单元质检六数列(B)(时间:45分钟满分:100分)单元质检卷第12页一、选择题(本大题共6小题,每小题7分,共42分)1.(2015太原二模)在单调递减的等比数列{an}中,若a3=1,a2+a4=,则a1=()A.2B.4C.D.2答案:B解析:由已知得:a1q2=1,a1q+a1q3=,∴,q2-q+1=0,∴q=(q=2舍去),∴a1=4.2.在等差数列{an}中,若a1+a2+a3=32,a11+a12+a13=118,则a4+a10=()A.45B.50C.75D.60答案:B解析:∵a1+a2+a3=3a2=32,a11+a12+a13=3a12=118,∴3(a2+a12)=150,即a2+a12=50,∴a4+a10=a2+a12=50.3.已知等差数列共有10项,其中奇数项之和为15,偶数项之和为30,则其公差是()A.5B.4C.3D.2导学号〚32470612〛答案:C解析:设等差数列为{an},公差为d,则∴5d=15,∴d=3.4.(2015石家庄二模)等比数列{an}的前n项和为Sn,已知S3=a2+5a1,a7=2,则a5=()A.B.-C.2D.-2导学号〚32470613〛答案:A解析:由条件得∴∴a5=a1q4=×42=.5.若{an}是等差数列,首项a1>0,a1007+a1008>0,a1007·a1008<0,则使前n项和Sn>0成立的最大自然数n是()A.2012B.2013C.2014D.2015导学号〚32470614〛答案:C解析:∵a1007+a1008>0,∴a1+a2014>0,∴S2014=>0.∵a1007·a1008<0,a1>0,∴a1007>0,a1008<0,∴2a1008=a1+a2015<0,∴S2015=<0,故选C.6.数列{an}中,已知对任意n∈N+,a1+a2+a3+…+an=3n-1,则+…+等于()A.(3n-1)2B.(9n-1)C.9n-1D.(3n-1)导学号〚32470615〛答案:B解析:∵a1+a2+a3+…+an=3n-1,∴a1+a2+a3+…+an-1=3n-1-1(n≥2),两式相减得an=3n-3n-1=2·3n-1(n≥2),又a1=2满足上式,∴an=2·3n-1.∴=4·32n-2=4·9n-1,∴+…+=4(1+9+92+…+9n-1)=(9n-1).二、填空题(本大题共2小题,每小题7分,共14分)7.若等差数列{an}前9项的和等于前4项的和,且ak+a4=0,则k=.答案:10解析:设等差数列{an}的前n项和为Sn,则S9-S4=0,即a5+a6+a7+a8+a9=0,5a7=0,故a7=0.而ak+a4=0=2a7,故k=10.8.已知a>0,且a≠1,则(a+1)+(a2+2)+…+(an+n)=.答案:解析:当a>0,且a≠1时,原式=(a+a2+…+an)+(1+2+…+n)=.三、解答题(本大题共3小题,共44分)9.(14分)(2015东北三校二模)已知数列{an}的前n项和为Sn,且a1=2,an+1=Sn+2,n∈N+.(1)求数列{an}的通项公式;(2)设bn=n·an,求数列{bn}的前n项和Tn.解:(1)当n=1时,a2=S1+2=4=2a1,∵an+1=Sn+2,∴当n≥2时,an=Sn-1+2,两式相减得an+1=2an(n≥2),综上可知,数列{an}满足an+1=2an(n∈N+),且a1=2,∴an=2n.(2)∵bn=n·an=n·2n,∴Tn=1×21+2×22+3×23+…+(n-1)·2n-1+n·2n,∴2Tn=1×22+2×23+3×24+…+(n-1)·2n+n·2n+1.两式相减,得-Tn=21+22+23+…+2n-1+2n-n·2n+1=-n·2n+1=2n+1-2-n·2n+1,Tn=2+(n-1)·2n+1.10.(15分)已知数列{an}是首项为a1,公比为q(q≠1)的等比数列,其前n项和为Sn,且有,设bn=2q+Sn.(1)求q的值.(2)数列{bn}能否为等比数列?若能,请求出a1的值;若不能,请说明理由.解:(1)因为q≠1,所以=1+q5=.所以q5=,则q=.(2)由(1)可知,bn=2q+Sn=1+=(2a1+1)-.若{bn}为等比数列,则=b1b3,即=(1+a1),解得a1=-(a1=0舍去),则bn=.因为bn≠0,且当n≥2时,,故当a1=-时,数列{bn}为等比数列.11.(15分)已知数列{an}的前n项和为Sn=3n,数列{bn}满足b1=-1,bn+1=bn+(2n-1)(n∈N+).(1)求数列{an}的通项公式an;(2)求数列{bn}的通项公式bn;(3)若cn=,求数列{cn}的前n项和Tn.解:(1)∵Sn=3n,∴Sn-1=3n-1(n≥2).∴an=Sn-Sn-1=3n-3n-1=2×3n-1(n≥2).当n=1时,2×31-1=2≠a1=S1=3,∴an=(2)∵bn+1=bn+(2n-1),∴b2-b1=1,b3-b2=3,b4-b3=5,…,bn-bn-1=2n-3.以上各式相加得bn-b1=1+3+5+…+(2n-3)==(n-1)2.∵b1=-1,∴bn=n2-2n.(3)由题意得cn=当n≥2时,Tn=-3+2×0×31+2×1×32+2×2×33+…+2(n-2)×3n-1,①∴3Tn=-9+2×0×32+2×1×33+2×2×34+…+2(n-2)×3n.②∴①-②得-2Tn=6+2×32+2×33+…+2×3n-1-2(n-2)×3n.∴Tn=(n-2)×3n-(3+32+33+…+3n-1)=(n-2)×3n-.∴Tn=又当n=1时,=-3,∴Tn=.导学号〚32470616〛
