《廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 單元質(zhì)檢六 數(shù)列(B) 文》由會(huì)員分享,可在線閱讀,更多相關(guān)《廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 單元質(zhì)檢六 數(shù)列(B) 文(6頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、單元質(zhì)檢六 數(shù)列(B)
(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}的公差和首項(xiàng)都不等于0,且a2,a4,a8成等比數(shù)列,則a1+a5+a9a2+a3=( )
A.2 B.3 C.5 D.7
答案B
解析設(shè){an}的公差為d.由題意,得a42=a2a8,
∴(a1+3d)2=(a1+d)(a1+7d),∴d2=a1d.
∵d≠0,∴d=a1,∴a1+a5+a9a2+a3=15a15a1=3.
2.在單調(diào)遞減的等比數(shù)列{an}中,若a3=1,a2+a4=52,則a1=
2、( )
A.2 B.4 C.2 D.22
答案B
解析設(shè){an}的公比為q.由已知,得a1q2=1,a1q+a1q3=52,
∴q+q3q2=52,q2-52q+1=0,
∴q=12(q=2舍去),∴a1=4.
3.(2018河北唐山期末)在數(shù)列{an}中,a1=1,an+1=2an,Sn為{an}的前n項(xiàng)和.若{Sn+λ}為等比數(shù)列,則λ=( )
A.-1 B.1 C.-2 D.2
答案B
解析由題意,得{an}是等比數(shù)列,公比為2,
∴Sn=2n-1,Sn+λ=2n-1+λ.
∵{Sn+λ}為等比數(shù)列,∴-1+λ=0,
∴λ=1,故選B.
4.(2018陜西西
3、安八校聯(lián)考)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若S6>S7>S5,則滿足SnSn+1<0的正整數(shù)n的值為( )
A.10 B.11 C.12 D.13
答案C
解析設(shè){an}的公差為d.
∵S6>S7>S5,
∴6a1+6×52d>7a1+7×62d>5a1+5×42d,
∴a7<0,a6+a7>0,
∴S13=13(a1+a13)2=13a7<0,
S12=12(a1+a12)2=6(a6+a7)>0,
∴滿足SnSn+1<0的正整數(shù)n的值為12,故選C.
5.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為Sn,若Sn=2,S3n=14,則S4n=( )
A.80
4、 B.26 C.30 D.16
答案C
解析設(shè)各項(xiàng)均為正數(shù)的等比數(shù)列{an}的首項(xiàng)為a1,公比為q.
∵Sn=2,S3n=14,∴a1(1-qn)1-q=2,a1(1-q3n)1-q=14,
解得qn=2,a11-q=-2.
∴S4n=a11-q(1-q4n)=-2×(1-16)=30.故選C.
6.(2018河南洛陽(yáng)一模)《九章算術(shù)》中的“竹九節(jié)”問(wèn)題:現(xiàn)有一根9節(jié)的竹子,自上而下各節(jié)的容積成等差數(shù)列,上面4節(jié)的容積共3升,下面3節(jié)的容積共4升,現(xiàn)自上而下取第1,3,9節(jié),則這3節(jié)的容積之和為( )
A.133升 B.176升 C.199升 D.2512升
答案B
解析設(shè)
5、自上而下各節(jié)的容積分別為a1,a2,…,a9,公差為d,
∵上面4節(jié)的容積共3升,下面3節(jié)的容積共4升,
∴a1+a2+a3+a4=4a1+6d=3,a9+a8+a7=3a1+21d=4,
解得a1=1322,d=766,
∴自上而下取第1,3,9節(jié),這3節(jié)的容積之和為a1+a3+a9=3a1+10d=3×1322+10×766=176(升).
二、填空題(本大題共2小題,每小題7分,共14分)
7.在3和一個(gè)未知數(shù)之間填上一個(gè)數(shù),使三數(shù)成等差數(shù)列.若中間項(xiàng)減去6,則三數(shù)成等比數(shù)列,則此未知數(shù)是 .?
答案3或27
解析設(shè)此三數(shù)為3,a,b,則2a=3+b,(a-
6、6)2=3b,解得a=3,b=3或a=15,b=27.故這個(gè)未知數(shù)為3或27.
8.已知數(shù)列{an}滿足:a1=1,an=an-12+2an-1(n≥2),若bn=1an+1+1an+2(n∈N*),則數(shù)列{bn}的前n項(xiàng)和Sn= .?
答案1-122n-1
解析當(dāng)n≥2時(shí),an+1=an-12+2an-1+1=(an-1+1)2>0,
兩邊取以2為底的對(duì)數(shù)可得log2(an+1)=log2(an-1+1)2=2log2(an-1+1),
則數(shù)列{log2(an+1)}是以1為首項(xiàng),2為公比的等比數(shù)列,log2(an+1)=2n-1,an=22n-1-1,
又an=an-1
7、2+2an-1(n≥2),可得an+1=an2+2an(n∈N*),
兩邊取倒數(shù)可得1an+1=1an2+2an=1an(an+2)
=121an-1an+2,
即2an+1=1an-1an+2,因此bn=1an+1+1an+2=1an-1an+1,
所以Sn=b1+…+bn=1a1-1an+1=1-122n-1,
故答案為1-122n-1.
三、解答題(本大題共3小題,共44分)
9.(14分)已知數(shù)列{an}的前n項(xiàng)和為Sn,首項(xiàng)為a1,且12,an,Sn成等差數(shù)列.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)數(shù)列{bn}滿足bn=(log2a2n+1)×(log2a2n+
8、3),求數(shù)列1bn的前n項(xiàng)和Tn.
解(1)∵12,an,Sn成等差數(shù)列,∴2an=Sn+12.
當(dāng)n=1時(shí),2a1=S1+12,即a1=12;
當(dāng)n≥2時(shí),an=Sn-Sn-1=2an-2an-1,即anan-1=2,
故數(shù)列{an}是首項(xiàng)為12,公比為2的等比數(shù)列,即an=2n-2.
(2)∵bn=(log2a2n+1)×(log2a2n+3)=(log222n+1-2)×(log222n+3-2)=(2n-1)(2n+1),
∴1bn=12n-1×12n+1=1212n-1-12n+1.
∴Tn=121-13+13-15+…+12n-1-12n+1
=121-12n+1=
9、n2n+1.
10.(15分)已知數(shù)列{an}和{bn}滿足a1=2,b1=1,2an+1=an,b1+12b2+13b3+…+1nbn=bn+1-1.
(1)求an與bn;
(2)記數(shù)列{anbn}的前n項(xiàng)和為T(mén)n,求Tn.
解(1)∵2an+1=an,∴{an}是公比為12的等比數(shù)列.
又a1=2,∴an=2·12n-1=12n-2.
∵b1+12b2+13b3+…+1nbn=bn+1-1,①
∴當(dāng)n=1時(shí),b1=b2-1,故b2=2.
當(dāng)n≥2時(shí),b1+12b2+13b3+…+1n-1bn-1=bn-1,②
①-②,得1nbn=bn+1-bn,得bn+1n+1=bnn,
10、故bn=n.
(2)由(1)知anbn=n·12n-2=n2n-2.
故Tn=12-1+220+…+n2n-2,
則12Tn=120+221+…+n2n-1.
以上兩式相減,得12Tn=12-1+120+…+12n-2-n2n-1=21-12n1-12-n2n-1,故Tn=8-n+22n-2.
11.(15分)設(shè){an}是等比數(shù)列,公比大于0,其前n項(xiàng)和為Sn(n∈N*),{bn}是等差數(shù)列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.
(1)求{an}和{bn}的通項(xiàng)公式;
(2)設(shè)數(shù)列{Sn}的前n項(xiàng)和為T(mén)n(n∈N*),
①求Tn;
②證明∑k=
11、1n(Tk+bk+2)bk(k+1)(k+2)=2n+2n+2-2(n∈N*).
(1)解設(shè)等比數(shù)列{an}的公比為q.由a1=1,a3=a2+2,可得q2-q-2=0.因?yàn)閝>0,可得q=2,故an=2n-1.
設(shè)等差數(shù)列{bn}的公差為d.由a4=b3+b5,可得b1+3d=4.
由a5=b4+2b6,可得3b1+13d=16,
從而b1=1,d=1,故bn=n.
所以,數(shù)列{an}的通項(xiàng)公式為an=2n-1,數(shù)列{bn}的通項(xiàng)公式為bn=n.
(2)①解由(1),有Sn=1-2n1-2=2n-1,
故Tn=∑k=1n(2k-1)=∑k=1n2k-n=2×(1-2n)1-2-n=2n+1-n-2.
②證明因?yàn)?Tk+bk+2)bk(k+1)(k+2)=(2k+1-k-2+k+2)k(k+1)(k+2)=k·2k+1(k+1)(k+2)=2k+2k+2-2k+1k+1,
所以,∑k=1n(Tk+bk+2)bk(k+1)(k+2)=233-222+244-233+…+2n+2n+2-2n+1n+1=2n+2n+2-2.
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