高中總復(fù)習(xí)理科數(shù)學(xué)配人教A版-課后習(xí)題Word-單元質(zhì)檢卷單元質(zhì)檢6　數(shù)列(A)

單元質(zhì)檢六數(shù)列(A)(時(shí)間:45分鐘滿分:100分)一、選擇題(本大題共6小題,每小題7分,共42分)1.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a6=15,S9=99,則等差數(shù)列{an}的公差是()A.14 B.4 C.-4 D.-答案:B解析:∵數(shù)列{an}是等差數(shù)列,a6=15,S9=99,∴a1+a9=22,∴2a5=22,a5=11.∴公差d=a6-a5=4.2.已知公比為32的等比數(shù)列{an}的各項(xiàng)都是正數(shù),且a3a11=16,則log2a16=(A.4 B.5 C.6 D.7答案:B解析:由等比中項(xiàng)的性質(zhì),得a3a11=a72=因?yàn)閿?shù)列{an}各項(xiàng)都是正數(shù),所以a7=4.所以a16=a7q9=32.所以log2a16=5.3.在等差數(shù)列{an}中,已知a4=5,a3是a2和a6的等比中項(xiàng),則數(shù)列{an}的前5項(xiàng)的和為()A.15 B.20 C.25 D.15或25答案:A解析:設(shè){an}的公差為d.∵在等差數(shù)列{an}中,a4=5,a3是a2和a6的等比中項(xiàng),∴a解得a∴S5=5a1+5×42d=5×(-1)+5×4=15.4.已知等差數(shù)列{an}和等比數(shù)列{bn}滿足3a1-a82+3a15=0,且a8=b10,則b3b17=(A.9 B.12 C.16 D.36答案:D解析:由3a1-a82+3a15=0,得a82=3a1+3a15=3(a1+a15)=3×即a82-6a8=0.因?yàn)閍8=b10≠0,所以a8=6,b10所以b3b17=b102=5.設(shè)公比為q(q>0)的等比數(shù)列{an}的前n項(xiàng)和為Sn,若S2=3a2+2,S4=3a4+2,則a1=()A.-2 B.-1 C.12 D.答案:B解析:∵S2=3a2+2,S4=3a4+2,∴S4-S2=3(a4-a2),即a1(q3+q2)=3a1(q3-q),q>0,解得q=32,代入a1(1+q)=3a1q+2,解得a1=-16.已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x≤0時(shí),f(x)=x(1-x).若數(shù)列{an}滿足a1=12,且an+1=11-an,則f(a11A.2 B.-2 C.6 D.-6答案:C解析:設(shè)x>0,則-x<0.因?yàn)閒(x)是定義在R上的奇函數(shù),所以f(x)=-f(-x)=-[-x(1+x)]=x(1+x).由a1=12,且an+1=1得a2=11-a3=11-a4=11……所以數(shù)列{an}是以3為周期的周期數(shù)列,即a11=a3×3+2=a2=2.所以f(a11)=f(a2)=f(2)=2×(1+2)=6.二、填空題(本大題共2小題,每小題7分,共14分)7.已知數(shù)列{an}滿足a1=1,an-an+1=2anan+1,則a6=.

答案:1解析:由an-an+1=2anan+1,得1an即數(shù)列1an是以1a1=1所以1a6=1a1+5×2=11,8.已知等比數(shù)列{an}滿足a2+8a5=0,設(shè)Sn是數(shù)列1an的前n項(xiàng)和,則S5答案:-11解析:設(shè){an}的公比為q.由a2+8a5=0,得a1q+8a1q4=0,解得q=-12.易知1an是等比數(shù)列,公比為-2,首項(xiàng)為1a1,所以S2=1a1[1-(-2)2三、解答題(本大題共3小題,共44分)9.(14分)在數(shù)列{an}和{bn}中,a1=1,an+1=an+2,b1=3,b2=7,等比數(shù)列{cn}滿足cn=bn-an.(1)求數(shù)列{an}和{cn}的通項(xiàng)公式;(2)若b6=am,求m的值.解:(1)因?yàn)閍n+1=an+2,且a1=1,所以數(shù)列{an}是首項(xiàng)為1,公差為2的等差數(shù)列.所以an=1+(n-1)·2=2n-1.因?yàn)閎1=3,b2=7,且a1=1,a2=3,所以c1=b1-a1=2,c2=b2-a2=4.因?yàn)閿?shù)列{cn}是等比數(shù)列,所以數(shù)列{cn}的公比q=c2c所以cn=c1·qn-1=2×2n-1=2n.(2)由(1)得bn-an=2n,an=2n-1,所以bn=2n+2n-1.所以b6=26+2×6-1=75.令2m-1=75,解得m=38.10.(15分)已知數(shù)列{an}的首項(xiàng)為a1=1,其前n項(xiàng)和為Sn,且數(shù)列Snn(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=(-1)nan,求數(shù)列{bn}的前n項(xiàng)和Tn.解:(1)∵數(shù)列Snn是公差為2的等差數(shù)列,且S11∴Snn=1+(n-1)×2=2n-∴Sn=2n2-n.∴當(dāng)n≥2時(shí),an=Sn-Sn-1=2n2-n-[2(n-1)2-(n-1)]=4n-3.∵a1符合an=4n-3,∴an=4n-3.(2)由(1)可得bn=(-1)nan=(-1)n·(4n-3).當(dāng)n為偶數(shù)時(shí),Tn=(-1+5)+(-9+13)+…+[-(4n-7)+(4n-3)]=4×n2=2n當(dāng)n為奇數(shù)時(shí),n+1為偶數(shù),Tn=Tn+1-bn+1=2(n+1)-(4n+1)=-2n+1.綜上所述,Tn=211.(15分)設(shè)數(shù)列{an}滿足a1=3,an+1=3an-4n.(1)計(jì)算a2,a3,猜想{an}的通項(xiàng)公式并加以證明;(2)求數(shù)列{2nan}的前n項(xiàng)和Sn.解:(1)a2=5,a3=7.猜想an=2n+1.由已知可得an+1-(2n+3)=3[an-(2n+1)],an-(2n+1)=3[an-1-(2n-1)],……a2-5=3(a1-3).因?yàn)閍1=3,所以an=2n+1.(2)由(1)得2nan=(2n+1)2n,所以Sn=3×2