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1、新編高考數(shù)學(xué)復(fù)習(xí)資料第五篇第3節(jié) 一、選擇題1已知數(shù)列an的前n項和Sn3nk(k為常數(shù)),那么下述結(jié)論正確的是()Ak為任意實數(shù)時,an是等比數(shù)列Bk1時,an是等比數(shù)列Ck0時,an是等比數(shù)列Dan不可能是等比數(shù)列解析:Sn3nk(k為常數(shù)),a1S13k,n2時,anSnSn13nk(3n1k)23n1,當(dāng)k1時,a12滿足an23n1,an是等比數(shù)列,當(dāng)k0時,a13不滿足an23n1,an不是等比數(shù)列故選B.答案:B2(2014河北石家莊一模)已知等比數(shù)列an,且a4a82,則a6(a22a6a10)的值為()A16B4C8D2解析:a6(a22a6a10)a6a22a6a6a6a1
2、0a2a4a8a(a4a8)24.故選B.答案:B3(2014湖北華中師大附中模擬)已知an是各項均為正數(shù)的等比數(shù)列,a1a21,a3a44,則a5a6a7a8等于()A80B20C32D解析:由等比數(shù)列前n項和性質(zhì)知,S2,S4S2,S6S4,S8S6也成等比數(shù)列,即1,4,a5a6,a7a8成等比數(shù)列,a5a616,a7a816464,a5a6a7a880.故選A.答案:A4(2014河北唐山市第三次模擬)若an為等比數(shù)列,a2a31,a3a42,則a5a6a7等于()A24B24C48D48解析:由已知得解得q2,a1,a5a6a7a5(1qq2)a1q4(1qq2)24.故選B.答案:
3、B5(2014亳州模擬)等比數(shù)列an中,a11,q2,則Tn的結(jié)果可化為()A1B1C.1D.1解析:由題意得an2n1,Tn1,故選C.答案:C6(2014安慶二模)已知等比數(shù)列an的公比為負數(shù),且an3an14a(nN*,n2),a22,則首項a1等于()A1B4C1D4解析:an3an14a(n2),a4a(n2),24(n2)又q0,q2,又a22,a11,故選C.答案:C二、填空題7(2014山東師大附中第三次模擬)已知等比數(shù)列an的公比為正數(shù),且a2a69a4,a21,則a1_.解析:由a2a69a4得a2(a2q4)9a2q2,解得q29,所以q3或q3(舍去),所以由a2a1q
4、,得a1.答案:8(2014河南省洛陽市高三檢測)已知等比數(shù)列an滿足an0,n1,2,3,且a5a2n522n(n3),則log2a1log2a3log2a2n1_.解析:a5a2n5a22n,且an0,an2n,log2a2n1log222n12n1,log2a1log2a3log2a2n1132n1n2.答案:n29(2012年高考遼寧卷)已知等比數(shù)列an為遞增數(shù)列,且aa10,2(anan2)5an1,則數(shù)列an的通項公式an_.解析:2(anan2)5an1,2an2anq25anq,即2q25q20,解得q2或q(舍去)又aa10a5q5,a5q52532,32a1q4,解得a12
5、,an22n12n,故an2n.答案:2n10(2013年高考遼寧卷)已知等比數(shù)列an是遞增數(shù)列,Sn是an的前n項和,若a1,a3是方程x25x40的兩個根,則S6_.解析:依題意a1a35,a1a34,又數(shù)列an為遞增數(shù)列,解得a11,a34,q24,q2,S663.答案:63三、解答題11一個項數(shù)為偶數(shù)的等比數(shù)列an,各項之和為偶數(shù)項之和的4倍,前3項之積為64,求此數(shù)列的通項公式解:設(shè)數(shù)列an的首項為a1,公比為q,全部奇數(shù)項、偶數(shù)項之和分別記為S奇、S偶,由題意知,S奇S偶4S偶,即S奇3S偶數(shù)列an的項數(shù)為偶數(shù),q.又a1a1qa1q264,aq364,即a112.故所求通項公式為an12n1.12(2014長春調(diào)研)已知數(shù)列an滿足a11,an12an1(nN*)(1)求證:數(shù)列an1是等比數(shù)列,并寫出數(shù)列an的通項公式;(2)若數(shù)列bn滿足4b114b214b314bn1(an1)n,求數(shù)列bn的前n項和Sn.(1)證明:an12an1,an112(an1),又a11,a1120,an10,2,數(shù)列an1是首項為2,公比為2的等比數(shù)列an12n,可得an2n1.(2)解:4b114b214b314bn1(an1)n,4b1b2b3bnn2n2,2(b1b2b3bn)2nn2,即2(b1b2b3bn)n22n,Snb1b2b3bnn2n.