第一篇：(新Ⅰ)2018年高考數(shù)學(xué)總復(fù)習(xí)專題06數(shù)列分項(xiàng)練習(xí)理

專題06 數(shù)列

一．基礎(chǔ)題組

1.【2013課標(biāo)全國(guó)Ⅰ，理7】設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn，若Sm－1＝－2，Sm＝0，Sm＋1＝3，則m＝()．

A．3 B．4 C．5 D．6 【答案】C 【解析】∵Sm－1＝－2，Sm＝0，Sm＋1＝3，∴am＝Sm－Sm－1＝0－(－2)＝2，am＋1＝Sm＋1－Sm＝3－0＝3.∴d＝am＋1－am＝3－2＝1.∵Sm＝ma1＋又∵am＋1＝a1＋m×1＝3，∴?m?m?1?m?1×1＝0，∴a1??.22m?1?m?3.∴m＝5.故選C.22.【2012全國(guó)，理5】已知{an}為等比數(shù)列，a4＋a7＝2，a5a6＝－8，則a1＋a10＝()A．7 B．5 C．－5 D．－7 【答案】D

3.【2008全國(guó)1，理5】已知等差數(shù)列?an?滿足a2?a4?4，a3?a5?10，則它的前10項(xiàng)的和S10?（）A．138 B．135

C．95

D．23 【答案】C.【解析】由a2?a4?4,a3?a5?10?a1??4,d?3,S10?10a1?45d?95.4.【2013課標(biāo)全國(guó)Ⅰ，理14】若數(shù)列{an}的前n項(xiàng)和Sn?＝__________.【答案】(－2)n－

121an?，則{an}的通項(xiàng)公式是an33 【解析】∵Sn?①－②，得an?2121an?，①∴當(dāng)n≥2時(shí)，Sn?1?an?1?.② 333322aan?an?1，即n＝－2.33an?1∵a1＝S1＝21a1?，∴a1＝1.33n－1∴{an}是以1為首項(xiàng)，－2為公比的等比數(shù)列，an＝(－2).5.【2009全國(guó)卷Ⅰ，理14】設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn.若S9=72,則a2+a4+a9=___________.【答案】24 【解析】∵S9?72?9(a1?a2),∴a1+a9=16.2∵a1+a9=2a5,∴a5=8.∴a2+a4+a9=a1+a5+a9=3a5=24.6.【2011全國(guó)新課標(biāo)，理17】等比數(shù)列{an}的各項(xiàng)均為正數(shù)，且2a1＋3a2＝1，a32?9a2a3.(1)求數(shù)列{an}的通項(xiàng)公式；

(2)設(shè)bn＝log3a1＋log3a2＋…＋log3an，求數(shù)列{1}的前n項(xiàng)和． bn(2)bn?log3a1?log3a2??log3an??(1?2??n)??n(n?1)故21211????2(?)，bnn(n?1)nn?111??b1b2?1111???2?(1?)?(?)?bn223?11?2n?(?)???.nn?1?n?1?1?2n所以數(shù)列??的前n項(xiàng)和為?.bn?1?n?7.【2010新課標(biāo)，理17】（12分）設(shè)數(shù)列{an}滿足a1＝2，an＋1－an＝3·2(1)求數(shù)列{an}的通項(xiàng)公式；

(2)令bn＝nan，求數(shù)列{bn}的前n項(xiàng)和Sn.【解析】(1)由已知，當(dāng)n≥1時(shí)，an＋1＝(an＋1－an)＋(an－an－1)＋…＋(a2－a1)]＋a1＝3(2

2n－1

2n－1

.＋

22n－

3＋…＋2)＋2＝2

2(n＋1)－1

.而a1＝2，所以數(shù)列{an}的通項(xiàng)公式為an＝2(2)由bn＝nan＝n·232n－1

2n－1

.知

2n－1Sn＝1·2＋2·2＋3·2＋…＋n·223

575

.①

2n＋1從而2·Sn＝1·2＋2·2＋3·2＋…＋n·2①－②，得

(1－2)Sn＝2＋2＋2＋…＋2即Sn＝235

2n－1

.②

－n·2

2n＋1，12n＋1(3n－1)2＋2]． 98.【2005全國(guó)1，理19】設(shè)等比數(shù)列{an}的公比為q，前n項(xiàng)和Sn>0（n=1，2，…）（1）求q的取值范圍；（2）設(shè)bn?an?2?3an?1,記{bn}的前n項(xiàng)和為Tn，試比較Sn和Tn的大小.2

解①式得q>1；解②，由于n可為奇數(shù)、可為偶數(shù)，得－1

試題分析：（Ⅰ）先用數(shù)列第項(xiàng)與前項(xiàng)和的關(guān)系求出數(shù)列{an}的遞推公式，可以判斷數(shù)列{an}是等差數(shù)列，利用等差數(shù)列的通項(xiàng)公式即可寫出數(shù)列{an}的通項(xiàng)公式；（Ⅱ）根據(jù)（Ⅰ）數(shù)列{bn}的通項(xiàng)公式，再用拆項(xiàng)消去法求其前項(xiàng)和.【考點(diǎn)定位】數(shù)列前n項(xiàng)和與第n項(xiàng)的關(guān)系；等差數(shù)列定義與通項(xiàng)公式；拆項(xiàng)消去法 10.【2016高考新課標(biāo)理數(shù)3】已知等差數(shù)列{an}前9項(xiàng)的和為27，a10=8，則a100=（A）100（B）99（C）98（D）97 【答案】C 【解析】 試題分析：由已知，?C.?9a1?36d?27,所以a1??1,d?1,a100?a1?99d??1?99?98,故選?a1?9d?8【考點(diǎn)】等差數(shù)列及其運(yùn)算

【名師點(diǎn)睛】等差、等比數(shù)列各有五個(gè)基本量,兩組基本公式,而這兩組公式可看作多元方程,利用這些方程可將等差、等比數(shù)列中的運(yùn)算問題轉(zhuǎn)化為解關(guān)于基本量的方程（組）,因此可以說數(shù)列中的絕大部分運(yùn)算題可看作方程應(yīng)用題,所以用方程思想解決數(shù)列問題是一種行之有效的方法.二．能力題組

1.【2011全國(guó)，理4】設(shè)Sn為等差數(shù)列{an}的前n項(xiàng)和，若a1＝1，公差d＝2，Sk＋2－Sk＝24，則k＝()A．8 B．7 C．6 D．5 【答案】D

2.【2006全國(guó)，理10】設(shè){an}是公差為正數(shù)的等差數(shù)列，若a1+a2+a3=15，a1a2a3=80則a11+a12+a13＝（）

（A）120（B）105（C）90（D）75 【答案】B 【解析】

3.【2012全國(guó)，理16】數(shù)列{an}滿足an＋1＋(－1)an＝2n－1，則{an}的前60項(xiàng)和為__________． 【答案】1 830 【解析】：∵an＋1＋(－1)an＝2n－1，nn∴a2＝1＋a1，a3＝2－a1，a4＝7－a1，a5＝a1，a6＝9＋a1，a7＝2－a1，a8＝15－a1，a9＝a1，a10＝17＋a1，a11＝2－a1，a12＝23－a1，…，a57＝a1，a58＝113＋a1，a59＝2－a1，a60＝119－a1，∴a1＋a2＋…＋a60＝(a1＋a2＋a3＋a4)＋(a5＋a6＋a7＋a8)＋…＋(a57＋a58＋a59＋a60)＝10＋26＋42＋…＋234＝

15?(10?234)?1830．

24.【2014課標(biāo)Ⅰ，理17】

已知數(shù)列?an?的前項(xiàng)和為Sn，a1?1，an?0，anan?1??Sn?1，其中?為常數(shù)，（I）證明：an?2?an??；

（II）是否存在?，使得?an?為等差數(shù)列？并說明理由.【答案】（I）詳見解析；（II）存在，??4.5.【2009全國(guó)卷Ⅰ，理20】 在數(shù)列{an}中，a1＝1,an+1＝（1?1n?1）an+n.n2（Ⅰ）設(shè)bn?an，求數(shù)列{bn}的通項(xiàng)公式； n（Ⅱ）求數(shù)列{an}的前n項(xiàng)和Sn.【解析】（Ⅰ）由已知得b1=a1=1,且

an?1an11??n,即bn?1?bn?n.n?1n22從而b2?b1?2n?11111于是bn?b1??2???n?1?2?n?1(n≥2).2222 bn?bn?1?11,b3?b2?2,…… 221(n≥2).又b1=1.故所求的通項(xiàng)公式bn?2?(Ⅱ)由（Ⅰ）知an?n(2?令Tn?n12n?1.12n)?2n?n?1n2n?1.?2k?1nkk?1,則2Tn??2k?1kk?2.于是Tn=2Tn-Tn=

?2k?0n?11k?1?n2n?1=4?n?2.又n?12?(2k)?n(n?1)，k?1所以Sn?n(n?1)?n?2?4.2n?1?an的最大值6.【2016高考新課標(biāo)理數(shù)1】設(shè)等比數(shù)列{an}滿足a1+a3=10，a2+a4=5，則a1a2鬃為.【答案】64

【考點(diǎn)】等比數(shù)列及其應(yīng)用

【名師點(diǎn)睛】高考中數(shù)列客觀題大多具有小、巧、活的特點(diǎn),在解答時(shí)要注意方程思想及數(shù)列相關(guān)性質(zhì)的應(yīng)用,盡量避免小題大做.7.【2017新課標(biāo)1，理4】記Sn為等差數(shù)列{an}的前項(xiàng)和．若a4?a5?24，S6?48，則{an}的公差為 A．1 【答案】C 【解析】

試題分析：設(shè)公差為d，a4?a5?a1?3d?a1?4d?2a1?7d?24，B．2

C．4

D．8 S6?6a1??2a1?7d?246?5,解得d?4，故選C.d?6a1?15d?48，聯(lián)立?6a?15d?482?1 7 【考點(diǎn)】等差數(shù)列的基本量求解

【名師點(diǎn)睛】求解等差數(shù)列基本量問題時(shí)，要多多使用等差數(shù)列的性質(zhì)，如{an}為等差數(shù)列，若

m?n?p?q，則am?an?ap?aq.三．拔高題組

1.【2013課標(biāo)全國(guó)Ⅰ，理12】設(shè)△AnBnCn的三邊長(zhǎng)分別為an，bn，cn，△AnBnCn的面積為Sn，n＝1,2,3，….若b1＞c1，b1＋c1＝2a1，an＋1＝an，bn＋1＝A．{Sn}為遞減數(shù)列 B．{Sn}為遞增數(shù)列

C．{S2n－1}為遞增數(shù)列，{S2n}為遞減數(shù)列 D．{S2n－1}為遞減數(shù)列，{S2n}為遞增數(shù)列 【答案】B 【解析】

cn?anb?an，cn＋1＝n，則()． 22 8

2.【2011全國(guó)，理20】設(shè)數(shù)列{an}滿足a1＝0且(1)求{an}的通項(xiàng)公式；(2)設(shè)bn?11??1.1?an?11?an1?an?1n，記Sn??bk?1nk，證明：Sn＜1.【解析】(1)由題設(shè)11??1，1?an?11?an即{1}是公差為1的等差數(shù)列． 1?an又11?1，故?n.1?an1?an1.n所以an?1?(2)由(1)得bn?nn1?an?1n?n?1?n11，??n?1?nnn?1Sn??bk??(k?1k?1111?)?1??1.kk?1n?13.【2006全國(guó)，理22】（本小題滿分12分）

設(shè)數(shù)列｛an｝的前n項(xiàng)和

S43an?13?2n?1?2n?3,n?1,2,3,…。（Ⅰ）求首項(xiàng)a1與通項(xiàng)an；

nn（Ⅱ）設(shè)T?2s,n?1,2,3,…，證明：?T3ni?.ni?12整理得 an?1n?2n?4(an?1?2),n?2,3,…，因而數(shù)列{an?2n}是首項(xiàng)為a1?2?4，公比為4的等比數(shù)列，即

a?1n?2n?4?4n?4n，n=1,2,3…，因而

an?4n?2n，n=1,2,3,…，（II）將ann?4?2n代入①得

S43?(4n?2n)?12n?3?2n?1?3 ?1?(2n?1?1)(2n?1?2)3

?2?(2n?1?1)(2n3?1).2nTn?S n

32n??n?12(2?1)?(2n?1)

311??（n?n?1,）22?12?13n11所以，?Ti??(i?i?1)

2i?12?12?1i?1311?(i?i?1)22?12?1

3?.2?4.【2017新課標(biāo)1，理12】幾位大學(xué)生響應(yīng)國(guó)家的創(chuàng)業(yè)號(hào)召，開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的

興趣，他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)學(xué)問題的答案：已知數(shù)列1，1，2，1，2，4，1，2，4，8，1，2，4，8，16，…，其中第一項(xiàng)是2，接下來的兩項(xiàng)是2，2，再接下來 的三項(xiàng)是2，2，2，依此類推.求滿足如下條件的最小整數(shù)N：N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么 該款軟件的激活碼是

A．440 【答案】A 【解析】

試題分析：由題意得，數(shù)列如下：

B．330

C．220

D．110 01

0

1n1,1,2,1,2,4,1,2,4，2k?1k(k?1)項(xiàng)和為 2

則該數(shù)列的前1?2??k??k(k?1)?S???1?(1?2)?2???(1?2??2k?1)?2k?1?k?2，【考點(diǎn)】等差數(shù)列、等比數(shù)列

【名師點(diǎn)睛】本題非常巧妙地將實(shí)際問題和數(shù)列融合在一起，首先需要讀懂題目所表達(dá)的具體含義，以及觀察所給定數(shù)列的特征，進(jìn)而判斷出該數(shù)列的通項(xiàng)和求和.另外，本題的難點(diǎn)在于數(shù)列里面套數(shù)列，第一個(gè)數(shù)列的和又作為下一個(gè)數(shù)列的通項(xiàng)，而且最后幾項(xiàng)并不能放在一個(gè)數(shù)列中，需要進(jìn)行判斷.12

第二篇：(新Ⅱ)2018年高考數(shù)學(xué)總復(fù)習(xí)專題06數(shù)列分項(xiàng)練習(xí)理!

專題06 數(shù)列

一．基礎(chǔ)題組

1.【2013課標(biāo)全國(guó)Ⅱ，理3】等比數(shù)列{an}的前n項(xiàng)和為Sn.已知S3＝a2＋10a1，a5＝9，則a1＝()． A．1111 B．? C． D．? 3399【答案】：C 2.【2012全國(guó)，理5】已知等差數(shù)列{an}的前n項(xiàng)和為Sn，a5＝5，S5＝15，則數(shù)列{和為()A．

1}的前100項(xiàng)anan?11009999101 B． C． D． 1011001001015(a1?a5)5(a1?5)??15，∴a1＝1.22【答案】 A 【解析】S5?∴d?a5?a15?1??1.5?15?1∴an＝1＋(n－1)×1＝n.∴11.?anan?1n(n?1)?1?設(shè)??的前n項(xiàng)和為Tn，aa?nn?1?則T100?＝1?＝1?111??…? 1?22?3100?10111111???…?? 2231001011100?.1011013.【2010全國(guó)2，理4】如果等差數(shù)列{an}中，a3＋a4＋a5＝12，那么a1＋a2＋?＋a7等于()A．14 B．21 C．28 D．35 【答案】：C

【解析】∵{an}為等差數(shù)列，a3＋a4＋a5＝12，∴a4＝4.∴a1＋a2＋?＋a7＝7(a1?a7)＝7a4＝28.24.【2006全國(guó)2，理14】已知△ABC的三個(gè)內(nèi)角A,B,C成等差數(shù)列,且AB=1,BC=4,則邊BC上的中線AD的長(zhǎng)為.【答案】:3

5.【2014新課標(biāo)，理17】（本小題滿分12分）已知數(shù)列?an?滿足a1=1，an?1?3an?1.（Ⅰ）證明an?1是等比數(shù)列，并求?an?的通項(xiàng)公式； ?2?（Ⅱ）證明：1?1?…+1?3.a1a2an21112?3，所以?a?1?是等比【解析】：（Ⅰ）證明：由an?1?3an?1得an?1??3(an?)，所以?n?1222??an?2an?1?1313n?13n?1數(shù)列，首項(xiàng)為a1??，公比為3，所以an???3，解得an?.222223n?112（Ⅱ）由（Ⅰ）知：an?，所以，?n2an3?1因?yàn)楫?dāng)n?1時(shí)，3n??1?n?12，3所以

11?3n?12?3n?1，于是11313111?1??L?n?1=(1?n)?，??L33232a1a2an所以3111?.??La1a2an26.【2011新課標(biāo)，理17】等比數(shù)列{an}的各項(xiàng)均為正數(shù)，且2a1＋3a2＝1，a32?9a2a3.(1)求數(shù)列{an}的通項(xiàng)公式；

(2)設(shè)bn＝log3a1＋log3a2＋?＋log3an，求數(shù)列{

1}的前n項(xiàng)和． bn2 7.【2015高考新課標(biāo)2，理16】設(shè)Sn是數(shù)列?an?的前n項(xiàng)和，且a1??1，則Sn?________． an?1?SnSn?1，【答案】?1 n【解析】由已知得an?1?Sn?1?Sn?Sn?1?Sn，兩邊同時(shí)除以Sn?1?Sn，得

?1?11???1，故數(shù)列??是Sn?1Sn?Sn?以?1為首項(xiàng)，?1為公差的等差數(shù)列，則

11??1?(n?1)??n，所以Sn??．

nSn【考點(diǎn)定位】等差數(shù)列和遞推關(guān)系．

8.【2017課標(biāo)II，理3】我國(guó)古代數(shù)學(xué)名著《算法統(tǒng)宗》中有如下問題：“遠(yuǎn)望巍巍塔七層，紅光點(diǎn)點(diǎn)倍加增，共燈三百八十一，請(qǐng)問尖頭幾盞燈？”意思是：一座7層塔共掛了381盞燈，且相鄰兩層中的下一層燈數(shù)是上一層燈數(shù)的2倍，則塔的頂層共有燈 A．1盞

【答案】B B．3盞

C．5盞

D．9盞

【考點(diǎn)】 等比數(shù)列的應(yīng)用、等比數(shù)列的求和公式

【名師點(diǎn)睛】用數(shù)列知識(shí)解相關(guān)的實(shí)際問題，關(guān)鍵是列出相關(guān)信息，合理建立數(shù)學(xué)模型——數(shù)列模型，判斷是等差數(shù)列還是等比數(shù)列模型；求解時(shí)要明確目標(biāo)，即搞清是求和、求通項(xiàng)、還是解遞推關(guān)系問題，所求結(jié)論對(duì)應(yīng)的是解方程問題、解不等式問題、還是最值問題，然后將經(jīng)過數(shù)學(xué)推理與計(jì)算得出的結(jié)果放回到實(shí)際問題中，進(jìn)行檢驗(yàn)，最終得出結(jié)論． 二．能力題組

1.【2013課標(biāo)全國(guó)Ⅱ，理16】等差數(shù)列{an}的前n項(xiàng)和為Sn，已知S10＝0，S15＝25，則nSn的最小值為__________． 【答案】：－49 【解析】：設(shè)數(shù)列{an}的首項(xiàng)為a1，公差為d，則S10＝10a1＋10?9d＝10a1＋45d＝0，① 2S15＝15a1?15?14d＝15a1＋105d＝25.② 22，3聯(lián)立①②，得a1＝－3，d?所以Sn＝?3n?n(n?1)21210??n?n.23331310220n?n，f'(n)?n2?n.33320.3令f(n)＝nSn，則f(n)?令f′(n)＝0，得n＝0或n?當(dāng)n?202020時(shí)，f′(n)＞0,0

【名師點(diǎn)睛】等差數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式，共涉及五個(gè)量a1，an，d，n，Sn，知其中三個(gè)就能求另外兩個(gè)，體現(xiàn)了用方程的思想解決問題．?dāng)?shù)列的通項(xiàng)公式和前n項(xiàng)和公式在解題中起到變量代換作用，而a1和d是等差數(shù)列的兩個(gè)基本量，用它們表示已知和未知是常用得方法．使用裂項(xiàng)法求和時(shí)，要注意正、負(fù)項(xiàng)相消時(shí)消去了哪些項(xiàng)，保留了哪些項(xiàng)，切不可漏寫未被消去的項(xiàng)，未被消去的項(xiàng)有前后對(duì)稱的特點(diǎn)．

3.【2005全國(guó)3，理20】(本小題滿分12分)在等差數(shù)列{an}中，公差d?0,a2是a1與a4的等差中項(xiàng).已知數(shù)列a1,a3,ak1,ak2,?,akn,?成等比數(shù)列，求數(shù)列{kn}的通項(xiàng)kn.【解析】：依題設(shè)得a2n?a1?(n?1)d, a2?a1a4

∴(a21?d)2?a1(a1?3d)，整理得d=a1d，∵d?0, ?d?a1，得an?nd, 所以，由已知得d，3d，k1d，k2d，?，kndn?是等比數(shù)列.由d?0,所以數(shù)列 1，3，k1，k2，?，kn，? 也是等比數(shù)列，首項(xiàng)為1，公比為q?31?3,由此得k1?9.等比數(shù)列{kn}的首項(xiàng)k1?9,公比q?3,所以kn?9?qn?1?3n?1(n?1,2,3,?)，即得到數(shù)列{kn}的通項(xiàng)kn?3n?1.4.【2005全國(guó)2，理18】(本小題滿分12分)已知?an?是各項(xiàng)為不同的正數(shù)的等差數(shù)列，lga11、lga2、lga4成等差數(shù)列．又bn?a，n?1,2,3,?．2n(Ⅰ)證明?bn?為等比數(shù)列；

(Ⅱ)如果無窮等比數(shù)列?b1n?各項(xiàng)的和S?3，求數(shù)列?an?的首項(xiàng)a1和公差d．

(注：無窮數(shù)列各項(xiàng)的和即當(dāng)n??時(shí)數(shù)列前n項(xiàng)和的極限)5

1[1?(1則S=limSn?lim2d2)n]n???n???1?1?1d 2由S?13，得公差d=3，首項(xiàng)a1=d=3 三．拔高題組

1.【2006全國(guó)2，理11】設(shè)SSn是等差數(shù)列{an}的前n項(xiàng)和,若

3=1,則S6S等于（） 63S12A.3110

B.3 C.1

D.9 【答案】:A

【解析】:由已知設(shè)a1+a2+a3=T,a4+a5+a6=2T,a7+a8+a9=3T, a10+a11+a12=4T.

∴S6t＋2t3S=?.12t?2t?3t?4t10∴選A.

2.【2005全國(guó)2，理11】如果a1,a2,?,a8為各項(xiàng)都大于零的等差數(shù)列，公差d?0，則（）

(A)a1a8?a4a5 【答案】B(B)a1a8?a4a5(C)a1?a8?a4?a5(D)a1a8?a4a5

3.【2012全國(guó)，理22】函數(shù)f(x)＝x－2x－3，定義數(shù)列{xn}如下：x1＝2，xn＋1是過兩點(diǎn)P(4,5)，Qn(xn，2f(xn))的直線PQn與x軸交點(diǎn)的橫坐標(biāo)．

(1)證明：2≤xn＜xn＋1＜3；(2)求數(shù)列{xn}的通項(xiàng)公式．

(2)由(1)及題意得xn?1?3?4xn.2?xn7 設(shè)bn＝xn－3，則15??1，bn?1bn1111??5(?)，bn?14bn4數(shù)列{311?}是首項(xiàng)為?，公比為5的等比數(shù)列．

4bn4因此

4113，????5n?1，即bn??n?13?5?1bn4443?5n?1?1.2所以數(shù)列{xn}的通項(xiàng)公式為xn＝3?4.【2006全國(guó)2，理22】設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且方程x-anx-an=0有一根為Sn-1,n= 1,2,3,?.(1)求a1,a2;(2)求{an}的通項(xiàng)公式.【解析】:(1)當(dāng)n=1時(shí),x-a1x-a1=0有一根為S1-1=a1-1,

21. 212當(dāng)n=2時(shí),x-a2x-a2=0有一根為S2-1=a2-,

21211于是(a2-)-a2(a2-)-a2=0,解得a2=.

226于是(a1-1)-a1(a1-1)-a1=0,解得a1=2(2)由題設(shè)(Sn-1)-an(Sn-1)-an=0,即Sn-2Sn+1-anSn=0. 當(dāng)n≥2時(shí),an=Sn-Sn-1,代入上式得 22Sn-1Sn-2Sn+1=0.由(1)知S1=a1=由①可得S3=

①

1112,S2=a1+a2=+=.22633. 4n由此猜想Sn=,n=1,2,3,?.

n?1下面用數(shù)學(xué)歸納法證明這個(gè)結(jié)論.

5.【2016高考新課標(biāo)2理數(shù)】Sn為等差數(shù)列?an?的前n項(xiàng)和，且a1=1，S7?28.記bn=?lgan?，其中?x?表示不超過x的最大整數(shù)，如?0.9?=0，?lg99?=1.（Ⅰ）求b1，b11，b101；（Ⅱ）求數(shù)列?bn?的前1 000項(xiàng)和.【答案】（Ⅰ）b1?0，b11?1，b101?2；（Ⅱ）1 893.【解析】

【考點(diǎn)】等差數(shù)列的通項(xiàng)公式、前n項(xiàng)和公式，對(duì)數(shù)的運(yùn)算

【名師點(diǎn)睛】解答新穎的數(shù)學(xué)題時(shí)，一是通過轉(zhuǎn)化，化“新”為“舊”；二是通過深入分析，多方聯(lián)想，以“舊”攻“新”；三是創(chuàng)造性地運(yùn)用數(shù)學(xué)思想方法，以“新”制“新”，應(yīng)特別關(guān)注創(chuàng)新題型的切入點(diǎn)和生長(zhǎng)點(diǎn).

第三篇：小學(xué)數(shù)學(xué)畢業(yè)總復(fù)習(xí)：數(shù)列求和考點(diǎn)

小學(xué)數(shù)學(xué)畢業(yè)總復(fù)習(xí)：數(shù)列求和考點(diǎn)

基礎(chǔ)教育一直是最受學(xué)校和家長(zhǎng)關(guān)注的，最為基礎(chǔ)教育重中之重的初等教育，更是得到更多的重視。查字典數(shù)學(xué)網(wǎng)小升初頻道為大家準(zhǔn)備了小學(xué)數(shù)學(xué)畢業(yè)總復(fù)習(xí)，希望能幫助大家做好小升初的復(fù)習(xí)備考，考入重點(diǎn)初中院校!小學(xué)數(shù)學(xué)畢業(yè)總復(fù)習(xí)：數(shù)列求和考點(diǎn) 數(shù)列求和

等差數(shù)列：在一列數(shù)中，任意相鄰兩個(gè)數(shù)的差是一定的，這樣的一列數(shù)，就叫做等差數(shù)列。

基本概念：首項(xiàng)：等差數(shù)列的第一個(gè)數(shù)，一般用a1表示;項(xiàng)數(shù)：等差數(shù)列的所有數(shù)的個(gè)數(shù)，一般用n表示;公差：數(shù)列中任意相鄰兩個(gè)數(shù)的差，一般用d表示;通項(xiàng)：表示數(shù)列中每一個(gè)數(shù)的公式，一般用an表示;數(shù)列的和：這一數(shù)列全部數(shù)字的和，一般用Sn表示.基本思路：等差數(shù)列中涉及五個(gè)量：a1 ,an, d, n, sn，通項(xiàng)公式中涉及四個(gè)量，如果己知其中三個(gè)，就可求出第四個(gè);求和公式中涉及四個(gè)量，如果己知其中三個(gè)，就可以求這第四個(gè)。

基本公式：通項(xiàng)公式：an = a1+(n-1)d;通項(xiàng)=首項(xiàng)+(項(xiàng)數(shù)一1)公差;數(shù)列和公式：sn,=(a1+ an)n 數(shù)列和=(首項(xiàng)+末項(xiàng))項(xiàng)數(shù)

第 1 頁 項(xiàng)數(shù)公式：n=(an+ a1)項(xiàng)數(shù)=(末項(xiàng)-首項(xiàng))公差+1;公差公式：d =(an-a1))(n-1);公差=(末項(xiàng)-首項(xiàng))(項(xiàng)數(shù)-1);關(guān)鍵問題：確定已知量和未知量，確定使用的公式;小升初考試是小學(xué)生進(jìn)入初等重點(diǎn)初中院校的一次重要考試，希望大家都能夠認(rèn)真復(fù)習(xí)，同時(shí)也希望我們準(zhǔn)備的小學(xué)數(shù)學(xué)畢業(yè)總復(fù)習(xí)能讓大家在小升初的備考過程助大家一臂之力!

第 2 頁

第四篇：【天津市2013屆高三數(shù)學(xué)總復(fù)習(xí)之綜合專題：數(shù)列(理)(學(xué)生版)

數(shù)列（理）

考查內(nèi)容：本小題主要考查等差數(shù)列與等比數(shù)列的通項(xiàng)公式及其前n項(xiàng)和公式、不等式證明等基礎(chǔ)知識(shí)，考查分類討論的思想方法，考查運(yùn)算能力、推理論證能力及綜合分析、解決問題的能力。

1、在數(shù)列?an?中，a1?1，an?1?2an?2n。（1）設(shè)bn?an。證明：數(shù)列?bn?是等差數(shù)列； n?12（2）求數(shù)列?an?的前n項(xiàng)和Sn。

2、設(shè)數(shù)列?an?的前n項(xiàng)和為Sn，已知ban?2n??b?1?Sn（1）證明：當(dāng)b?2時(shí)，?an?n?2n?1?是等比數(shù)列；（2）求?an?的通項(xiàng)公式

3、已知數(shù)列{an}的首項(xiàng)a1?22an，an?1?，n?1,2,3,…。3an?1?1?（1）證明：數(shù)列??1?是等比數(shù)列；

?an??n?（2）數(shù)列??的前n項(xiàng)和Sn。

?an?

4、已知數(shù)列?an?滿足：an??1，a1?22?cn?an?1?an，n?N。

1222，31?an?1?21?an，記數(shù)列bn?1?an，2????（1）證明數(shù)列?bn?是等比數(shù)列；（2）求數(shù)列{cn}的通項(xiàng)公式；

（3）是否存在數(shù)列{cn}的不同項(xiàng)ci,cj,ck，i?j?k，使之成為等差數(shù)列？若存在請(qǐng)求出這樣的不同項(xiàng)ci,cj,ck，i?j?k；若不存在，請(qǐng)說明理由。

5、已知數(shù)列{an}、{bn}中，對(duì)任何正整數(shù)n都有：

a1bn?a2bn?1?a3bn?2??an?1b2?anb1?2n?1?n?2。

（1）若數(shù)列{an}是首項(xiàng)和公差都是1的等差數(shù)列，求證：數(shù)列{bn}是等比數(shù)列；（2）若數(shù)列{bn}是等比數(shù)列，數(shù)列{an}是否是等差數(shù)列，若是請(qǐng)求出通項(xiàng)公式，若不是請(qǐng)說明理由；

（3）若數(shù)列{an}是等差數(shù)列，數(shù)列{bn}是等比數(shù)列，求證：?i?1n13?。aibi2)。數(shù)列{bn}

16、設(shè)數(shù)列{an}滿足a1?1，a2?2，an?(an?1?2an?2)，(n?3,4,3滿足b1?1,bn(n?2,3,)是非零整數(shù)，且對(duì)任意的正整數(shù)m和自然數(shù)k，都有?1?bm?bm?1??bm?k?1。

（1）求數(shù)列{an}和{bn}的通項(xiàng)公式；（2）記cn?nanbn(n?1,2,)，求數(shù)列{cn}的前n項(xiàng)和Sn。

7、有n個(gè)首項(xiàng)都是1的等差數(shù)列，設(shè)第m個(gè)數(shù)列的第k項(xiàng)為amk，(m,k?1,2,3，n, n≥3)，公差為dm，并且a1n,a2n,a3n，ann成等差數(shù)列。

（1）證明dm?p1d1?p2d2，3?m?n，p1,p2是m的多項(xiàng)式，并求p1?p2的值；（2）當(dāng)d1?1, d2?3時(shí)，將數(shù)列{dm}分組如下：(d1),(d2,d3,d4),(d5,d6,d7,d8,d9),（每組數(shù)的個(gè)數(shù)構(gòu)成等差數(shù)列），設(shè)前m組中所有數(shù)之和為(cm)4(cm?0)，求數(shù)列{2cmdm}的前n項(xiàng)和Sn。

（3）設(shè)N是不超過20的正整數(shù)，當(dāng)n?N時(shí)，對(duì)于（2）中的Sn，求使得不等式1(Sn?6)?dn成立的所有N的值。50

n?n???

8、數(shù)列{an}的通項(xiàng)公式為an?n2?cos2?sin2?，其前n項(xiàng)和為Sn。

33??（1）求Sn；

S3n，求數(shù)列{bn}的前n項(xiàng)和Tn。n?4nn?n?滿足a1?1,a2?2,an?2?(1?cos2)an?sin2,n?1,2,3,9、數(shù)列{an}?滿足

22（2）設(shè)bn?.。

（1）求a3,a4,并求數(shù)列?an?的通項(xiàng)公式；（2）設(shè)bn?a2n?1,Sn?b1?b2?a2n1?bn.。證明：當(dāng)n?n6?時(shí)，6時(shí)，Sn?2?。.n10、已知數(shù)列{an}和{bn}的通項(xiàng)公式分別為an?3n?6，bn?2n?7，n?N*，若將**集合{x|x?an,n?N}{x|x?bn,n?N}中的元素從小到大依次排列，構(gòu)成一個(gè)新的數(shù)列{cn}。（1）求c1,c2,c3,c4；

（2）求證：在數(shù)列{cn}中，但不在數(shù)列{bn}中的項(xiàng)恰為a2,a4,（3）求數(shù)列{cn}的通項(xiàng)公式。

11、在數(shù)列?an?中，a1?2，an?1??an??n?1?(2??)2n(n?N?)，其中??0。（1）求數(shù)列?an?的通項(xiàng)公式；（2）求數(shù)列?an?的前n項(xiàng)和Sn。,a2n,；

an?1ak?1?（3）證明：存在k?N，使得對(duì)任意n?N*均成立。anak*

12、在數(shù)列?an?與?bn?中，a1?1,b1?4，數(shù)列?an?的前n項(xiàng)和Sn滿足nSn?1?(n?3)Sn?0，且2an?1為bn與bn?1的等比中項(xiàng)，n?N*。

（1）求a2，b2的值；

（2）求數(shù)列?an?與?bn?的通項(xiàng)公式；

*2n?N（3）設(shè)Tn?(?1)1b1?(?1)2b2?…?(?1)nbn，證明n≥?3。NT?2n，nn，aaa*

13、已知等差數(shù)列?an?的公差為d?d?0?，等比數(shù)列?bn?的公比為q，且q?1。設(shè)Sn?a1b1?a2b2??anbn，Tn?a1b1?a2b2??(?1)n?1anbn，n?N*。

（1）若a1?b1?1，d?2,q?3求S3的值；

2dq(1?q2n)*n?N（2）若b1?1，證明?1?q?S2n??1?q?T2n?，； 21?q（3）若正整數(shù)n滿足2?n?q，設(shè)k1,k2，kn和l1,l2,，2，,n ,ln是1的兩個(gè)不同的排列，c1?ak1b1?ak2b2?...?aknbn，c2?al1b1?al2b2?...?alnbn，證明c1?c2。

14、在數(shù)列?an?中，a1?0，且對(duì)任意k?N*，a2k?1,a2k,a2k?1成等差數(shù)列，其公差為dk。

（1）若dk?2k，證明a2k,a2k?1,a2k?2成等比數(shù)列；

（2）若對(duì)任意k?N*，a2k,a2k?1,a2k?2成等比數(shù)列，其公比為qk。

?1?

①設(shè)q1?1，證明??是等差數(shù)列；

q?1?k?n3k2?2?n?2?。

②若a2?2，證明?2n??2k?2ak15、已知數(shù)列{an}與{bn}滿足：bnan?an?1?bn?1an?2且a1?2,a2?4。（1）求a3,a4,a5的值；

3?(?1)n，n?N*，?0，bn?2（2）設(shè)cn?a2n?1?a2n?1，n?N*，證明?cn?是等比數(shù)列；

Sk7?(n?N*)。（3）設(shè)Sk?a2?a4?????a2k，k?N，證明?6k?1ak*4n

第五篇：高考英語總復(fù)習(xí)Unit1Womenofachievement練習(xí)新人教版必修4

Unit 1 Women of achievement

A級(jí):基礎(chǔ)鞏固

Ⅰ.語境填詞

1.I felt a great sense of

(achieve)when I reached the top of the mountain.答案:achievement 2.A good student must

(connection)what he reads with what he sees around him.答案:connect 3.The

(organize)of such a large-scale party takes a lot of time and energy.答案:organization 4.I like her attitude very much,and the

(behave)of the other students shows that they like her,too.答案:behaviour 5.She

(observation)a man breaking into the bank and she reported it to the police at once.答案:observed 6.After a long time of discussion they accepted the agreement without

(argue).答案:argument 7.Rough seas caused much

(sick)among the passengers.答案:sickness 8.Books offered an excellent

(entertain)for idle hours.答案:entertainment 9.It is

(consider)of you to call on me from time to time.答案:considerate 10.I think it would be a

(kind)to tell him the bad news straight away.答案:kindness Ⅱ.單句改錯(cuò)

1.I am writing in respect to the complaint you made last week.答案:to→of

2.A crowd of children was passing my house,singing and laughing.答案:was→were

3.No matter how low you consider yourself,there is always someone looking up you wishing they were that high.答案:up后加on 4.Tom came late for the meeting.That was why he was ill.答案:why→because

5.How did it come across that humans speak so many different languages?

答案:across→about

B級(jí):能力提升

Ⅲ.完形填空

閱讀下面短文,從短文后各題所給的四個(gè)選項(xiàng)(A、B、C和D)中,選出可以填入空白處的最佳選項(xiàng)。

A Young Man Learns What’s the Most Important in Life

In his busy life,Jack had little time to think about the past and little to spend with his wife and son.教育資料 One day,his mother phoned him and told him that his old 1 ,Mr.Belser,had died.She asked if Jack would attend the funeral.Jack remembered 2 some of his childhood days with his old neighbor.It had been so long since Jack had thought of him.He 3 thought Mr.Belser had died years before.Jack’s mother said,“He didn’t forget you.When I saw him,he’d ask 4 you were doing.He’d remember the many days you spent at his home.After your father died,Mr.Belser stepped in to make sure you had a man’s 5 in your life.”

“He taught me carpentry(木工手藝),”Jack said.“I wouldn’t be in this business if it weren’t for him.He spent a lot of time 6 me important things.I’ll be there for the funeral.”

Mr.Belser’s funeral was 7.He had no children and most of his relatives had died.The night after he returned home,Jack and his mother 8 the old house next door.The houses was 9 as Jack remembered.Jack told his mother that there was a small gold box that Mr.Belser kept 10 on top of his desk.He had asked a thousand times what was inside, 11 Mr.Belser only said “The thing I value most.”It was 12.The house was exactly how Jack remembered it,except for the box.He figured someone from the Belser 13 had taken it.“Now I’ll never know what was so 14 to him,”Jack said.Two weeks after Mr.Belser died,Jack discovered a note in his mailbox.“Signature requested on a package.Please 15 by the main post office.”

Next day Jack collected the package.The return address 16 his attention:“Mr.Harold Belser”.Jack opened the package.Inside was the gold box and an envelope.“Upon my death,please 17 this box and its contents to Jack Bennett.It’s the thing I valued most in my life.”Jack 18 opened the box.Inside he found a simple pocket watch and also these words 19 to it,“Jack,Thanks for your time!—Harold Belser.”

“My god!The thing he valued most was...my time.”He couldn’t believe it.Immediately he called 20 his appointments for the next two days,because he needed some time to spend with his family.1.A.friend B.neighbor C.relative D.classmate 解析:friend “朋友”;neighbor “鄰居”;relative “親戚”;classmate “同學(xué)”。根據(jù)本段第三句中的“...some of his childhood days with his old neighbor”可知,此處應(yīng)該選neighbor,指杰克的鄰居,故B項(xiàng)正確。答案:B 2.A.working B.playing C.spending D.talking 解析:句意:杰克想起了小時(shí)候與他的老鄰居一起度過的時(shí)光。由第三段中的“He’d remember the many days you spent at his home.”可知此處選spending,意為“度過”。答案:C 3.A.honestly B.actively C.foolishly D.carefully 解析:honestly “真地;老實(shí)說”;actively “積極地”;foolishly “愚笨地”;carefully “小心地”。根據(jù)上一句“It had been so long since Jack had thought of him.”可知,杰克真的以為貝爾瑟先生幾年前就死了。故A項(xiàng)正確。答案:A 4.A.when B.where

教育資料 C.how D.why 解析:當(dāng)杰克的媽媽見到貝爾瑟先生的時(shí)候,他會(huì)問起杰克過得怎樣。how“如何”,符合語境。How is sb.doing?意為“某人過得怎么樣?” 答案:C 5.A.help B.influence C.shadow D.attitude 解析:help“幫助”;influence “影響”;shadow“陰影,影子”;attitude “態(tài)度”。此處指貝爾瑟先生來確認(rèn)杰克是否能在生活中像個(gè)男人一樣有所擔(dān)當(dāng),故B項(xiàng)正確。答案:B 6.A.giving B.teaching C.helping D.assisting 解析:由本段開頭的“He taught me carpentry...”可知,貝爾瑟先生花時(shí)間教杰克重要的事情,故B項(xiàng)正確。答案:B 7.A.big B.wonderful C.small D.moving 解析:big “大的”;wonderful “奇妙的;極好的”;small “小的”;moving “令人感動(dòng)的”。從下一句“He had no children and most of his relatives had died.”可推斷此處表示貝爾瑟先生的葬禮規(guī)模很小,故C項(xiàng)正確。答案:C 8.A.came B.returned C.painted D.visited 解析:根據(jù)語境可知此處指杰克和母親去看了看隔壁的老房子,故選D項(xiàng)visited,意為“拜訪;參觀”。答案:D 9.A.possibly B.strangely C.differently D.completely 解析:possibly “可能地”;strangely “奇怪地”;differently “不同地”;completely “完全地”。從下一段倒數(shù)第二句The house was exactly how Jack remembered it...可知,這個(gè)房子和杰克記憶中的完全一樣,故D項(xiàng)正確。答案:D 10.A.buried B.discovered C.locked D.reached 解析:bury “埋葬”;discover “發(fā)現(xiàn)”;lock “鎖”;reach “夠到”。從下一句中的He had asked a thousand times what was inside...可推知此處表示桌子上的小金盒子是上了鎖的,故C項(xiàng)正確。答案:C 11.A.so B.but C.or D.when 解析:根據(jù)前面的a thousand times和后面的only可知,前后是轉(zhuǎn)折關(guān)系,故用but,故B項(xiàng)正確。答案:B 12.A.dear B.gone C.old D.clear 解析:由本段最后一句“He figured someone from the Belser

had taken it.”可知,那個(gè)金盒子不見了。be gone “不見了”,符合語境。

答案:B 13.A.wife B.son C.family D.neighbor 解析:此處是指貝爾瑟先生的家人,故用family。the Belser family 意為“貝爾瑟一家人”。答案:C 14.A.valuable B.necessary

教育資料 C.important D.expensive 解析:valuable “有價(jià)值的,貴重的”;necessary “有必要的”;important “重要的”;expensive “昂貴的”。根據(jù)上一段中的“The thing I value most.”可知,盒子里的東西很貴重,故A項(xiàng)正確。答案:A 15.A.stop B.begin C.start D.hurry 解析:從下段中的“Next Day Jack collected the package.”可知,紙條上給杰克指出了取包裹的地址,stop by “順便來訪”,為固定短語,符合語境。答案:A 16.A.gathered B.visited C.greeted D.caught 解析:gather “收集;收割”;visit “訪問;拜訪”;greet “歡迎,迎接”;catch “抓住”。根據(jù)下一段“Jack opened the package.”可推斷此處表示吸引了他的注意,catch one’s attention “引起某人的注意”,為固定短語,符合語境。答案:D 17.A.give B.improve C.return D.pay 解析:杰克收到了貝爾瑟先生寄給他的盒子,信中說貝爾瑟先生要把盒子給杰克。give sth.to sb.“把某物給某人”,為固定結(jié)構(gòu),符合語境。答案:A 18.A.casually B.sadly C.nervously D.carefully 解析:casually “隨便地”;sadly “悲哀地”;nervously “緊張地”;carefully “仔細(xì)地,小心地”。杰克收到盒子后,應(yīng)該是小心地打開了盒子,故D項(xiàng)正確。答案:D 19.A.attached B.writing C.reading D.printed 解析:attach “附加,附屬”;write “寫”;read “閱讀”;print “印刷”。根據(jù)此空前的“...and also these words”可知,在那個(gè)手表上還附帶著一些話,attach...to...“把……附在……上”,為固定短語,符合語境。答案:A 20.A.on B.at C.in D.off 解析:call on “探望;拜訪”;call at “訪問”;call in “召集”;call off “取消”。貝爾瑟先生送給杰克的東西和對(duì)他的感謝讓杰克領(lǐng)悟到,即使工作再忙也要抽出時(shí)間與家人待在一起。此處表示“他立刻取消了接下來兩天的預(yù)約”,故D項(xiàng)正確。答案:D Ⅳ.七選五

根據(jù)短文內(nèi)容,從短文后的選項(xiàng)中選出能填入空白處的最佳選項(xiàng)。選項(xiàng)中有兩項(xiàng)為多余選項(xiàng)。If you’re finding it tough to find a job,try expanding your job-hunting plan to include the following key points: Set your target.1 A specific job hunt will be more efficient than an ordinary one.Plan enough interviews.Use every possible method to get interviews—answering ads,using search firms,contacting companies directly,surfing the Web,and networking.2

Follow up.3 Then,some weeks later,send another brief letter to explain that you still have not found the perfect position and that you will be available to interview again if the original position you applied for—or any other position,for that matter—is open.4

教育資料 5 You can’t find a job by looking at times.You have to make time for it.If you’re unemployed and looking,devote as much time as you would to a full-time job.A.Do this with every position you interview for,and you may just catch a break.B.Even if a job is not perfect for you,every interview can be approached as a positive experience.C.Value the chance of work.D.You should also be sure to target exactly what you want in a job.E.If you are well-dressed,it’s more likely that you can get the job.F.Even if someone does not hire you,write them a thank-you note for the interview.G.Make it your full-time job.答案:1~5 DBFAG

教育資料