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“数列”双基过关检测
一、选择题1.2017·全国卷Ⅰ记Sn为等差数列{an}的前n项和.若a4+a5=24,S6=48,则{an}的公差为 A.1 B.2C.4D.8解析选C 设等差数列{an}的公差为d,由得即解得d=
4.2.2018·江西六校联考在等比数列{an}中,若a3a5a7=-3,则a2a8= A.3B.C.9D.13解析选A 由a3a5a7=-3,得a=-3,即a5=-,故a2a8=a=
3.3.在数列{an}中,已知a1=2,a2=7,an+2等于anan+1n∈N*的个位数,则a2018= A.8B.6C.4D.2解析选D 由题意得a3=4,a4=8,a5=2,a6=6,a7=2,a8=2,a9=4,a10=
8.所以数列中的项从第3项开始呈周期性出现,周期为6,故a2018=a335×6+8=a8=
2.4.已知数列{an}满足a1=1,an=an-1+2nn≥2,n∈N*,则a7= A.53B.54C.55D.109解析选C a2=a1+2×2,a3=a2+2×3,……,a7=a6+2×7,各式相加得a7=a1+22+3+4+…+7=
55.5.设数列{an}的前n项和为Sn,若a1=1,an+1=3Snn∈N*,则S6= A.44B.45C.×46-1D.×45-1解析选B 由an+1=3Sn,得a2=3S1=
3.当n≥2时,an=3Sn-1,则an+1-an=3an,n≥2,即an+1=4an,n≥2,则数列{an}从第二项起构成等比数列,所以S6===
45.6.等差数列{an}和{bn}的前n项和分别为Sn,Tn,对一切自然数n,都有=,则等于 A.B.C.D.解析选C ∵S9==9a5,T9==9b5,∴==.7.已知数列{an}是首项为1的等比数列,Sn是其前n项和,若5S2=S4,则log4a3的值为 A.1B.2C.0或1D.0或2解析选C 由题意得,等比数列{an}中,5S2=S4,a1=1,所以5a1+a2=a1+a2+a3+a4,即51+q=1+q+q2+q3,q3+q2-4q-4=0,即q+1q2-4=0,解得q=-1或±2,当q=-1时,a3=1,log4a3=
0.当q=±2时,a3=4,log4a3=
1.综上所述,log4a3的值为0或
1.8.设数列{an}是公差为dd0的等差数列,若a1+a2+a3=15,a1a2a3=80,则a11+a12+a13= A.75B.90C.105D.120解析选C 由a1+a2+a3=15得3a2=15,解得a2=5,由a1a2a3=80,得a2-da2a2+d=80,将a2=5代入,得d=3d=-3舍去,从而a11+a12+a13=3a12=3a2+10d=3×5+30=
105.
二、填空题9.若数列{an}满足a1+3a2+32a3+…+3n-1an=,则数列{an}的通项公式为________.解析当n≥2时,由a1+3a2+32a3+…+3n-1an=,得a1+3a2+32a3+…+3n-2an-1=,两式相减得3n-1an=-=,则an=.当n=1时,a1=满足an=,所以an=.答案an=10.数列{an}的前n项和为Sn,若Sn=2an-1,则an=________.解析∵Sn=2an-1,
①∴Sn-1=2an-1-1n≥2,
②①-
②得an=2an-2an-1,即an=2an-
1.∵S1=a1=2a1-1,即a1=1,∴数列{an}为首项是1,公比是2的等比数列,故an=2n-
1.答案2n-111.已知数列{an}中,a2n=a2n-1+-1n,a2n+1=a2n+n,a1=1,则a20=________.解析由a2n=a2n-1+-1n,得a2n-a2n-1=-1n,由a2n+1=a2n+n,得a2n+1-a2n=n,故a2-a1=-1,a4-a3=1,a6-a5=-1,…,a20-a19=
1.a3-a2=1,a5-a4=2,a7-a6=3,…,a19-a18=
9.又a1=1,累加得a20=
46.答案4612.数列{an}为正项等比数列,若a3=3,且an+1=2an+3an-1n≥2,n∈N*,则此数列的前5项和S5=________.解析设公比为qq0,由an+1=2an+3an-1,可得q2=2q+3,所以q=3,又a3=3,则a1=,所以此数列的前5项和S5==.答案
三、解答题13.已知在等差数列{an}中,a3=5,a1+a19=-
18.1求公差d及通项an;2求数列{an}的前n项和Sn及使得Sn取得最大值时n的值.解1∵a3=5,a1+a19=-18,∴∴∴an=11-2n.2由1知,Sn===-n2+10n=-n-52+25,∴n=5时,Sn取得最大值.14.已知数列{an}满足+++…+=n2+n.1求数列{an}的通项公式;2若bn=,求数列{bn}的前n项和Sn.解1∵+++…+=n2+n,∴当n≥2时,+++…+=n-12+n-1,两式相减得=2nn≥2,∴an=n·2n+1n≥2.又∵当n=1时,=1+1,∴a1=4,满足an=n·2n+
1.∴an=n·2n+
1.2∵bn==n-2n,∴Sn=1×-21+2×-22+3×-23+…+n×-2n.-2Sn=1×-22+2×-23+3×-24+…+n-1×-2n+n-2n+1,∴两式相减得3Sn=-2+-22+-23+-24+…+-2n-n-2n+1=-n-2n+1=-n-2n+1=-,∴Sn=-.。