高三数学《专题四等差、等比数列的综合运用》VIP免费

等差、等比数列的综合运用石港中学赵建兵第一课时：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:,211...