20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：63-等比數(shù)列及其前n項(xiàng)和-【含解析】-全文預(yù)覽
【正文】 ,S20S10,S30S20,S40S30成等比數(shù)列,因此(S20S10)2=S10(S30S20),即(S2010)2=10(70S20),故S20=20或S20={an}的各項(xiàng)都為正數(shù),即S200,因此S20=30,S20S10=20,所以S40S30=S10S20S10S103=80,S40=S30+(S40S30)=70+80=.對點(diǎn)訓(xùn)練3(1)B (2)A (3)2 (1)由a5,a7是函數(shù)f(x)=x24x+3的兩個(gè)零點(diǎn),即a5,a7是方程x24x+3=a6,即(a1+2d)2=a1(a1+5d),整理,得4d2=≠0,所以a1d=4.(2)數(shù)列{an}滿足a1=2,an+1n+1ann=2,則數(shù)列ann是以a11=2為首項(xiàng),=2+2(n1)=2n.由于首項(xiàng)符合通項(xiàng),故an=2n2,所以bn=22an=22n=4n.因?yàn)閎n+1bn=4n+14n=4,所以數(shù)列{bn}是以b1=4為首項(xiàng),公比為4的等比數(shù)列,所以Sn=4(4n1)41=4(4n1)(4n1)3.8。a9=a5(m+2),即m25m6=0,解得m=1(舍去),m=6.對點(diǎn)訓(xùn)練1(1)C (2)B (1)設(shè)等比數(shù)列{an}的公比為q(q0),則a1(1q4)1q=15,a1q4=3a1q2+4a1,解得a1=1,q=2,所以a3=a1q2=122=.(2)設(shè)等比數(shù)列{an}的公比為q.∵a5a3=12,a6a4=24,∴a6a4a5a3=q=2.又a5a3=a1q4a1q2=12a1=12,∴a1=1.∴an=a1an+2≠0,n∈N*)?{an}是等比數(shù)列.(1)方程思想:如求等比數(shù)列中的基本量。a9=( ) (2)在等比數(shù)列{an}中,如果a1+a2=40,a3+a4=60,那么a7+a8=( ) (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= .(2)等比中項(xiàng)的推廣與變形:ap2=aman+1≠0,n≥2,n∈N*)。(2)求{an}和{bn}的通項(xiàng)公式.思考判斷或證明一個(gè)數(shù)列是等比數(shù)列有哪些方法?{an}是等比數(shù)列常用的方法:(1)定義法,證明anan1=q(n≥2,q為常數(shù))。an=Tn,則Tn,T2nTn,T3nT2n,…成等比數(shù)列. (6)若數(shù)列{an}的項(xiàng)數(shù)為2n,則S偶S奇=q。若2s=p+r,則apar=as2,其中m,n,p,q,s,r∈N*.(2)ak,ak+m,ak+2m,…仍是等比數(shù)列,公比為qm(k,m∈N*).(3)若數(shù)列{an},{bn}是兩個(gè)項(xiàng)數(shù)相同的等比數(shù)列,則數(shù)列{ban},{pan通項(xiàng)公式的推廣an=amqnm.設(shè)等比數(shù)列{an}的首項(xiàng)為a1,公比為q,則它的通項(xiàng)公式為an= 。當(dāng)q≠1時(shí),Sn=a1(1qn)1q=a1anq1q.設(shè)數(shù)列{an}是等比數(shù)列,Sn是其前n項(xiàng)和.(1
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