《新編高考數(shù)學(xué)復(fù)習(xí):第五章 :第三節(jié) 等比數(shù)列及其前n項(xiàng)和演練知能檢測(cè)》由會(huì)員分享,可在線閱讀,更多相關(guān)《新編高考數(shù)學(xué)復(fù)習(xí):第五章 :第三節(jié) 等比數(shù)列及其前n項(xiàng)和演練知能檢測(cè)(5頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、新編高考數(shù)學(xué)復(fù)習(xí)資料第三節(jié)等比數(shù)列及其前n項(xiàng)和全盤鞏固1設(shè)Sn是等比數(shù)列an的前n項(xiàng)和,a3,S3,則公比q()A. B C1或 D1或解析:選C當(dāng)q1時(shí),a1a2a3,S3a1a2a3,符合題意;當(dāng)q1時(shí),由題意得解得q.故q1或q.2各項(xiàng)都為正數(shù)的等比數(shù)列an中,首項(xiàng)a13,前三項(xiàng)和為21,則a3a4a5()A33 B72 C84 D189解析:選Ca1a2a321,a1a1qa1q221,33q3q221,來(lái)源:即1qq27,解得q2或q3.an0,q2,a3a4a521q221484.3已知等比數(shù)列an滿足an0(nN*),且a5a2n522n(n3),則當(dāng)n1時(shí),log2a1log2
2、a3log2a5log2a2n1等于()A(n1)2 Bn2 Cn(2n1) D(n1)2解析:選B由等比數(shù)列的性質(zhì)可知a5a2n5a,又a5a2n522n,所以an2n.又log2a2n1log222n12n1,所以log2a1log2a3log2a5log2a2n1135(2n1)n2.4已知數(shù)列an滿足a15,anan12n,則()A2 B4 C5 D.解析:選B依題意得2,即2,故數(shù)列a1,a3,a5,a7,是一個(gè)以5為首項(xiàng)、2為公比的等比數(shù)列,因此4.5數(shù)列an中,已知對(duì)任意nN*,a1a2a3an3n1,則aaaa()A(3n1)2 B.(9n1) C9n1 D.(3n1)解析:選
3、Ba1a2a3an3n1,a1a2a3an13n11.由,得an3n3n123n1.當(dāng)n2時(shí),an3n3n123n1,又n1時(shí),a12適合上式,an23n1,故數(shù)列a是首項(xiàng)為4,公比為9的等比數(shù)列因此aaa(9n1)6已知an為等比數(shù)列,下面結(jié)論中正確的是()Aa1a32a2Baa2aC若a1a3,則a1a2D若a3a1,則a4a2解析:選B設(shè)an的首項(xiàng)為a1,公比為q,則a2a1q,a3a1q2.a1a3a1(1q2),又1q22q,當(dāng)a10時(shí),a1(1q2)2a1q,即a1a32a2;當(dāng)a10,aa2a,故B正確;若a1a3,則q21.q1.當(dāng)q1時(shí),a1a2;當(dāng)q1時(shí),a1a2,故C不正
4、確;D項(xiàng)中,若q0,則a3qa1q,即a4a2;若q0,則a3qa1q,此時(shí)a4a1a2an的最大正整數(shù)n的值為_解析:設(shè)等比數(shù)列的首項(xiàng)為a1,公比為q0,由得a1,q2.所以an2n6.a1a2an2n525,a1a2an2.由a1a2ana1a2an,得2n5252,由2n52,得n213n100,解得na1a2an,n13時(shí)不滿足a1a2ana1a2an,故n的最大值為12.來(lái)源:答案:1210數(shù)列an中,Sn1kan(k0,k1)(1)證明:數(shù)列an為等比數(shù)列;(2)求通項(xiàng)an;(3)當(dāng)k1時(shí),求和aaa.解:(1)證明:Sn1kan,Sn11kan1,得SnSn1kankan1(n2
5、),(k1)ankan1,為常數(shù),n2.an是公比為的等比數(shù)列(2)S1a11ka1,a1.ann1.(3)an中a1,q,a是首項(xiàng)為2,公比為2的等比數(shù)列當(dāng)k1時(shí),等比數(shù)列a的首項(xiàng)為,公比為,aaa.11已知函數(shù)f(x)的圖象過(guò)原點(diǎn),且關(guān)于點(diǎn)(1,2)成中心對(duì)稱(1)求函數(shù)f(x)的解析式;(2)若數(shù)列an滿足a12,an1f(an),證明數(shù)列為等比數(shù)列,并求出數(shù)列an的通項(xiàng)公式解:(1)f(0)0,c0.f(x)的圖象關(guān)于點(diǎn)(1,2)成中心對(duì)稱,f(x)f(2x)4,解得b2.f(x).(2)an1f(an),當(dāng)n2時(shí),2.又20,數(shù)列是首項(xiàng)為2,公比為2的等比數(shù)列,2n,an.來(lái)源:12
6、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(xiàng)公式;(2)若數(shù)列bn滿足4b114b214b314bn1(an1)n,求數(shù)列bn的前n項(xiàng)和Sn.解:(1)證明:an12an1,an112(an1),又a11,a1120,an10,2,數(shù)列an1是首項(xiàng)為2,公比為2的等比數(shù)列an12n,可得an2n1.(2)4b114b214b314bn1(an1)n,4b1b2b3bnn2n2,2(b1b2b3bn)2nn2,即2(b1b2b3bn)n22n,Snb1b2b3bnn2n.沖擊名校1設(shè)f(x)是定義在R上恒不為零的函數(shù),且對(duì)任意的實(shí)數(shù)x,y
7、R,都有f(x)f(y)f(xy),若a1,anf(n)(nN*),則數(shù)列an的前n項(xiàng)和Sn的取值范圍是_解析:由已知可得a1f(1),a2f(2)f(1)22,a3f(3)f(2)f(1)f(1)33,anf(n)f(1)nn,所以Sn23n1n.nN*,Sn1,且nN*)an1an3(SnSn1)3an,an14an(n1,nN*),a23S113a113t1,當(dāng)t1時(shí),a24a1,數(shù)列an是等比數(shù)列(2)在(1)的結(jié)論下,an14an,an14n,bnlog4an1n,cnanbn4n1n,Tnc1c2cn(401)(412)(4n1n)(14424n1)(123n).高頻滾動(dòng)1已知等差
8、數(shù)列an的前n項(xiàng)和為Sn,S440,Sn210,Sn4130,則n()A12 B14 C16 D18解析:選BSnSn4anan1an2an380,S4a1a2a3a440,所以4(a1an)120,a1an30,由Sn210,得n14.2已知數(shù)列an滿足a11,且an2an12n(n2,nN*)(1)求證:數(shù)列是等差數(shù)列,并求出數(shù)列an的通項(xiàng)公式;(2)求數(shù)列an的前n項(xiàng)和Sn.解:(1)證明:因?yàn)閍n2an12n,所以1,即1,所以數(shù)列是等差數(shù)列,且公差d1,其首項(xiàng),來(lái)源:所以(n1)1n,解得an2n(2n1)2n1.(2)Sn120321522(2n1)2n1,2Sn121322523(2n3)2n1(2n1)2n,得Sn12022122222n1(2n1)2n1(2n1)2n(32n)2n3.所以Sn(2n3)2n3.