第三節(jié)等比數(shù)列【教材·知識(shí)梳理】1.等比數(shù)列定義與等比中項(xiàng)(1)定義：①文字語(yǔ)言：從______起，每一項(xiàng)與它的前一項(xiàng)的___都等于___一個(gè)常數(shù)；②符號(hào)語(yǔ)言：_________(n∈N*，q為非零常數(shù)).(2)等比中項(xiàng)：如果a，G，b成等比數(shù)列，那么__叫做a與b的等比中項(xiàng).即：G是a與b的等比中項(xiàng)?a，G，b成等比數(shù)列?G2=___.第2項(xiàng)比同Gab2.等比數(shù)列的通項(xiàng)公式及前n項(xiàng)和公式(1)通項(xiàng)公式：an=_____.(2)前n項(xiàng)和公式：Sn=a1qn-13.等比數(shù)列的性質(zhì)(1)通項(xiàng)公式的推廣：an=am·qn-m(m，n∈N*).(2)對(duì)任意的正整數(shù)m，n，p，q，若m+n=p+q，則______=______.特別地，若m+n=2p，則_______________.am·anap·aq

【知識(shí)點(diǎn)辨析】

(正確的打“√”,錯(cuò)誤的打“×”)(1)等比數(shù)列{an}的公比q>1,則該數(shù)列單調(diào)遞增. (

)(2)數(shù)列{an},{bn}都是等比數(shù)列,則數(shù)列{anbn},仍然是等比數(shù)列. (

)(3)數(shù)列{an}的通項(xiàng)an=an,則其前n項(xiàng)和為Sn=. (

)(4)設(shè)Sn是等比數(shù)列{an}的前n項(xiàng)和,則Sn,S2n-Sn,S3n-S2n是等比數(shù)列. (

)提示:(1)×.在等比數(shù)列{an}中,若公比q>1,當(dāng)a1>0時(shí),該數(shù)列為遞增數(shù)列,a1<0時(shí),該數(shù)列為遞減數(shù)列,所以(1)錯(cuò)誤.(2)√.設(shè)等比數(shù)列{an},{bn}的公比分別為q1,q2,則=q1q2(與n無(wú)關(guān)的常數(shù)),(與n無(wú)關(guān)的常數(shù)),所以(2)正確.(3)×.對(duì)于數(shù)列{an},當(dāng)a=1時(shí),Sn=n,當(dāng)a≠1時(shí),則其前n項(xiàng)和為Sn=,所以(3)錯(cuò)誤.(4)×.在公比q=-1,n為偶數(shù)時(shí),Sn=S2n-Sn=S3n-S2n=0,此時(shí),Sn,S2n-Sn,S3n-S2n不是等比數(shù)列,所以(4)錯(cuò)誤.【易錯(cuò)點(diǎn)索引】序號(hào)易錯(cuò)警示典題索引1運(yùn)算錯(cuò)誤考點(diǎn)一、T52q=1的特殊性是否討論考點(diǎn)一、T23不能進(jìn)行正確轉(zhuǎn)化考點(diǎn)二、T24等比數(shù)列性質(zhì)應(yīng)用錯(cuò)誤考點(diǎn)三、角度1,2【教材·基礎(chǔ)自測(cè)】1.(必修5P48練習(xí)BT1改編)對(duì)任意等比數(shù)列{an},下列說(shuō)法一定正確的是(

)A.a1,a3,a9成等比數(shù)列 B.a2,a3,a6成等比數(shù)列C.a2,a4,a8成等比數(shù)列 D.a3,a6,a9成等比數(shù)列【解析】選D.因?yàn)閿?shù)列{an}為等比數(shù)列,設(shè)其公比為q,則a3·a9=a1·q2·a1·q8=

所以a3,a6,a9一定成等比數(shù)列.2.(必修5P51習(xí)題2-3AT1改編)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且a1+a3=,a2+a4=,則公比q= (

)【解析】選C.由等比數(shù)列可得,a2+a4=(a1+a3)q=q=,解得q=2.3.(必修5P46例4改編)在3與192中間插入兩個(gè)數(shù),使它們同這兩個(gè)數(shù)成等比數(shù)列,則這兩個(gè)數(shù)為_(kāi)_______.

【解析】設(shè)該數(shù)列的公比為q,由題意知,192=3×q3,q3=64,所以q=4.所以插入的兩個(gè)數(shù)分別為3×4=12,12×4=48.答案:12,484.(必修5P51練習(xí)BT1改編)等比數(shù)列{an}的首項(xiàng)a1=-1,前n項(xiàng)和為Sn,若,則{an}的通項(xiàng)公式an=________.

【解析】因?yàn)?所以,因?yàn)镾5,S10-S5,S15-S10成等比數(shù)列,且公比為q5,所以q5=-,q=-,則an=.答案:

【解題新思維】活用等比數(shù)列前n項(xiàng)和的性質(zhì)解題

【結(jié)論】在等比數(shù)列{an}中,其前n項(xiàng)和為Sn,當(dāng)公比q≠-1時(shí),Sn,S2n-Sn,S3n-S2n,…成等比數(shù)列(n∈N*).【典例】設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,若=3,則=________.

【解析】由等比數(shù)列的性質(zhì)S3,S6-S3,S9-S6仍成等比數(shù)列,由已知得S6=3S3,所以即S9-S6=4S3,S9=7S3,所以.答案:

【一題多解】因?yàn)閧an}為等比數(shù)列,由=3,設(shè)S6=3a,S3=a(a≠0),所以S3,S6-S3,S9-S6為等比數(shù)列,即a,2a,S9-S6成等比數(shù)列,所以S9-S6=4a,解得S9=7a,所以答案:

【遷移應(yīng)用】已知數(shù)列{an}是等比數(shù)列,Sn為其前n項(xiàng)和,若a1+a2+a3=4,a4+a5+a6=8,則S12=(

)A.40 B.60 C.32 D.50【解析】選B.數(shù)列S3,S6-S3,S9-S6,S12-S9