(2)1(2,21.,02)2)(1(2nnnnnSqbSnnnnn则若故1,2nnnSbSn时当.,11;,10;,92,,4)10)(1(,21nnnnnnnnnbSnbSnbSnNnnnSbSn时当时当时当故对于时当第二课时：等差、等比数列的综合运用[课前导引]第二课时：等差、等比数列的综合运用)(],1)32[()32(}{.111则下列叙述正确的是的通项已知数列nnnnaa413131,D.,C.,B.,A.aaaaaa最小项为最大项为最小项为最大项不存在最小项不存在最大项为最小项为最大项为第二课时：等差、等比数列的综合运用[课前导引].,3,212782194;278)32(,4;94)32(,3,0),1(],1,0(,,)32(,32,1,)32(11121最小时而时当时又当项为故最大则且则令nnnnnannnattattt[解析].lim,}{)2(;}{)1(.1)()(,1}{,5)(,13)(.221112nnnnnnnnnnnSSnaaaaagaafaacxxgbxxxf求项和为的前若的通项公式求满足正数数列是奇函数函数是偶已知数列,023,1)(51)(3)()(.5)(,0),()(,)(13)(,0),()(,)((1)212121212112nnnnnnnnnnnnnnaaaaaaaaaaaagaafxxgcxgxgxgxxfbxfxfxf即为奇函数即为偶函数[解析].)32(.32,1}{,32,023,0,0)23)((1111111nnnnnnnnnnnnnaaaaaaaaaaaaa的等比数列公比为为首项是以即.)32(.32,1}{,32,023,0,0)23)((1111111nnnnnnnnnnnnnaaaaaaaaaaaaa的等比数列公比为为首项是以即.33211lim)2(nnS[链接高考][链接高考].1111)2(;}{)1(.9,3,*))}(1({log12312312nnnnaaaaaaaaaNna证明的通项公式求数列且为等比数列已知数列[例1].12,)1(1)1(log.1,8log2log)2(log2:9,3.)}1({log1)(2222312nnnnannaddaada即即得由的公差为设等差数列[解析].121121121212121212121111,212211)2(3211231211nnnnnnnnnnaaaaaaaa.15)3(;}{)2(;)1(.,,3,2,1,)25()85(,11,6,1,}{1321都成立、对任何正整数证明不等式为等差数列证明数列的值和求为常数、其中且已知项和为的前设数列nmaaaaBABAnBAnSnSnaaaSnanmmnnnnnn[例2].8,20:48228,212273)25()85(,18,7,11)(23121321321211BABABABASSBASSBAnSnSnaaaSaaSaSnn解得即知由由已知得[解析]320)25()110()35(:1222820)75()35(1820)25()85()1()2(12121nnnnnnnSnSnSnnSnSnnSnSn得得：由0)25()410()25(,0)25()615()615()25(:34420)75(2)910()25(1