新高考一輪復(fù)習(xí)導(dǎo)學(xué)案第42講 等比數(shù)列（解析版）

第42講等比數(shù)列1、等比數(shù)列的定義一般地，如果一個(gè)數(shù)列從第2項(xiàng)起，每一項(xiàng)與它的前一項(xiàng)的比都等于同一個(gè)常數(shù)，那么這個(gè)數(shù)列就叫做等比數(shù)列，這個(gè)常數(shù)叫做等比數(shù)列的公比，公比通常用字母__q__表示．2、等比數(shù)列的通項(xiàng)公式一般地，對(duì)于等比數(shù)列{an}的第n項(xiàng)an，有公式an＝a1qn－1，這就是等比數(shù)列{an}的通項(xiàng)公式，其中a1為首項(xiàng)，q為公比．第二通項(xiàng)公式為：an＝amqn－m．3、等比數(shù)列的前n項(xiàng)和公式等比數(shù)列{an}的前n項(xiàng)和公式：Sn＝eq\f(a1（1－qn）,1－q)(q≠1)或Sn＝eq\f(a1－anq,1－q)(q≠1)．注意：(1)當(dāng)q＝1時(shí)，該數(shù)列是各項(xiàng)不為零的常數(shù)列，Sn＝na1；(2)有關(guān)等比數(shù)列的求和問(wèn)題，當(dāng)q不能確定時(shí)，應(yīng)分q＝1，q≠1來(lái)討論．4、等比數(shù)列的性質(zhì)(1)若a，G，b成等比數(shù)列，則稱G為a和b的等比中項(xiàng)，則G2＝ab.(2)等比數(shù)列{an}中，若m＋n＝k＋l(m，n，k，l∈N*)，則有am·an＝ak·al，特別地，當(dāng)m＋n＝2p時(shí)，am·an＝aeq\o\al(2,p).(3)設(shè)Sm是等比數(shù)列{an}的前n項(xiàng)和，則Sm，S2m－Sm，S3m－S2m滿足關(guān)系式(S2m－Sm)2＝Sm·(S3m－S2m)．(4)等比數(shù)列的單調(diào)性，若首項(xiàng)a1＞0，公比q>1或首項(xiàng)a1＜0，公比0<q<1，則數(shù)列為遞增數(shù)列；若首項(xiàng)a1＞0，公比0<q<1或首項(xiàng)a1＜0，公比q＞1，則數(shù)列為遞減數(shù)列；若公比q＝1，則數(shù)列為常數(shù)列；公比q＜0，則數(shù)列為擺動(dòng)數(shù)列．(5)若{an}和{bn}均為等比數(shù)列，則{λan}(λ≠0)、{|an|}、eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(1,an)))、{aeq\o\al(2,n)}、eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(an,bn)))、{manbn}(m≠0)仍為等比數(shù)列．1、（2022?乙卷（文））已知等比數(shù)列SKIPIF1<0的前3項(xiàng)和為168，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．14 B．12 C．6 D．3【答案】SKIPIF1<0【解析】設(shè)等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0，SKIPIF1<0，由題意，SKIPIF1<0．SKIPIF1<0前3項(xiàng)和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故選：SKIPIF1<0．2、（2021?甲卷（文））記SKIPIF1<0為等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．7 B．8 C．9 D．10【答案】SKIPIF1<0【解析】SKIPIF1<0為等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，SKIPIF1<0，SKIPIF1<0，由等比數(shù)列的性質(zhì)，可知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列，SKIPIF1<0，2，SKIPIF1<0成等比數(shù)列，SKIPIF1<0，解得SKIPIF1<0．故選：SKIPIF1<0．3、（2023?甲卷（文））記SKIPIF1<0為等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．若SKIPIF1<0，則SKIPIF1<0的公比為．【答案】SKIPIF1<0．【解析】等比數(shù)列SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0．4、（2021?上海）已知SKIPIF1<0為無(wú)窮等比數(shù)列，SKIPIF1<0，SKIPIF1<0的各項(xiàng)和為9，SKIPIF1<0，則數(shù)列SKIPIF1<0的各項(xiàng)和為．【答案】SKIPIF1<0．【解析】設(shè)SKIPIF1<0的公比為SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0的各項(xiàng)和為9，可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，可得數(shù)列SKIPIF1<0是首項(xiàng)為2，公比為SKIPIF1<0的等比數(shù)列，則數(shù)列SKIPIF1<0的各項(xiàng)和為SKIPIF1<0．故答案為：SKIPIF1<0．5、（2023?乙卷（理））已知SKIPIF1<0為等比數(shù)列，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0．【解析】SKIPIF1<0等比數(shù)列SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，而SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0．6、（2021?甲卷（理））等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0，前SKIPIF1<0項(xiàng)和為SKIPIF1<0．設(shè)甲：SKIPIF1<0，乙：SKIPIF1<0是遞增數(shù)列，則SKIPIF1<0SKIPIF1<0A．甲是乙的充分條件但不是必要條件 B．甲是乙的必要條件但不是充分條件 C．甲是乙的充要條件 D．甲既不是乙的充分條件也不是乙的必要條件【答案】SKIPIF1<0【解析】若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0是遞減數(shù)列，不滿足充分性；SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0是遞增數(shù)列，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足必要性，故甲是乙的必要條件但不是充分條件，故選：SKIPIF1<0．7、（2023?天津）已知SKIPIF1<0為等比數(shù)列，SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，SKIPIF1<0，則SKIPIF1<0的值為SKIPIF1<0SKIPIF1<0A．3 B．18 C．54 D．152【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0為等比數(shù)列，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由等比數(shù)列的性質(zhì)可得，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0（舍SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．故選：SKIPIF1<0．8、（2023?甲卷（理））已知等比數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0前SKIPIF1<0項(xiàng)和，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．7 B．9 C．15 D．30【答案】SKIPIF1<0【解析】等比數(shù)列SKIPIF1<0中，設(shè)公比為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0前SKIPIF1<0項(xiàng)和，SKIPIF1<0，顯然SKIPIF1<0，（如果SKIPIF1<0，可得SKIPIF1<0矛盾，如果SKIPIF1<0，可得SKIPIF1<0矛盾），可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．沒(méi)有選項(xiàng)．故選：SKIPIF1<0．9、（2022?上海）已知等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，前SKIPIF1<0項(xiàng)積為SKIPIF1<0，則下列選項(xiàng)判斷正確的是SKIPIF1<0SKIPIF1<0A．若SKIPIF1<0，則數(shù)列SKIPIF1<0是遞增數(shù)列 B．若SKIPIF1<0，則數(shù)列SKIPIF1<0是遞增數(shù)列 C．若數(shù)列SKIPIF1<0是遞增數(shù)列，則SKIPIF1<0 D．若數(shù)列SKIPIF1<0是遞增數(shù)列，則SKIPIF1<0【答案】SKIPIF1<0【解析】如果數(shù)列SKIPIF1<0，公比為SKIPIF1<0，滿足SKIPIF1<0，但是數(shù)列SKIPIF1<0不是遞增數(shù)列，所以SKIPIF1<0不正確；如果數(shù)列SKIPIF1<0，公比為SKIPIF1<0，滿足SKIPIF1<0，但是數(shù)列SKIPIF1<0不是遞增數(shù)列，所以SKIPIF1<0不正確；如果數(shù)列SKIPIF1<0，公比為SKIPIF1<0，SKIPIF1<0，數(shù)列SKIPIF1<0是遞增數(shù)列，但是SKIPIF1<0，所以SKIPIF1<0不正確；數(shù)列SKIPIF1<0是遞增數(shù)列，可知SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0正確，所以SKIPIF1<0正確；故選：SKIPIF1<0．10、（2023?新高考Ⅱ）記SKIPIF1<0為等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．120 B．85 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】等比數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，顯然公比SKIPIF1<0，設(shè)首項(xiàng)為SKIPIF1<0，則SKIPIF1<0①，SKIPIF1<0②，化簡(jiǎn)②得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（不合題意，舍去），代入①得SKIPIF1<0，所以SKIPIF1<0．故選：SKIPIF1<0．1、若等比數(shù)列{an}的各項(xiàng)均為正數(shù)，a2＝3,4aeq\o\al(2,3)＝a1a7，則a5等于()A.eq\f(3,4)B.eq\f(3,8)C．12D．24【答案】：D【解析】：數(shù)列{an}是等比數(shù)列，各項(xiàng)均為正數(shù)，4aeq\o\al(2,3)＝a1a7＝aeq\o\al(2,4)，所以q2＝eq\f(a\o\al(2,4),a\o\al(2,3))＝4，所以q＝2.所以a5＝a2·q3＝3×23＝24，故選D.2、等比數(shù)列{an}的前n項(xiàng)和為Sn＝32n－1＋r，則r的值為()A.eq\f(1,3)B．－eq\f(1,3)C.eq\f(1,9)D．－eq\f(1,9)【答案】：B【解析】：當(dāng)n＝1時(shí)，a1＝S1＝3＋r，當(dāng)n≥2時(shí)，an＝Sn－Sn－1＝32n－1－32n－3＝32n－3(32－1)＝8·32n－3＝8·32n－2·3－1＝eq\f(8,3)·9n－1，所以3＋r＝eq\f(8,3)，即r＝－eq\f(1,3)，故選B.3、已知遞增的等比數(shù)列{an}中，a2＝6，a1＋1，a2＋2，a3成等差數(shù)列，則該數(shù)列的前6項(xiàng)和S6等于()A．93B．189C.eq\f(189,16)D．378【答案】：B【解析】：設(shè)數(shù)列{an}的公比為q，由題意可知，q>1，且2(a2＋2)＝a1＋1＋a3，即2×(6＋2)＝eq\f(6,q)＋1＋6q，整理可得2q2－5q＋2＝0，則q＝2eq\b\lc\(\rc\)(\a\vs4\al\co1(q＝\f(1,2)舍去))，則a1＝eq\f(6,2)＝3，∴數(shù)列{an}的前6項(xiàng)和S6＝eq\f(3×\b\lc\(\rc\)(\a\vs4\al\co1(1－26)),1－2)＝189.4、（2022年廣州附屬中學(xué)高三模擬試卷）已知等比數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0______．【答案】SKIPIF1<0【解析】設(shè)等比數(shù)列的公比為SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0；故答案為：SKIPIF1<05、（2023·云南紅河·統(tǒng)考一模）在數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0為等比數(shù)列，則SKIPIF1<0____________．【答案】127【詳解】設(shè)等比數(shù)列SKIPIF1<0的公比為q，則SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，故SKIPIF1<0．故答案為：127.考向一等比數(shù)列的基本運(yùn)算例1、（1）設(shè)正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn，若S2＝3，S4＝15，則公比q等于()A．5B．4C．3D．2（2）已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前4項(xiàng)和為15，且a5＝3a3＋4a1，則a3等于()A．16B．8C．4D．2（3）（2021·吉林延邊朝鮮族自治州·高三月考（文））已知各項(xiàng)均為正數(shù)且單調(diào)遞減的等比數(shù)列SKIPIF1<0滿足SKIPIF1<0、SKIPIF1<0、SKIPIF1<0成等差數(shù)列.其前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】．（1）D（2）C（3）C【解析】（1）：因?yàn)镾2＝3，S4＝15，S4－S2＝12，所以eq\b\lc\{\rc\(\a\vs4\al\co1(a1＋a2＝3，,a3＋a4＝12，))兩個(gè)方程左右兩邊分別相除，得q2＝4，因?yàn)閿?shù)列是正項(xiàng)等比數(shù)列，所以q＝2，故選D.：設(shè)等比數(shù)列{an}的公比為q，由a5＝3a3＋4a1得q4＝3q2＋4，得q2＝4，因?yàn)閿?shù)列{an}的各項(xiàng)均為正數(shù)，所以q＝2，又a1＋a2＋a3＋a4＝a1(1＋q＋q2＋q3)＝a1(1＋2＋4＋8)＝15，所以a1＝1，所以a3＝a1q2＝4.（3）【解析】：由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列，得：SKIPIF1<0，設(shè)SKIPIF1<0的公比為SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0單調(diào)遞減，SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0數(shù)列SKIPIF1<0的通項(xiàng)公式為：SKIPIF1<0，SKIPIF1<0.故選：C．方法總結(jié)：(1)等比數(shù)列基本量的運(yùn)算是等比數(shù)列中的一類基本問(wèn)題，等比數(shù)列中有五個(gè)量a1，n，q，an，Sn，一般可以“知三求二”，通過(guò)列方程(組)便可迎刃而解；（2）等比數(shù)列的前n項(xiàng)和公式涉及對(duì)公比q的分類討論，當(dāng)q＝1時(shí)，{an}的前n項(xiàng)和Sn＝na1；當(dāng)q≠1時(shí)，{an}的前n項(xiàng)和Sn＝eq\f(a11－qn,1－q)＝eq\f(a1－anq,1－q)?？枷蚨缺葦?shù)列的性質(zhì)例2、(1)已知等比數(shù)列{an}的各項(xiàng)為正數(shù)，且a5a6＋a4a7＝18，則log3a1＋log3a2＋…＋log3a10＝()A．12 B．10C．8 D．2＋log35(2)設(shè)等比數(shù)列{an}中，前n項(xiàng)和為Sn，已知S3＝8，S6＝7，則a7＋a8＋a9等于()A.eq\f(1,8) B．－eq\f(1,8)C.eq\f(57,8) D.eq\f(55,8)(3)已知等比數(shù)列{an}共有2n項(xiàng)，其和為－240，且奇數(shù)項(xiàng)的和比偶數(shù)項(xiàng)的和大80，則公比q＝________.【答案】(1)B(2)A(3)2【解析】(1)由a5a6＋a4a7＝18，得a5a6＝9，所以log3a1＋log3a2＋…＋log3a10＝log3(a1a2…a10)＝log3(a5a6)5＝5log39＝10.(2)因?yàn)閍7＋a8＋a9＝S9－S6，且S3，S6－S3，S9－S6也成等比數(shù)列，即8，－1，S9－S6成等比數(shù)列，所以8(S9－S6)＝1，即S9－S6＝eq\f(1,8)，所以a7＋a8＋a9＝eq\f(1,8).(3)由題意，得eq\b\lc\{\rc\(\a\vs4\al\co1(S奇＋S偶＝－240，,S奇－S偶＝80，))解得eq\b\lc\{\rc\(\a\vs4\al\co1(S奇＝－80，,S偶＝－160，))所以q＝eq\f(S偶,S奇)＝eq\f(－160,－80)＝2.變式1、(1)在等比數(shù)列{an}中，若a1a2a3a4＝1，a13a14a15a16＝8，則a41a42a43a44＝________；【答案】1024【解析】由等比數(shù)列的性質(zhì)可知，依次4項(xiàng)的積為等比數(shù)列，設(shè)其公比為q，T1＝a1a2a3a4＝1，T4＝a13a14a15a16＝8，所以T4＝T1q3＝1·q3＝8，即q＝2，所以T11＝a41a42a43a44＝T1·q10＝210＝1024.(2)已知數(shù)列{an}為等比數(shù)列，Sn為其前n項(xiàng)和，n∈N*，若a1＋a2＋a3＝3，a4＋a5＋a6＝6，則S12＝________；【答案】45【解析】設(shè)等比數(shù)列{an}的公比為q，則eq\f(a4＋a5＋a6,a1＋a2＋a3)＝q3＝eq\f(6,3)＝2.因?yàn)镾6＝a1＋a2＋a3＋a4＋a5＋a6＝9，S12－S6＝a7＋a8＋a9＋a10＋a11＋a12，所以eq\f(S12－S6,S6)＝eq\f(a7＋a8＋a9＋a10＋a11＋a12,a1＋a2＋a3＋a4＋a5＋a6)＝q6＝4，所以S12＝5S6＝45.(3)已知{an}是等比數(shù)列，a2＝2，a5＝eq\f(1,4)，則a1a2＋a2a3＋…＋anan＋1＝________．【答案】eq\f(32,3)(1－4－n)【解析】因?yàn)閍2＝2，a5＝eq\f(1,4)，所以a1＝4，q＝eq\f(1,2)，所以a1a2＋a2a3＋…＋anan＋1＝eq\f(8\b\lc\[\rc\](\a\vs4\al\co1(1－\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,4)))\s\up12(n))),1－\f(1,4))＝eq\f(32,3)(1－4－n).方法總結(jié)：(1)在解決等比數(shù)列的有關(guān)問(wèn)題時(shí)，要注意挖掘隱含條件，利用性質(zhì)，特別是性質(zhì)“若m＋n＝p＋q(m，n，p，q∈N*)，則am·an＝ap·aq”，可以減少運(yùn)算量，提高解題速度．(2)在應(yīng)用相應(yīng)性質(zhì)解題時(shí)，要注意性質(zhì)成立的前提條件，有時(shí)需要進(jìn)行適當(dāng)變形．此外，解題時(shí)注意設(shè)而不求思想的運(yùn)用考向三等比數(shù)列的判定與證明例3、（1）設(shè)等比數(shù)列{an}的公比為q，則下列結(jié)論正確的是()A．?dāng)?shù)列{anan＋1}是公比為q的等比數(shù)列B．?dāng)?shù)列{an＋an＋1}是公比為q的等比數(shù)列C．?dāng)?shù)列{an－an＋1}是公比為q的等比數(shù)列D．?dāng)?shù)列eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(1,an)))是公比為eq\f(1,q)的等比數(shù)列【答案】：D【解析】：對(duì)于A，由eq\f(anan＋1,an－1an)＝q2(n≥2)知其是公比為q2的等比數(shù)列；對(duì)于B，若q＝－1，則{an＋an＋1}項(xiàng)中有0，不是等比數(shù)列；對(duì)于C，若q＝1，則數(shù)列{an－an＋1}項(xiàng)中有0，不是等比數(shù)列；對(duì)于D，eq\f(\f(1,an＋1),\f(1,an))＝eq\f(an,an＋1)＝eq\f(1,q)，所以數(shù)列eq\b\lc\{\rc\}(\a\vs4\al\co1(\f(1,an)))是公比為eq\f(1,q)的等比數(shù)列，故選D.（2）（2023·安徽蚌埠·統(tǒng)考三模）（多選）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)積為SKIPIF1<0，則下列結(jié)論正確的是(

)A．?dāng)?shù)列SKIPIF1<0是等差數(shù)列 B．?dāng)?shù)列SKIPIF1<0是等差數(shù)列C．?dāng)?shù)列SKIPIF1<0是等比數(shù)列 D．?dāng)?shù)列SKIPIF1<0是等差數(shù)列【答案】ABC【詳解】設(shè)等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0．對(duì)于A選項(xiàng)，SKIPIF1<0，∴SKIPIF1<0為等差數(shù)列，A正確；對(duì)于B選項(xiàng)，令SKIPIF1<0，∴SKIPIF1<0，故數(shù)列SKIPIF1<0是等差數(shù)列，B正確；設(shè)等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0，對(duì)于C選項(xiàng)，令SKIPIF1<0，則SKIPIF1<0，故數(shù)列SKIPIF1<0是等比數(shù)列，C正確；對(duì)于D選項(xiàng)，∵SKIPIF1<0不一定為常數(shù)，故數(shù)列SKIPIF1<0不一定是等差數(shù)列，故D錯(cuò)誤；故選：ABC．變式1、(1)已知數(shù)列{an}的前n項(xiàng)和為Sn，且an＋Sn＝n.若數(shù)列{bn}滿足b1＝a1，bn＝an－an－1(n≥2)，求證：數(shù)列{bn}是等比數(shù)列；(2)已知數(shù)列{an}滿足a1＝1，a2＝2，an＋2＝eq\f(an＋an＋1,2)，n∈N*.①令bn＝an＋1－an，求證：{bn}是等比數(shù)列；②求數(shù)列{an}的通項(xiàng)公式．【解析】(1)因?yàn)橛衫?(2)知an＝1－eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))eq\s\up12(n)，所以當(dāng)n≥2時(shí)，bn＝an－an－1＝1－eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))eq\s\up12(n)－eq\b\lc\[\rc\](\a\vs4\al\co1(1－\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))\s\up12(n－1)))＝eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))eq\s\up12(n－1)－eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))eq\s\up12(n)＝eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))eq\s\up12(n).又b1＝a1＝eq\f(1,2)也符合上式，所以bn＝eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))eq\s\up12(n).因?yàn)閑q\f(bn＋1,bn)＝eq\f(1,2)，所以數(shù)列{bn}是等比數(shù)列．(2)①由題意，得b1＝a2－a1＝1.當(dāng)n≥2時(shí)，bn＝an＋1－an＝eq\f(an－1＋an,2)－an＝－eq\f(1,2)(an－an－1)＝－eq\f(1,2)bn－1，則eq\f(bn,bn－1)＝－eq\f(1,2).故{bn}是以1為首項(xiàng)，－eq\f(1,2)為公比的等比數(shù)列．②由①知bn＝an＋1－an＝eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))eq\s\up12(n－1)，當(dāng)n≥2時(shí)，an＝a1＋(a2－a1)＋(a3－a2)＋…＋(an－an－1)＝1＋1＋eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))＋…＋eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))eq\s\up12(n－2)＝1＋eq\f(1－\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))\s\up12(n－1),1－\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2))))＝1＋eq\f(2,3)[1－eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))eq\s\up12(n－1)]＝eq\f(5,3)－eq\f(2,3)eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))eq\s\up12(n－1).當(dāng)n＝1時(shí)，eq\f(5,3)－eq\f(2,3)×eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))eq\s\up12(1－1)＝1＝a1，故an＝eq\f(5,3)－eq\f(2,3)eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)))eq\s\up12(n－1)(n∈N*).變式2、（2022年河北省高三大聯(lián)考模擬試卷）已知數(shù)列SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0（1）求SKIPIF1<0，SKIPIF1<0的值，并證明數(shù)列SKIPIF1<0是等比數(shù)列；（2）求數(shù)列SKIPIF1<0，SKIPIF1<0的通項(xiàng)公式．【解析】（1）∵SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0.∵SKIPIF1<0，∴SKIPIF1<0＝SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0是SKIPIF1<0為首項(xiàng)，SKIPIF1<0為公比的等比數(shù)列（2）由（1）知SKIPIF1<0是SKIPIF1<0為首項(xiàng)，SKIPIF1<0為公比的等比數(shù)列.∴SKIPIF1<0，∴SKIPIF1<0∵SKIPIF1<0，∴SKIPIF1<0∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0也適合上式所以數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0.方法總結(jié)：證明一個(gè)數(shù)列為等差數(shù)列或者等比數(shù)列常用定義法與等差、等比中項(xiàng)法，其他方法只用于選擇、填空題中的判定；若證明某數(shù)列不是等差或等比數(shù)列，則只要證明存在連續(xù)三項(xiàng)不成等差或等比數(shù)列即可．而研究數(shù)列中的取值范圍問(wèn)題，一般都是通過(guò)研究數(shù)列的單調(diào)性來(lái)進(jìn)行求解1、（2022年河北省張家口高三模擬試卷）已知等比數(shù)列SKIPIF1<0各項(xiàng)均為正數(shù)，且SKIPIF1<0，則SKIPIF1<0（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0或SKIPIF1<0【答案】A【解析】【詳解】根據(jù)等比數(shù)列的通項(xiàng)公式可知SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0（舍），故選：A2、（2022年河北省衡水中學(xué)高三模擬試卷）等比數(shù)列{an}中，每項(xiàng)均為正數(shù)，且a3a8＝81，則log3a1＋log3a2＋…＋log3a10等于（）A.5 B.10 C.20 D.40【答案】C【解析】【詳解】SKIPIF1<0是等比數(shù)列，則SKIPIF1<0，所以log3a1＋log3a2＋…＋log3a10SKIPIF1<0．故選：C．3、（2023·吉林長(zhǎng)春·統(tǒng)考三模）已知等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），若SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．4【答案】C【詳解】已知等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故選：C.4、（2023·黑龍江·黑龍江實(shí)驗(yàn)中學(xué)?？家荒＃ǘ噙x題）已知SKIPIF1<0是等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，且SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】AD【分析】根據(jù)SKIPIF1<0與SKIPIF1<0的關(guān)系以及SKIPIF1<0是等比數(shù)列，可求得SKIPIF