数列经典综合题66例(9)

an?1?an?1 ? an?1?an?d?1? an?a1??n?1??1?n?5直线l:y?2x?1, ? bn?2n?1??n?5, ?n为奇数?（Ⅱ）f?n??? 2n?1, n为偶数????

当k为偶数时，k?27为奇数， ? f?k?27??4f?k?? k?27?5?4?2k?1?, ? k?4当k为奇数时，k?27为偶数，? 2?k?27??1?4?k?5?, ? k?综上，存在唯一的k?4符合条件。（Ⅲ）由

35?舍去?2an?1?1??1??1??1???1????1???b1??b2??bn??ann?2?an?0即a??1??1??1?1?1??1???????2n?3?b1??b2??bn?1?1??1??1??1???1????1??2n?3?b1??b2??bn?1?1??1??1??1?1?1??1?1?????????2n?5?b1??b2??bn??bn?1?12n?3?1?2n?32n?42n?4??1?????2n?5?bn?1?2n?52n?32n?5?2n?3 ?1记f?n??? f?n?1??? ?f?n?1?f?n?2?4n2?16n?164n?16n?15? f?n?1??f?n?, 即f?n?递增，? f?n?min?f?1??? 0?a?4515n?1445?,5315?例41 已知等比数列?an?的前n项和为Sn?2?3?k(k?R,n?N) （Ⅰ）求数列?an?的通项公式； （Ⅱ）设数列?bn?满足an?4(5?k)并证明你的结论．

解：（Ⅰ）由Sn?2?3n?k(k?R,n?N?)得:n?2时，an?Sn?Sn?1?4?3n?1anbn，Tn为数列?bn? 的前n项和，试比较3?16Tn 与 4(n?1)bn?1的大小，

??an?是等比数列，?a1?S1?6?k?4?k??2，得 an?4?3n?1(n?N?)

（Ⅱ）由an?4(5?k)anbn和an?4?3n?1得bn?n?1

4?3n?1?Tn?b1?b2?b3??bn?1?bn?12n?2n?1??????(1)2n?2n?14?34?34?34?3

123n?2n?13Tn??????????(2)2n?3n?244?34?34?34?311111n?1??????? 44?34?324?3n?34?3n?24?3n?111111n?132n?1?Tn????????????10分

88?38?328?3n?38?3n?28?3n?11616?3n?1n(n?1)2n?1n(n?1)?3(2n?1)4(n?1)bn?1?(3?16Tn)??n?1? nn333?(2)?(1):2Tn??n(n?1)?3(2n?1)?n2?5n?3?当n?5?375?37?0时 或n?22?有n(n?1)?3(2n?1)，所以当n?5(n?N)时有3?16Tn?4(n?1)bn?1那么同理可得：

当

5?375?37??n?时有n(n?1)?3(2n?1)，所以当1?n?5(n?N)时有3?16Tn?4(n?1)bn?1 22??综上：当n?5(n?N)时有3?16Tn?4(n?1)bn?1；当1?n?5(n?N)时有3?16Tn?4(n?1)bn?1 例42 已知数列?an?中，a1?3，a2?5，其前n项和Sn满足Sn?Sn?2?2Sn?1?2n?1?n≥3?.令bn?（Ⅰ）求数列?an?的通项公式；

（Ⅱ）若f?x??2x?1，求证：Tn?b1f?1??b2f?2????bnf?n??（Ⅲ）令Tn?1.

an?an?1

1（n≥1）； 61b1a?b2a2?b3a3???bnan?（a?0），求同时满足下列两个条件的所有a的值：①对于任意正?21?1?整数n，都有Tn?；②对于任意的m??0,?，均存在n0?N?，使得n≥n0时，Tn?m.

6?6?

n?1n?1解：（Ⅰ）由题意知Sn?Sn?1?Sn?1?Sn?2?2?n≥3?即an?an?1?2?n≥3?

∴an??an?an?1???an?1?an?2?????a3?a2??a2

?2n?1?2n?2???22?5?2n?1?2n?2???22?2?1?2?2n?1?n≥3?

检验知n?1、2时，结论也成立，故an?2n?1.

n?1n11?2?1???2?1?1?11?n?1?2????（Ⅱ）由于bnf?n??n?n? n?1n?1nn?1222?12?12?12?12?12?1??????????故Tn?b1f?1??b2f?2????bnf?n??1??11??11?1???1??????????n?? 2?23?n?12?1?21?21?21?22?12?1????????1?11?111???n?1?. ???2?1?22?1?21?2611，即条件①满足；又0?m?， 661?11?3?3?n?1?n?1?m?2??1?n?log?1??1?0. ∴Tn?m??2??2?1?22?1?1?6m?1?6m??3??1?的最大整数，则当n≥n0时，Tn?m.?9′ 取n0等于不超过log2??1?6m?（Ⅲ）（ⅰ）当a?2时，由（Ⅱ）知：Tn?an?a?aaaa（ⅱ）当a?2时，∵n≥1，n???≥，∴an≥?2n，∴bn?an≥bn??2n??bn?2n.

2?2?2222na1?11??1i?an?n?1?. ∴Tn???bia?≥??bi?2i?1????22?1?22?1??2i?1i?1?21?11?1?n?1由（ⅰ）知存在n0?N?，当n≥n0时，?， ??2?1?22?1?3a故存在n0?N?，当n≥n0时，Tn?na1?11?a11???n?1?，不满足条件. ???22?1?22?1?23a6nan?a?aaaa（ⅲ）当0?a?2时，∵n≥1，n???≤，∴an≤?2n，∴bn?an≤bn??2n??bn?2n.

2?2?2222nn1aa1?11?i?n?1∴Tn???bia?≤??bi2i?1?????.

22221?22?1??i?1i?1a?1?a1?11?a?n?1取m???0,?，若存在n0?N?，当n≥n0时，Tn?m，则????.

12?6?22?1?22?1?12111∴?n?1?矛盾. 故不存在n0?N?，当n≥n0时，Tn?m.不满足条件. 1?22?13综上所述：只有a?2时满足条件，故a?2.

例43 已知数列?an?满足a1?（1）求a2,a3,a4； （2）已知存在实数?，使??n?1??2an?n?n?N?. 1,an?1???2an?4n?an??n??为公差为?1的等差数列，求?的值；

?an?n?23?1. 12（3）记bn?13n?22?n?N??，数列?bn?的前n项和为Sn，求证：Sn??an?2

解：（1）?a1?（2）

138，由数列?an?的递推公式得a2?0，a3??，a4?? 245(n?1)(2an?n)??(n?1)an?4na??n(??2)an?(4??1)nan??n??1an?1??(n?1)an??n?n===???(n?1)(2a?n)an?nan?1?n?1an?n3an?3nan?n3 n?n?1an?4n?a??n???1??1??1，得???2 ?数列?n为公差是的等差数列.由题意，令?3a?n3?n??n2?2nan??na1?2（3）由（2）知 ??(n?1)(?1)??n，所以an?n?1an?na1?1此时bn?1?n?3111==[?]， n?22n?2n?2n2?(n?1)?2(n?2)(3)n(n?2)(3)(n?2)(3)n32?n?31111111???Sn?[????????423532(3)?33(3)?4(3)?2(3)?5(3)?3 …… 此处隐藏：1346字，全部文档内容请下载后查看。喜欢就下载吧 ……