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】(參考版)
2025-04-03 03:37本頁(yè)面
【正文】 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當(dāng)a5=3,a8=6時(shí),q3=a8a5=2,∴a2+a11=a5q3+a8q3=32+(6)(2)=.(2)a4(a6+2a8)+a3a11=a4a6+2a4a8+a3a11=a52+2a5a7+a72=(a5+a7)2=.例4(1)B (2)A (1)由等比數(shù)列的性質(zhì)可知,數(shù)列S3,S6S3,S9S6,S12S9是等比數(shù)列,即數(shù)列4,8,S9S6,S12S9是等比數(shù)列,因此S12=4+8+16+32=60,故選B.(2)依題意,S10,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(duì)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=(m+2),即m25m6=0,解得m=1(舍去),m=6.對(duì)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=a1預(yù)案自診知識(shí)梳理 同一個(gè) 公比(a1≠0,q≠0),G,b G2=ab考點(diǎn)自診1.(1) (2) (3)√ (4) (5) (6) 因?yàn)樵诘缺葦?shù)列{an}中,a1=1,a3+4S2=0,所以q2+4(1+q)=0,解得q=2,則a10=512. 因?yàn)閍1a3=a22=4,a4=4,所以a2=2,所以q2=a4a2=2,所以a6=a2q4=24=8,故a6的所有可能值構(gòu)成的集合是{8}. 設(shè)等比數(shù)列{an}的公比為q,因?yàn)閍1+a2+a3=1,a2+a3+a4=2,所以q(a1+a2+a3)=2,解得q=+a7+a8=q5(a1+a2+a3)=25=32. 因?yàn)閿?shù)列{an}是等比數(shù)列,且公比q=4,a1+a2+a3=21,所以a1+4a1+16a1=21,解得a1=1,所以an=4n1,故答案為4n1.關(guān)鍵能力
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