课时训练11等比数列的前n项和1.设Sn为等比数列{an}的前n项和,已知3S3=a4-2,3S2=a3-2,则公比q等于()A.3B.4C.5D.6答案:B解析:由题意,得3S3-3S2=(a4-2)-(a3-2),则3a3=a4-a3,即a4=4a3,故q==4.2.设Sn为等比数列{an}的前n项和,8a2+a5=0,则等于()A.11B.5C.-8D.-11答案:D解析:设{an}的公比为q.∵{an}为等比数列,且8a2+a5=0,∴8a2+a2q3=0.∵a2≠0,∴q3=-8.∴q=-2,=-11.4.在等比数列{an}中,若a1=,a4=-4,则公比q=;|a1|+|a2|+…+|an|=.答案:-22n-1-解析:∵a4=q3=-4,∴q=-2.∴an=×(-2)n-1.∴|an|=2n-2.∴|a1|+|a2|+…+|an|==2n-1-.5.已知Sn为等比数列{an}的前n项和,a2=3,a6=243,Sn=364,则n=.答案:6解析:解得:当a1=1,q=3时,Sn==364,解得:n=6;当a1=-1,q=-3时,Sn==364,无整数解.综上n=6.6.已知等比数列{an}中,a1=3,a4=81.若数列{bn}满足bn=log3an,则数列的前2013项的和为.答案:解析:=q3=27,∴q=3.∴an=a1·qn-1=3×3n-1=3n.∴bn=log3an=n.∴.∴数列的前2013项的和为:+…+=1-.7.设{an}是公比为正数的等比数列,a1=2,a3=a2+4.(1)求{an}的通项公式;(2)设{bn}是首项为1,公差为2的等差数列,求数列{an+bn}的前n项和Sn.解:(1)设q为等比数列{an}的公比,则由a1=2,a3=a2+4,得2q2=2q+4,即q2-q-2=0.解得:q=2或q=-1(舍去),∴q=2.∴{an}的通项公式an=2·2n-1=2n(n∈N*).(2)Sn=+n×1+×2=2n+1+n2-2.18.(2016课标全国高考乙卷)已知{an}是公差为3的等差数列,数列{bn}满足b1=1,b2=,anbn+1+bn+1=nbn.(1)求{an}的通项公式;(2)求{bn}的前n项和.(导学号51830100)解:(1)由已知,a1b2+b2=b1,b1=1,b2=,得a1=2.所以数列{an}是首项为2,公差为3的等差数列,通项公式为an=3n-1.(2)由(1)和anbn+1+bn+1=nbn得bn+1=,因此{bn}是首项为1,公比为的等比数列.记{bn}的前n项和为Sn,则Sn=.9.已知数列{an}的前n项和为Sn,且Sn=2an-2,数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=0上.(1)求数列{an},{bn}的通项公式;(2)记Tn=a1b1+a2b2+…+anbn,求Tn.解:(1)由Sn=2an-2得Sn-1=2an-1-2(n≥2),两式相减得an=2an-2an-1,即=2(n≥2),又a1=2a1-2,∴a1=2,∴{an}是以2为首项,以2为公比的等比数列,∴an=2n.∵点P(bn,bn+1)在直线x-y+2=0上,∴bn-bn+1+2=0,即bn+1-bn=2,∴{bn}是等差数列.∵b1=1,∴bn=2n-1.(2)∵Tn=1×2+3×22+5×23+…+(2n-3)2n-1+(2n-1)2n,①∴2Tn=1×22+3×23+5×24+…+(2n-3)2n+(2n-1)2n+1.②①-②得:-Tn=1×2+2(22+23+…+2n)-(2n-1)·2n+1=2+2·-(2n-1)2n+1=2+4·2n-8-(2n-1)2n+1=(3-2n)·2n+1-6.∴Tn=(2n-3)·2n+1+6.10.已知公差不为0的等差数列{an}的首项a1为a(a∈R),且成等比数列.(1)求数列{an}的通项公式;(2)对n∈N*,试比较+…+的大小.(导学号51830101)解:(1)设等差数列{an}的公差为d,由题意可知,即(a1+d)2=a1(a1+3d),从而a1d=d2,因为d≠0,∴d=a1=a.故通项公式an=na.(2)记Tn=+…+,因为=2na,所以Tn==.2从而,当a>0时,Tn<;当a<0时,Tn>.3
