发布时间 : 星期一 文章2019届浙江省高三数学理一轮复习专题突破训练：数列更新完毕开始阅读9a5c09d048fe04a1b0717fd5360cba1aa8118c1c
为2的等差数列. （2）显然an?0,an?1?2an?2an?3?1,Qf?x??x2?2x?3?1在x??0,1?上单调递减,
??f?x????2?1,3?1?,故当0?an?1时,0?an?1?f?an??1, 即当0?a1?1时,an?1?1 与an?1同号
?0?an?1,
22an?2?an?an?2a?3?a?1n?1n?1?2an?1?3??an?1?an?1?2??an?1?an?1?a2n?1?2an?1?3?a2n?1, ?2an?1?3 Qan?1?an?1?2?0,an?2?an与an?1?an?1异号，且a3?a1?0,
?a2n?2?a2n?0,a2n?1?a2n?1?0,??a2n?1?单调递减,?a2n? 单调递增, 3??1??a?a???n?n11?112??2??2Qan?1??an?2an?3??,?an?1?与an?异号,
322222an?2an?3?2Qa1?1111?0,?a2n?1??0,a2n??0,?a2n??a2n?1,n?N?. 222223??1?3??3?5????a?a?a?a?2???2n??2n??2n?1??2n1?1??1?2??2?2??2?2??????Qa2n?1???a??a?2n?1??2?2n?13?23??23??2222????3??a2n?2a2n?3?a2n?2a2n?3???a2n?1?2a2n?1?3??2????2?2??2?2??1?1???a2n?1??4?2???1?n?1?????n?1?1??1?1?1111?1?1??4??????2,??a1????a3???...??a2n?1????g?...?g???2??2?2?2242?4?21?13 ??43n?4?a1?a3?...?a2n?1?68、 证明：（1）Qan?1?又f(x)?x?1?1???an???an?0， an?1?an?1在(0,1)单调递减，0?an?1， x?an?1?an. ………………………5分
（2）Qan?1?11?2an?， an?1an11?. an?1an1115??an?1??. ………………………8分 an?1a1an?12?an?an?1?an??Sn?an?1?a1?2又Qan?1?2?112， ?4a?4?n22ana?1n?1122??2?4a?ann?1. ………………………10分 22an?1an由0?an?1?an可知?11112?2???2?4a??3， ………………………14分 12222ananaa?1nn即2?1111，??3?2n???3n， 2222an?1anan?1a1111.，， 0?a??3n?4?2n?4??3n?4n?122ana?1n?15?Sn?3n?4?2 ………………………15分 2?2n?4??2n?4?
22229、解：（Ⅰ）令m?1，得an从而a12a3所以a3??3，?1an?1?n?1，
3 ………………
2分
22令n?m?2，得a2m?2?a2?4m?4
从而a4?28a2，a6?12，又a4a6?52?1?24， a22 …………………4分
所以a2?2，a2?从而a2m?2?2m?2 可知当n为偶数时，an?n；
2m?1，可知当n为奇数时，an?n
令n?m?1，得a2m?1?综上可得an?分
n (n?N?)． …………………6
（Ⅱ）（i）
a2n?1?a2n?1?2a2n?(2n?1?2n)?(2n?1?2n) ?12n?1?2n?12n?1?2n
?0所以a2n?1?a2n?1?2a2n …………………9分 （ii）即证明2?4???2n?n(1?3?5???2n?1) n?1由（i）得1?3?22， 3?5?24，…，2n?1?将上述的n个式子相加，得
2n?1?22n
2(1?3???2n?1?2n?1)?(1?2n?1)?2(2?4???2n)
所以2?4???2n?(1?3?5???2n?1)?1?2n?1 2所以，只需证1?3???2n?1?1?2n?1n?(1?3???2n?1)
2n?1即1?3???2n?1?(n?1)(1?2n?1) ……………………………12分
2事实上，当k?0,1,2,?,n时
1?2k?2n?1?2k?1?2n?1?（因为1?2k?1?2n，1?所以1?2k?从而
2k1?2k?1?2k2n?1?2k?2n?1?0
2n?1?2k）
2n?1?2k?1?2n?1
11?3???2n?1?[(1?2n?1)?(3?2n?1)???(2n?1?3)?(2n?1?1)]2
?15分
1…………………………………………(n?1)(1?2n?1)．
223320.10、（Ⅰ）∵Sn?a13?a2?L?ann?N* 2333 ∴Sn?a?a?L?a?112n?1
??22332两式相减得Sn?Sn?1?an?anSn?Sn?1?an?Sn?Sn?1?an 22则Sn?Sn?1?an,Sn?1?Sn?2?an?1????????
?a两式相减得
所以
n22?an?1??an?an?1??an?an?1??1an?n 4分
（Ⅱ）根据（Ⅰ）知，??1?1? ?ann?n?2322?k?2n?2?k?∵k?2n?2?k??? ?n?1???2??∴1kk?321(2n?2?k)(2n?2?k)3232?2k?2n?2?k?k?2n?2?k??2?n?1?n?1
?1??1??1?即??????2??
aaa?k??2n?2?k??n?1?令k?1,2,3,L,n，累加后再加?323232?1??得 ?an?1?32323232323232?1??1??1??1??1??1??1??1??1????L??2?2?2?L2???????????????????aaaaaaaa?1??2??3??2n?1??n?1??n?1??n?1??n?1??an?1?
?1?2n?1??2n?1??? 9分 ?(n?1)n?1?an?1?又∵?而
321111111??L??3???L??2
12233(2n?1)2n?12233(2n?1)2n?11kk
?111?11?1k?k?1?11??????????kkkkkk?1k?k?1k?k?k?1kk??k?1?2k?11?1??1??2?????
k?k?1k?k??k?1