2022年高考數(shù)學(xué)一輪復(fù)習(xí) 單元質(zhì)檢（,2022年高考數(shù)學(xué)一輪復(fù)習(xí),單元質(zhì)檢（,2022,年高,數(shù)學(xué),一輪,復(fù)習(xí),單元,質(zhì)檢
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=( )
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.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.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.(2021浙江,10)已知數(shù)列{an}滿足a1=1,an+1=an1+an(n∈N*),記數(shù)列{an}的前n項(xiàng)和為Sn,則( )
A.32
a1+a2=1+12=32.
因?yàn)閍n+1=an1+an,所以an+1+an+1an=an,
所以an+1=an-an+1an.
因?yàn)閧an}為遞減數(shù)列,所以an+an+1<2an,
所以an+10,
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-12+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+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=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Tn,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,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)已知{an}為等差數(shù)列,{bn}為等比數(shù)列,a1=b1=1,a5=5(a4-a3),b5=4(b4-b3).
(1)求{an}和{bn}的通項(xiàng)公式;
(2)記{an}的前n項(xiàng)和為Sn,求證:SnSn+2
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