【全程復(fù)習(xí)方略】高考數(shù)學(xué) 5.3等比數(shù)列課時(shí)提升作業(yè) 理 北師大版.doc

【全程復(fù)習(xí)方略】2014版高考數(shù)學(xué) 5.3等比數(shù)列課時(shí)提升作業(yè) 理 北師大版一、選擇題1.已知等比數(shù)列an的公比q=2,其前4項(xiàng)和s4=60,則a2等于()(a)8(b)6(c)-8(d)-62.(2013吉安模擬)已知a1,a2a1,a3a2,anan-1,是首項(xiàng)為1,公比為2的等比數(shù)列,則數(shù)列an的第100項(xiàng)等于()(a)25050(b)24950(c)2100(d)2993.在正項(xiàng)等比數(shù)列an中,a1,a19分別是方程x2-10x+16=0的兩根,則a8a10a12等于()(a)16(b)32(c)64(d)2564.設(shè)等比數(shù)列an的公比q=2,前n項(xiàng)和為sn,則s4a3的值為()(a)154(b)152(c)74(d)725.(2013沈陽(yáng)模擬)已知數(shù)列an滿足log3an+1=log3an+1(nn+)且a2+a4+a6=9,則log13(a5+a7+a9)的值是()(a)-5(b)-15(c)5(d)156.設(shè)等比數(shù)列an的前n項(xiàng)和為sn,若a2011=3s2010+2012,a2010=3s2009+2012,則公比q=()(a)4(b)1或4(c)2(d)1或27.公差不為零的等差數(shù)列an的前n項(xiàng)和為sn,若a4是a3與a7的等比中項(xiàng),s8=32,則s10等于()(a)18(b)24(c)60(d)908.(2013漢中模擬)在等比數(shù)列an中,a6與a7的等差中項(xiàng)等于48,a4a5a6a7a8a9a10=1286.如果設(shè)數(shù)列an的前n項(xiàng)和為sn,那么sn=()(a)5n-4(b)4n-3(c)3n-2(d)2n-1二、填空題9.(2012廣東高考)若等比數(shù)列an滿足a2a4=12,則a1a5=.10.已知等比數(shù)列an的首項(xiàng)為2,公比為2,則aan+1aa1aa2aa3aan=.11.(能力挑戰(zhàn)題)設(shè)數(shù)列an的前n項(xiàng)和為sn,已知a1=1,sn+1=2sn+n+1(nn+),則數(shù)列an的通項(xiàng)公式an=.三、解答題12.(2013寶雞模擬)已知數(shù)列an滿足:a1=2,an+1=2an+1.(1)證明:數(shù)列an+1為等比數(shù)列.(2)求數(shù)列an的通項(xiàng)公式.13.(2013西安模擬)已知數(shù)列an的首項(xiàng)為a1=1,其前n項(xiàng)和為sn,且對(duì)任意正整數(shù)n有n,an,sn成等差數(shù)列.(1)求證:數(shù)列sn+n+2成等比數(shù)列.(2)求數(shù)列an的通項(xiàng)公式.14.(能力挑戰(zhàn)題)已知an是各項(xiàng)均為正數(shù)的等比數(shù)列,且a1+a2=2(1a1+1a2),a3+a4+a5=64(1a3+1a4+1a5),(1)求an的通項(xiàng)公式.(2)設(shè)bn=(an+1an)2,求數(shù)列bn的前n項(xiàng)和tn.15.(能力挑戰(zhàn)題)設(shè)一元二次方程anx2-an+1x+1=0(n=1,2,3,)有兩根和,且滿足6-2+6=3.(1)試用an表示an+1.(2)求證:數(shù)列an-23是等比數(shù)列.(3)當(dāng)a1=76時(shí),求數(shù)列an的通項(xiàng)公式.答案解析1.【解析】選a.s4=60,q=2a1(1-q4)1-q=60a1=4,a2=a1q=42=8.2.【解析】選b.假設(shè)a0=1,數(shù)列anan-1的通項(xiàng)公式是anan-1=2n-1.所以a100=a1a2a1a3a2a100a99=20+1+99=24950.3.【解析】選c.根據(jù)根與系數(shù)的關(guān)系得a1a19=16,由此得a10=4,a8a12=16,故a8a10a12=64.4.【解析】選a.s4a3=a1(1-q4)1-qa1q2=1-q4(1-q)q2=-15-4=154.5.【思路點(diǎn)撥】根據(jù)數(shù)列滿足log3an+1=log3an+1(nn+)且a2+a4+a6=9可以確定數(shù)列是公比為3的等比數(shù)列,再根據(jù)等比數(shù)列的通項(xiàng)公式即可通過(guò)a2+a4+a6=9求出a5+a7+a9的值.【解析】選a.由log3an+1=log3an+1(nn+),得an+1=3an,又因?yàn)閍n0,所以數(shù)列an是公比為3的等比數(shù)列,a5+a7+a9=(a2+a4+a6)33=35,所以log13(a5+a7+a9)=-log335=-5.6.【解析】選a.由a2011=3s2010+2012,a2010=3s2009+2012兩式相減得a2011-a2010=3a2010,即q=4.7.【解析】選c.由a42=a3a7得(a1+3d)2=(a1+2d)(a1+6d),又因?yàn)楣畈粸榱?所以2a1+3d=0,再由s8=8a1+562d=32得2a1+7d=8,則d=2,a1=-3,所以s10=10a1+902d=60.故選c.8.【解析】選d.設(shè)等比數(shù)列an的公比為q,由a6與a7的等差中項(xiàng)等于48,得a6+a7=96,即a1q5(1+q)=96,由等比數(shù)列的性質(zhì),得a4a10=a5a9=a6a8=a72,因?yàn)閍4a5a6a7a8a9a10=1286,則a77=1286=(26)7,即a1q6=26,由解得a1=1,q=2,sn=1-2n1-2=2n-1,故選d.9.【思路點(diǎn)撥】本題考查了等比數(shù)列的性質(zhì):已知m,n,pn+,若m+n=2p,則aman=ap2.【解析】a2a4=12,a32=12,a1a32a5=a34=14.答案：1410.【解析】由題意知an=2n,所以aan+1aa1aa2aa3aan=2an+12a1+a2+an=22n+122n+1-2=22=4.答案：411.【解析】sn+1=2sn+n+1,當(dāng)n2時(shí)sn=2sn-1+n,兩式相減得:an+1=2an+1,an+1+1=2(an+1),即an+1+1an+1=2.又s2=2s1+1+1,a1=s1=1,a2=3,a2+1a1+1=2,an+1是首項(xiàng)為2,公比為2的等比數(shù)列,an+1=2n即an=2n-1(nn+).答案：2n-1【方法技巧】含sn,an問(wèn)題的求解策略當(dāng)已知含有sn+1,sn之間的等式時(shí),或者含有sn,an的混合關(guān)系的等式時(shí),可以采用降級(jí)角標(biāo)或者升級(jí)角標(biāo)的方法再得出一個(gè)等式,兩個(gè)等式相減就把問(wèn)題轉(zhuǎn)化為數(shù)列的通項(xiàng)之間的遞推關(guān)系式.12.【解析】(1)an+1+1an+1=2an+1+1an+1=2,所以an+1是以2為公比的等比數(shù)列.(2)由(1)知an+1=(a1+1)2n-1,所以an=32n-1-1.13.【解析】(1)因?yàn)閚,an,sn成等差數(shù)列,所以2an=sn+n,當(dāng)n2時(shí),an=sn-sn-1,所以2(sn-sn-1)=sn+n,即sn=2sn-1+n(n2),所以sn+n+2=2sn-1+2n+2=2sn-1+(n-1)+2,又s1+2-1+2=40,所以sn+n+2sn-1+(n-1)+2=2,所以數(shù)列sn+n+2成等比數(shù)列.(2)由(1)知sn+n+2是以4為首項(xiàng),2為公比的等比數(shù)列,所以sn+n+2=42n-1=2n+1,又2an=n+sn,所以2an+2=2n+1,所以an=2n-1.14.【思路點(diǎn)撥】(1)設(shè)出公比q,根據(jù)條件列出關(guān)于a1與q的方程組求得a1與q,即可求得數(shù)列的通項(xiàng)公式.(2)由(1)中求得數(shù)列的通項(xiàng)公式,可求出bn的通項(xiàng)公式,由其通項(xiàng)公式可知分開求和即可.【解析】(1)設(shè)公比為q,則an=a1qn-1.由已知得a1+a1q=2(1a1+1a1q),a1q2+a1q3+a1q4=64(1a1q2+1a1q3+1a1q4).化簡(jiǎn)得a12q=2,a12q6=64.又a10,故q=2,a1=1,所以an=2n-1.(2)由(1)得bn=(an+1an)2=an2+2+1an2=4n-1+14n-1+2.所以tn=(1+4+4n-1)+(1+14+14n-1)+2n=1-4n1-4+1-(14)n1-14+2n=13(4n-41-n)+2n+1.15.【解析】(1)一元二次方程anx2-an+1x+1=0(n=1,2,3,)有兩根和,由根與系數(shù)的關(guān)系易得+=,=1an,6-2+6=3,-2an=3,即an+1=12an+13.(2)an+1=12an+13,an+1-23=12(an-23),當(dāng)an-230時(shí),an+1-23an-23=12,當(dāng)an-23=0,即an=23時(shí),此時(shí)一元二次方程為23x2-23x+1=0,即2x2-2x+3=0,=4-240,不合題意,即數(shù)列an-23是等比數(shù)列.(3)由(2)知:數(shù)列an-23是以a1-23=76-23=12為首項(xiàng),公比為12的等比數(shù)列,an-23=12(12)n-1=(12)n,即an=(12)n+23,數(shù)列an的通項(xiàng)公式是an=(12)n+23.【變式備選】定義:若數(shù)列an滿足an+1=an2,則稱數(shù)列an為“平方遞推數(shù)列”.已知數(shù)列an中,a1=2,點(diǎn)(an,an+1)在函數(shù)f(x)=2x2+2x的圖像上,其中n為正整數(shù).(1)證明:數(shù)列2an+1是“平方遞推數(shù)列”,且數(shù)列l(wèi)g(2an+1)為等比數(shù)列.(2)設(shè)(1)中“平方遞推數(shù)列”的前n項(xiàng)之積為tn,即tn=(2a1+1)(2a2+1)(2an+1),求數(shù)列an的通項(xiàng)公式及tn關(guān)于n的表達(dá)式.【解析】(1)由條件得:an+1=2an2+2an,2an+1+1=4an2+4an+1=(2an+1)2,2an+1是“平方遞推數(shù)