2023高考数学一轮复习课时规范练31数列求和文含解析北师大版202303232140

课时规范练31　数列求和基础巩固组1.数列112,314,518,7116,…,(2n-1)+12n,…的前n项和Sn的值等于(　　)A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n2.(2020山东滨州模拟)若数列{an}的通项公式为an=2n+2n-1,则该数列的前10项和为(　　)　　　　　　　　　　　　　　　　　A.2146B.1122C.2148D.11243.已知函数f(n)=n2,n为奇数,-n2,n为偶数,且an=f(n)+f(n+1),则a1+a2+a3+…+a100等于(　　)A.0B.100C.-100D.102004.(2020德州调研)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.10235.设等差数列{an}满足a2=5,a6+a8=30,则数列1an2-1的前n项的和等于　　　　　. 6.已知在等比数列{an}中,a1=2,且a1,a2,a3-2成等差数列.(1)求数列{an}的通项公式;(2)若数列{bn}满足:bn=1an+2log2an-1,求数列{bn}的前n项和Sn.7.(2020河北保定二模,文17)已知数列{an}的前n项和为Sn,且满足2Sn+an-n=0(n∈N*).(1)求证:数列an-12为等比数列;(2)求数列{an-n}的前n项和Tn.综合提升组\n8.已知数列{an}满足a1=1,且对于任意的n∈N+都有an+1=an+a1+n,则1a1+1a2+…+1a2017等于(　　)A.20162017B.40322017C.20172018D.403420189.(2020江苏,11)设{an}是公差为d的等差数列,{bn}是公比为q的等比数列.已知数列{an+bn}的前n项和Sn=n2-n+2n-1(n∈N+),则d+q的值是　　　　. 10.(2020全国1,文16)数列{an}满足an+2+(-1)nan=3n-1,前16项和为540,则a1=　　　　　. 11.(2020河北衡水中学三模,理17)已知数列{an}满足a1=4,且当n≥2时,(n-1)an=n(an-1+2n-2).(1)求证:数列ann是等差数列;(2)记bn=2n+1an2,求数列{bn}的前n项和Sn.12.(2020山东烟台一模,18)已知数列{an}的前n项和Sn=3n2+8n,{bn}是等差数列,且an=bn+bn+1.(1)求数列{bn}的通项公式;(2)令cn=(an+1)n+1(bn+2)n,求数列{cn}的前n项和Tn.创新应用组13.(2020江西九江一模,理12)在平面直角坐标系xOy中,已知An,Bn是圆x2+y2=n2上两个动点,且满足OAn·OBn=-n22(n∈N+),设An,Bn到直线x+3y+n(n+1)=0的距离之和的最大值为an,若数列1an的前n项和Sn<m恒成立,则实数m的取值范围是(　　)A.34,+∞B.34,+∞C.32,+∞D.32,+∞14.(2020新高考全国1,18)已知公比大于1的等比数列{an}满足a2+a4=20,a3=8.(1)求{an}的通项公式;(2)记bm为{an}在区间(0,m](m∈N+)中的项的个数,求数列{bm}的前100项和S100.\n参考答案课时规范练31　数列求和1.A　该数列的通项公式为an=(2n-1)+12n,则Sn=[1+3+5+…+(2n-1)]+12+122+…+12n=n2+1-12n.2.A　因为an=2n+2n-1,所以前n项和Sn=2(1-2n)1-2+n(2n-1+1)2=2n+1+n2-2,所以前10项和S10=211+102-2=2146.3.B　由题意,得a1+a2+a3+…+a100=12-22-22+32+32-42-42+52+…+992-1002-1002+1012=-(1+2)+(3+2)-(4+3)+…-(99+100)+(101+100)=-(1+2+…+99+100)+(2+3+…+100+101)=-50×101+50×103=100.故选B.4.C　∵2n+12n=1+12n,∴Tn=n+1-12n,∴T10+1013=11-1210+1013=1024-1210.又m>T10+1013恒成立,∴整数m的最小值为1024.5.n4(n+1)　设等差数列{an}的公差为d,由等差数列的性质可得a6+a8=30=2a7,a7=15,a7-a2=5d,即15=5+5d,d=2,an=a2+(n-2)d=2n+1,1an2-1=14n(n+1)=141n-1n+1,∴前n项和为141-12+12-13+…+1n-1n+1=141-1n+1=n4(n+1).6.解(1)设等比数列{an}的公比为q,∵a1,a2,a3-2成等差数列,∴2a2=a1+(a3-2)=2+(a3-2)=a3,∴q=a3a2=2,∴an=a1qn-1=2n(n∈N+).(2)由(1)及bn=1an+2log2an-1,可知bn=12n+2log22n-1=12n+2n-1,∴Sn=12+1+122+3+123+5+…+12n+(2n-1)=12+122+123+…+12n+[1+3+5+…+(2n-1)]=121-12n1-12+n·[1+(2n-1)]2=n2-12n+1(n∈N+).7.(1)证明当n=1时,2S1+a1-1=0,解得a1=13.因为2Sn+an-n=0(n∈N+),①当n≥2时,2Sn-1+an-1-(n-1)=0,②①-②,得3an=an-1+1,即an=13an-1+13,当n≥2时,an-12an-1-12=13an-1+13-12an-1-12=13,\n又a1-12=-16,所以an-12是以-16为首项,以13为公比的等比数列.(2)解由(1)可得an=-12×13n+12,所以an-n=-12×13n-n+12,所以数列{an-n}的前n项和Tn=-12×13[1-13n]1-13-(n+1)n2+n2,化简得Tn=1413n-1-n22.8.D　由题意可得an+1-an=n+1,则a1=1,a2-a1=2,a3-a2=3,…,an-an-1=n,以上各式相加可得an=n(n+1)2,则1an=21n-1n+1,1a1+1a2+…+1a2017=2×1-12+12-13+…+12017-12018=40342018.9.4　由等差数列的前n项和公式和等比数列的前n项和公式得Sn=na1+n(n-1)2d+b1(1-qn)1-q=d2n2+a1-d2n+-b11-qqn+b11-q.对照已知条件Sn=n2-n+2n-1,得d=2,q=2,所以d+q=4.10.7　当n为偶数时,有an+2+an=3n-1,则(a2+a4)+(a6+a8)+(a10+a12)+(a14+a16)=5+17+29+41=92,因为前16项和为540,所以a1+a3+a5+a7+a9+a11+a13+a15=448.当n为奇数时,有an+2-an=3n-1,由累加法得an+2-a1=3(1+3+5+…+n)-1+n2=34n2+n+14,所以an+2=34n2+n+14+a1,所以a1+34×12+1+14+a1+34×32+3+14+a1+34×52+5+14+a1+34×72+7+14+a1+34×92+9+14+a1+34×112+11+14+a1+34×132+13+14+a1=448,解得a1=7.11.(1)证明当n≥2时,(n-1)an=n(an-1+2n-2),将上式两边都除以n(n-1),得ann=an-1+2n-2n-1,即ann-an-1n-1=2,所以数列ann是以4为首项,2为公差的等差数列.(2)解由(1)得ann=4+2(n-1)=2n+2,即an=2n·(n+1),所以bn=2n+1an2=141n2-1(n+1)2.所以Sn=141-122+122-132+…+1n2-1(n+1)2=141-1(n+1)2=n2+2n4(n+1)2.12.解(1)由题意,当n≥2时,an=Sn-Sn-1=6n+5,当n=1时,a1=S1=11,所以an=6n+5.设数列{bn}的公差为d,\n由a1=b1+b2,a2=b2+b3,即11=2b1+d,17=2b1+3d,解得b1=4,d=3,所以bn=3n+1.(2)由(1)得cn=(6n+6)n+1(3n+3)n=3(n+1)·2n+1,所以Tn=3×[2×22+3×23+4×24+…+(n+1)×2n+1],2Tn=3×[2×23+3×24+4×25+…+(n+1)×2n+2],两式作差,得-Tn=3×[2×22+23+24+…+2n+1-(n+1)×2n+2]=3×4+4×(2n-1)2-1-(n+1)×2n+2=-3n·2n+2.所以Tn=3n·2n+2.13.B　由OAn·OBn=-n22,得n·ncos∠AnOBn=-n22,所以cos∠AnOBn=-12,所以∠AnOBn=120°.设线段AnBn的中点为Cn,则|OCn|=n2,所以Cn在圆x2+y2=n24上,An,Bn到直线x+3y+n(n+1)=0的距离之和等于点Cn到该直线的距离的两倍.点Cn到直线距离的最大值为圆心到直线的距离与圆的半径之和,而圆x2+y2=n24的圆心(0,0)到直线x+3y+n(n+1)=0的距离为d=|n(n+1)|1+3=n(n+1)2,所以an=2n(n+1)2+n2=n2+2n,所以1an=1n2+2n=121n-1n+2.则Sn=1a1+1a2+1a3+…+1an=121-13+12-14+13-15+…+1n-1n+2=121+12-1n+1-1n+2<34,所以m≥34.故选B.14.解(1)设{an}的公比为q.由题设得a1q+a1q3=20,a1q2=8.解得q=12(舍去),q=2.因为a1q2=8,所以a1=2.所以{an}的通项公式为an=2n.(2)由题设及(1)知b1=0,且当2n≤m<2n+1时,bm=n.所以S100=b1+(b2+b3)+(b4+b5+b6+b7)+…+(b32+b33+…+b63)+(b64+b65+…+b100)=0+1×2+2×22+3×23+4×24+5×25+6×(100-63)=480.