PAGEPAGE1習(xí)題課——等比數(shù)列習(xí)題課課時過關(guān)·實(shí)力提升1已知在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1a2a3=5,a7a8a9=10,則a4a5a6等于()A.52 B.7 C.6 D.42解析數(shù)列{an}為等比數(shù)列,由a1a2a3=5,得a23=5.由a7a8a9=10,得a83=10.所以a23a83=50,即(a2a8)3=50,即a56=50,所以a53=52(因?yàn)閍n>0).答案A2在等比數(shù)列{an}中,已知|a1|=1,a5=-8a2,a5>a2,則an=()A.(-2)n-1 B.-(-2)n-1C.(-2)n D.-(-2)n解析由a5=-8a2,得公比q=-2.又由a5>a2,知a5>0,∴a1>0,∴a1=1.∴an=a1qn-1=(-2)n-1.答案A3已知等差數(shù)列{an}的前n項(xiàng)和為Sn,S5=15,S9=18,在等比數(shù)列{bn}中,b3=a3,b5=a5,則b7的值為()A.23 B.43 C.2 D解析在等差數(shù)列{an}中,由S5=15,S9=18,得5解得a故a3=3,a5=2.于是b3=3,b5=2,所以b7=b5答案B4已知數(shù)列{an}的前n項(xiàng)和Sn=an-1(a是不為零的常數(shù)),則數(shù)列{an}()A.肯定是等差數(shù)列B.肯定是等比數(shù)列C.或者是等差數(shù)列,或者是等比數(shù)列D.既不行能是等差數(shù)列,也不行能是等比數(shù)列解析∵Sn=an-1,∴a1=S1=a-1.當(dāng)n≥2時,an=Sn-Sn-1=(an-1)-(an-1-1)=(a-1)an-1.由于上式對n=1也適合,故此數(shù)列的通項(xiàng)公式為an=(a-1)an-1(n∈N+).當(dāng)a=1時,an=0,此時{an}為等差數(shù)列;當(dāng)a≠1時,因an+1an=a,故此時數(shù)列{答案C5已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),且a5a6+a4a7=18,則log3a1+log3a2+…+log3a10=()A.12 B.10 C.8 D.2解析由a5a6=a4a7,得a4a7=a5a6=9.則log3a1+log3a2+…+log3a10=log3a1a2…a10=log3(a5a6)5=10.答案B6若數(shù)列{an}是等比數(shù)列,數(shù)列{bn}是等差數(shù)列,且b1=0,cn=an+bn,當(dāng)數(shù)列{cn}的前三項(xiàng)為1,1,2時,數(shù)列{cn}的前9項(xiàng)的和是()A.456 B.475 C.547 D.978解析設(shè)數(shù)列{an}的首項(xiàng)為a1,公比為q,數(shù)列{bn}的公差為d,依據(jù)題意cn=an+bn,b1=0,得a解得a則an=a1qn-1=2n-1,bn=b1+(n-1)d=(n-1)(-1)=1-n.故數(shù)列{cn}的前9項(xiàng)和為c1+c2+…+c9=(a1+b1)+(a2+b2)+…+(a9+b9)=(1+0)+[2+(-1)]+…+[28+(-8)]=(1+2+…+28)+(0-1-…-8)=29-1-36=475.答案B7已知等比數(shù)列{an}是遞增數(shù)列,Sn是數(shù)列{an}的前n項(xiàng)和.若a1,a3是方程x2-5x+4=0的兩個根,則S6=.

解析因?yàn)閤2-5x+4=0的兩根為1和4,又?jǐn)?shù)列{an}是遞增數(shù)列,所以a1=1,a3=4,所以q=2.所以S6=1×(1答案638若an=92n-1-9n-1,則數(shù)列{an}的前n項(xiàng)和Sn=.

解析Sn=a1+a2+…+an=(9+93+…+92n-1)-(1+9+…+9n-1)=9(答案99設(shè)1=a1≤a2≤…≤a7,其中a1,a3,a5,a7成公比為q的等比數(shù)列,a2,a4,a6成公差為1的等差數(shù)列,則q的最小值是.

解析由題意,得1≤a2≤q≤a2+1≤q2≤a2+2≤q3,所以q3≥a2+2≥3答案310設(shè)數(shù)列{an}是公比為正數(shù)的等比數(shù)列,a1=2,a3=a2+4.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)數(shù)列{bn}是首項(xiàng)為1,公差為2的等差數(shù)列,求數(shù)列{an+bn}的前n項(xiàng)和Sn.解(1)設(shè)q為等比數(shù)列{an}的公比,則由a1=2,a3=a2+4,得2q2=2q+4,即q2-q-2=0,解得q=2(q=-1不合題意,舍去),因此q=2.所以數(shù)列{an}的通項(xiàng)為an=2·2n-1=2n(n∈N+).(2)bn=1+(n-1)×2=2n-1,所以數(shù)列{an+bn}的前n項(xiàng)和Sn=2(1-2n)1-211已知數(shù)列{an}滿意a1=0,且Sn+1=2Sn+12n(n+1)(n∈N+).(1)求a2,a3,并證明:an+1=2an+n(n∈N+);(2)設(shè)bn=an+1-an(n∈N+),求證:bn+1=2bn+1;(3)求數(shù)列{an}(n∈N+)的通項(xiàng)公式.(1)解由已知,得S2=2S1+1,即a1+a2=2a1+1,所以a2=1.S3=2S2+3,即a1+a2+a3=2(a1+a2)+3,a3=4.證明:由Sn+1=2Sn+12n(n+有Sn=2Sn-1+12(n-1)n(n≥2)所以Sn+1-Sn=2(Sn-Sn-1)+12n(n+1)-12n(即an+1=2an+n(n≥2).同時,a2=2a1+1=1,也滿意上式.所以an+1=2an+n(n∈N+).(2)證明由(1)an+1=2an+n,有an+2=2an+1+n+1,所以an+2-an+1=2(an+1-an)+1,即bn+1=2bn+1.(3)解由(2),得bn+1+1=2(bn+1),而b1+1=a2-a1+1=2,所以數(shù)列{bn+1}是以2為首項(xiàng),2為公比的等比數(shù)列,所以bn+1=2×2n-1=2n,所以bn=2n-1,即an+1-an=2n-1,而an+1=2an+n,所以2an+n-an=2n-1,所以an=2n-n-1(n∈N+).★12已知數(shù)列{an}滿意a1=0,a2=2,且對隨意m,n∈N+都有a2m-1+a2n-1=2am+n-1+2(m-n)2.(1)求a3,a5;(2)設(shè)bn=a2n+1-a2n-1(n∈N+),證明:數(shù)列{bn}是等差數(shù)列;(3)設(shè)cn=(an+1-an)qn-1(q≠0,n∈N+),求數(shù)列{cn}的前n項(xiàng)和Sn.(1)解由題意,令m=2,n=1,得a3=2a2-a1+2=6,再令m=3,n=1,得a5=2a3-a1+8=20.(2)證明當(dāng)n∈N+時,由已知(以n+2代替m),得a2n+3+a2n-1=2a2n+1+8.于是(a2(n+1)+1-a2(n+1)-1)-(a2n+1-a2n-1)=8,即bn+1-bn=8.所以數(shù)列{bn}是公差為8的等差數(shù)列.(3)解由(1)(2)的解答可知,數(shù)列{bn}是首項(xiàng)b1=a3-a1=6,公差為8的等差數(shù)列,則bn=8n-2,即a2n+1-a2n-1=8n-2.另由已知(令m=1),得an=a2n-1+a那么,an+1-an=a2n+1-a2n-12-2n+1于是,cn=2nqn-1.當(dāng)q=1時,Sn=2+4+6+…+2n=n(n+1).當(dāng)