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項和-【含解析】-資料下載頁
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【正文】 所以數(shù)列anbn是以1為首項,1為公比的等比數(shù)列.例3(1)C (2)B (1)∵{an}為等比數(shù)列,∴a4a9=a5a8=18,聯(lián)立a5a8=18,a5+a8=3,得a5=6,a8=3或a5=3,a8=6.設(shè)等比數(shù)列{an}的公比為q,則當(dāng)a5=6,a8=3時,q3=a8a5=12,∴a2+a11=a5q3+a8q3=612+312=212。當(dāng)a5=3,a8=6時,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的兩個零點,即a5,a7是方程x24x+3=a7=3,所以在等比數(shù)列{an}中,a3a9=a5a7=.(2)由等比數(shù)列的性質(zhì)知,a1+a2,a3+a4,a5+a6,a7+a8成等比數(shù)列,其首項為40,公比為6040=32,所以a7+a8=40323=135.(3)由題意,得S奇+S偶=240,S奇S偶=80,解得S奇=80,S偶=160,所以q=S偶S奇=16080=2.例5(1)B (2)B (1)由已知,a32=a1a9,a1=2,故(a1+2d)2=a1(a1+8d),解得d=2或d=0(舍),故an=2+(n1)2=2n,S8=8(a1+a8)2=4(2+28)=.(2)設(shè)等比數(shù)列{an}的公比為q,∵a3a11=4a7,∴a72=4a7≠0,解得a7=4,數(shù)列{bn}是等差數(shù)列,且b7=a7.∴S13=13(b1+b13)2=13b7=13a7=52,故選B.對點訓(xùn)練4(1)D (2)4(4n1)3 (1)由a1,a3,a6成等比數(shù)列,得a32=a1a6,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,=2+2(n1)=2n.由于首項符合通項,故an=2n2,所以bn=22an=22n=4n.因為bn+1bn=4n+14n=4,所以數(shù)列{bn}是以b1=4為首項,公比為4的等比數(shù)列,所以Sn=4(4n1)41=4(4n1)(4n1)3.8
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