2019高三數(shù)學(xué)（人教B文）一輪單元質(zhì)檢第六章數(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.2.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.73.設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,a1>0,且a6a5=911,當(dāng)Sn取最大值時(shí)A.9 B.10 C.11 D.124.已知等差數(shù)列{an}和等比數(shù)列{bn}滿足:3a1a82+3a15=0,且a8=b10,則b3b17=(A.9 B.12 C.16 D.365.設(shè)Sn是數(shù)列{an}的前n項(xiàng)和,當(dāng)n≥2時(shí),點(diǎn)(an1,2an)在直線y=2x+1上,且{an}的首項(xiàng)a1是二次函數(shù)y=x22x+3的最小值,則S9的值為()A.6 B.7 C.36 D.326.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(1x).若數(shù)列{an}滿足a1=12,且an+1=11-an,則f(a11A.2 B.2 C.6 D.6二、填空題(本大題共2小題,每小題7分,共14分)7.(2017湖南長(zhǎng)沙一模)若等比數(shù)列{an}的公比為2,則ln(a2017)2ln(a2016)2=.

8.(2017遼寧沈陽(yáng)一模)若等比數(shù)列{an}的公比q>0,且a2=1,an+2+an+1=6an,則{an}的前4項(xiàng)和S4=.

三、解答題(本大題共3小題,共44分)9.(14分)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a2=5,S4=28.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=(1)nan,求數(shù)列{bn}的前2n項(xiàng)和T2n.10.(15分)數(shù)列{an}滿足an=69an-1(n∈N+,(1)求證:數(shù)列1an(2)若a1=6,求數(shù)列{lgan}的前999項(xiàng)的和.11.(15分)(2017天津部分區(qū)一模)已知正項(xiàng)數(shù)列{an}中,a1=1,a2=2,前n項(xiàng)和為Sn,且滿足Sn+1Sn-1+Sn-(1)求數(shù)列{an}的通項(xiàng)公式;(2)記cn=1Sn·Sn+1,數(shù)列{cn}的前n項(xiàng)和為Tn,求證:1參考答案單元質(zhì)檢六數(shù)列(A)1.B解析∵{an}是等差數(shù)列,a6=15,S9=99,∴a1+a9=22,∴2a5=22,a5=11.∴公差d=a6a5=4.2.B解析由等比中項(xiàng)的性質(zhì)得a3a11=a72=16,又?jǐn)?shù)列{an}各項(xiàng)為正,所以a7=所以a16=a7q9=32.所以log2a16=5.3.B解析不妨設(shè)a6=9t,則a5=11t,故公差d=2t,其中t>0.因此a10=t,a11=t,即當(dāng)n=10時(shí),Sn取最大值,故選B.4.D解析由3a1a82+3a15=0得a82=3a1+3a15=3(a1+a15)=3×2a8,即a826a8=0,因?yàn)閍8=b10≠0,所以a8=6,b10=6,所以b3b5.C解析由點(diǎn)(an1,2an)在直線y=2x+1上,得2an=2an1+1,anan1=12,故數(shù)列{an}是公差為12由函數(shù)y=x22x+3的最小值為2,得a1=2,故S9=9×2+12×9×8×12=6.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-a1=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)=6.7.ln2解析ln(a2017)2ln(a2016)2=lna2017a20162=ln(2)8.152解析∵{an}是等比數(shù)列,∴an+2+an+1=6an可化為a1qn+1+a1qn=6a1qn1,∴q2+q6=0.∵q>0,∴q=2.a2=a1q=∴a1=12.∴S4=a9.解(1)由已知條件可知a2=故an=a1+(n1)×d=4n3.(2)由(1)可得bn=(1)nan=(1)n(4n3),則T2n=1+59+1317+…+(8n3)=4×n=4n.10.(1)證明∵1an-3-1an(2)解∵1an-3是等差數(shù)列,且1∴1an-3=1a1-3+1∴l(xiāng)gan=lg(n+1)lgn+lg3.設(shè)數(shù)列{lgan}的前999項(xiàng)的和為S,則S=999lg3+(lg2lg1+lg3lg2+…+lg1000lg999)=999lg3+lg1000=3+999lg3.11.(1)解由Sn+1Sn-1+Sn-1Sn+1=4Sn2Sn+1Sn-12(n≥2,n∈N+),得Sn+12+2Sn+1Sn1+Sn-12=4Sn2,即(Sn+1+Sn1由a1=1,a2=2,得S1=a1=1,S2=a1+a2=3,則數(shù)列{Sn}的公差為d=S2S1=2,所以Sn=1+(n1)×2=2n1.當(dāng)n≥2時(shí),an=SnSn1=(2n1)(2n3