等差、等比数列的综合运用第一课时: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*,}{ }{ .1 483759bbabbannTSNnTSnbannnnnn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 ].4119112112222 1111111111666396369483759TSbbaababaabababbabba[ 解析 ].4119112112222 1111111111666396369483759TSbbaababaabababbabba[ 答案 ] 4119.}{,12}{ .2 nnnnnnaanaSSna的通项求数列满足项和的前已知数列[ 解 ]242,422,22:,,1)1(2,12 11111nnnnnnnnnnaaaaaanaSnaS即得两式相减.}{,12}{ .2 nnnnnnaanaSSna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) 44332211babababa故故故故[ 解析 ]).1(81625 ),2( 2.2,32}34{:)34()34)(34(,)34()34(,)34(3832)34)(34( (2) 12231222231naaaaaqbbbbbbbnnnnnn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 ,34 36162038212)34(2,3616203436816 34211341111bbaaabaaaaabnnnnnnnnnnn.,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)(2121nnnbbbnnn.320,4,38,21,342,0364,052168,211:211 (1) 4321111111bbbbabbbbbbaaaabaabnnnnnnnnnnnnnn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)111nbbqbbbbbbnnnnn...