等比数列1.(2008年高考全国卷Ⅰ)已知等比数列{an}满足a1+a2=3,a2+a3=6,则a7=()A.64B.81C.128D.2432.(2009年高考广东卷)已知等比数列{an}满足an>0,n=1,2,…,且a5·a2n-5=22n(n≥3),则当n≥1时,log2a1+log2a3+…+log2a2n-1=()A.n(2n-1)B.(n+1)2C.n2D.(n-1)23.在正项数列{an}中,a1=2,点(,)(n≥2)在直线x-y=0上,则数列{an}的前n项和Sn等于()A.2n-1B.2n+1-2C.2-D.2-4.已知{an}是公比为常数q的等比数列,若a4,a5+a7,a6成等差数列,则q等于________.5.已知数列{an}是等比数列,a2=2,a5=16,则a1a2+a2a3+…+anan+1=________.6.(2009年高考陕西卷)已知数列{an}满足a1=1,a2=2,an+2=,n∈N*.(1)令bn=an+1-an,证明:{bn}是等比数列.(2)求{an}的通项公式.1/2参考答案1.解析:选A.设首项为a1,公比为q,则⇒,∴a7=a1q6=64.2.解析:选C.由题知an=2n,log2a2n-1=2n-1,log2a1+log2a3+…+log2a2n-1=1+3+…+(2n-1)=n2.3.解析:选B.由点(,)(n≥2)在直线x-y=0上得,-=0,即an=2an-1.又a1=2,所以当n≥2时,=2,故数列{an}是以2为首项,以2为公比的等比数列.所以Sn==2n+1-2,故选B.4.解析:由题知a4+a6=2(a5+a7)=2(a4q+a6q)=2q(a4+a6),由a4+a6≠0得q=.答案:5.解析:根据a2=2和a5=16,可求得等比数列{an}的首项a1=1,公比q=2,而所求的和式可看成是数列bn=anan+1的前n项和,而bn=anan+1=a12q2n-1=(a12q)(q2)n-1,所以{bn}是首项为b1=a12q=2,公比为q2=4的等比数列,故其前n项和为Sn==(4n-1).答案:(4n-1)6.解:(1)证明:b1=a2-a1=1.当n≥2时,bn=an+1-an=-an=-(an-an-1)=-bn-1,∴{bn}是以1为首项,-为公比的等比数列.(2)由(1)知bn=an+1-an=(-)n-1,当n≥2时,an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+1+(-)+…+(-)n-2=1+=1+[1-(-)n-1]=-(-)n-1当n=1时,-(-)1-1=1=a1,∴an=-(-)n-1(n∈N*).2/2