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...1专题强化练九数列的求和及综合应用一、选择题1.已知等差数列{an}的前n项和为Sn,且a1=1,S3=a5.令bn=(-1)n-1an,则数列{bn}的前2n项和T2n为()A.-nB.-2nC.nD.2n解析:设等差数列{an}的公差为d,由S3=a5得3a2=a5,即3(1+d)=1+4d,解得d=2,所以an=2n-1,所以bn=(-1)n-1(2n-1),所以T2n=1-3+5-7+…+(4n-3)-(4n-1)=-2n.答案:B2.已知Tn为数列2n+12n的前n项和,若m>T10+1013恒成立,则整数m的最小值为()A.1026B.1025C.1024D.1023解析:因为2n+12n=1+12n,所以Tn=n+121-12n1-12=n+1-12n,所以T10+1013=11-1210+1013=1024-1210,又m>T10+1013恒成立,所以整数m的最小值为1024.答案:C3.已知数列{an}满足an+1-an=2,a1=-5,则|a1|+|a2|+…+|a6|=()A.9B.15C.18D.30解析:因为an+1-an=2,a1=-5,所以数列{an}是公差为2的等差数列.所以an=-5+2(n-1)=2n-7.令an=2n-7≥0,解得n≥72.所以n≤3时,|an|=-an;n≥4时,|an|=an.则|a1|+|a2|+…+|a6|=5+3+1+1+3+5=18.答案:C4.(2018·衡水中学月考)数列an=1n(n+1),其前n项之和为910,则在平面直角坐标系中,直线(n+1)x+y+n=0在y轴上的截距为()...2A.-10B.-9C.10D.9解析:由于an=1n(n+1)=1n-1n+1.所以Sn=1-12+12-13+…+1n-1n+1=1-1n+1.因此1-1n+1=910,所以n=9.所以直线方程为10x+y+9=0.令x=0,得y=-9,所以在y轴上的截距为-9.答案:B5.(2018·河南商丘第二次模拟)已知数列{an}满足a1=1,an+1-an≥2(n∈N*),且Sn为{an}的前n项和,则()A.an≥2n+1B.Sn≥n2C.an≥2n-1D.Sn≥2n-1解析:因为a2-a1≥2,a3-a2≥2,…,an-an-1≥2,且a1=1.各式相加,得an-a1≥2(n-1),则an≥2n-1(n≥2).则Sn=a1+a1+…+an≥1+3+5+…+2n-1=n2.答案:B二、填空题6.(2018·江西名校联考)若{an},{bn}满足anbn=1,an=n2+3n+2,则{bn}的前2018项和为________.解析:因为anbn=1,且an=n2+3n+2,所以bn=1n2+3n+2=1n+1-1n+2,故b1+b2+…+b2018=12-13+13-14+…+12019-12020=12-12020=10092020.答案:100920207.(2018·衡水中学质检)已知[x]表示不超过x的最大整数,例如:[2.3]=2,[-1.5]=-2.在数列{an}中,an=[lgn],n∈N*,记Sn为数列{an}的前n项和,则S2018=________.解析:当1≤n≤9时,an=[lgn]=0,当10≤n≤99时,an=[lgn]=1,当100≤n≤999时,an=[lgn]=2,当1000≤n≤2018时,an=[lgn]=3.故S2018=9×0+90×1+900×2+1019×3=4947.答案:49478.(2018·河北邯郸第一次模拟)已知数列{an},{bn}的前n项和分别为Sn,Tn,bn-an=2n+1,且Sn+Tn=2n+1+...3n2-2,则2Tn=________.解析:因为Tn-Sn=b1-a1+b2-a2+…+bn-an=2+22+…+2n+n=2n+1+n-2.又Sn+Tn=2n+1+n2-2.相加,得2Tn=2n+2+n2+n-4=2n+2+n(n+1)-4.答案:2n+2+n(n+1)-4三、解答题9.记Sn为数列{an}的前n项和,已知Sn=2n2+n,n∈N*.(1)求数列{an}的通项公式;(2)设bn=1anan+1,求数列{bn}的前n项和Tn.解:(1)由Sn=2n2+n,得当n=1时,a1=S1=3;当n≥2时,an=Sn-Sn-1=2n2+n-[2(n-1)2+(n-1)]=4n-1.又a1=3满足上式,所以an=4n-1(n∈N*).(2)bn=1anan+1=1(4n-1)(4n+3)=1414n-1-14n+3所以Tn=14[13-17+(17-111)+…+(14n-1-14n+3)]=14(13-14n+3)=n12n+9.10.(2018·日照调研)已知递增的等比数列{an}满足a2+a3=12,a1·a4=27.(1)求数列{an}的通项公式;(2)设bn=(n+1)an,求{bn}的前n项和Sn.解:(1)因为数列{an}是等比数列,且a2·a3=a1·a4=27,由a2+a3=12,a2·a3=27,得a2=3,a3=9,或a2=9,a3=3,(舍去)所以q=3,an=a2·3n-2=3n-1.(2)bn=(n+1)3n-1,所以Sn=b1+b2+b3+…+bn=2×30+3×31+4×32+…+(n+1)×3n-1,①所以3Sn=2×31+3×32+4×33+…+(n+1)×3n,②由①-②得-2Sn=2+31+32+33+…+3n-1-(n+1)×3n=2+3(1-3n-1)1-3-(n+1)×3n=12-12(2n+1)·3n.故Sn=(2n+1)·3n4-14.11.已知数列{an}的前n项和为Sn,点(n,Sn)(n∈N*)均在函数f(x)=3x2-2x的图象上.(1)求数列{an}的通项公式.(2)设bn=3anan+1,Tn是数列{bn}的前n项和,求使得2Tn≤λ-2018对任意n∈N*都成立的实数λ的取值范围....4解:(1)因为点(n,Sn)均在函数f(x)=3x2-2x的图象上,所以Sn=3n2-2n.当n=1时,a1=S1=3-2=1;当n≥2时,an=Sn-Sn-1=(3n2-2n)-[3(n-1)2-2(n-1)]=6n-5.又a1=1也满足an=6n-5,所以an=6n-5(n∈N*).(2)因为bn=3anan+1=3(6n-5)[6(n+1)-5]=1216n-5-16n+1,所以Tn=12[1-17+17-113+…+(16n-5-16n+1)]=12(1-16n+1)=3n6n+1,所以2Tn=6n6n+1=1-16n+1<1.又2Tn≤λ-2018对任意n∈N*都成立,所以1≤λ-2018,即λ≥2019.故实数λ的取值范围是[2019,+∞).
本文标题:2019高考数学二轮复习-第二部分-专题三-数列-专题强化练九-数列的求和及综合应用-文
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