1、课时规范练28数列求和基础巩固组1.(2021广东汕头二中高三模拟)设数列an为等比数列,数列bn为等差数列,且Sn为数列bn的前n项和,若a2=1,a10=16,且a6=b6,则S11=()A.20B.30C.44D.882.(2021江西南昌高三月考)Sn=12+222+38+n2n等于()A.2n+1-n-22nB.2n-n2nC.2n-n+12n+1D.2n+1-n+22n3.(2021广东肇庆高三月考)在数列an中,a1=2,(1-an)an+1=1,Sn是数列an的前n项和,则S2 021=()A.4 042B.2 021C.20232D.202124.(2021河北衡水高三期末)
2、在数列an中,an+an+1=2n,Sn为其前n项和,若a1=a4,则S101=()A.4 882B.5 100C.5 102D.5 2125.(2021江苏南京高三月考)已知数列an的通项公式为an=n(n+1)!,则其前n项和为()A.1-1(n+1)!B.1-1n!C.2-1n!D.2-1(n+1)!6.(2021天津和平高三期中)函数y=x2在点(n,n2)(nN*)处的切线记为ln,直线ln,ln+1及x轴围成的三角形的面积记为Sn,则1S1+1S2+1S3+1Sn=.7.(2021湖南长郡中学高三模拟)在数列an中,a1=1,a2=2,且an+2=2an+1+3an,设bn=an+
3、1+an.(1)求证:数列bn是等比数列,并求数列bn的通项公式;(2)若数列bn的前n项和为Sn,数列94bnSnSn+1的前n项和为Tn,求证:Tn0,S2=2a2-2,S3=a4-2,数列an满足a2=4b1,nbn+1-(n+1)bn=n2+n(nN*).(1)求数列an的通项公式;(2)证明数列bnn为等差数列;(3)设数列cn的通项公式为cn=-anbn2,n为奇数,anbn4,n为偶数,其前n项和为Tn,求T2n.课时规范练28数列求和1.C解析:数列an为等比数列,a62=a2a10=16.又a6=a2q40,b6=a6=4.又数列bn为等差数列,S11=a1+a11211=1
4、1a6=44.故选C.2.A解析:由Sn=12+222+323+n2n,可得12Sn=122+223+n-12n+n2n+1.两式相减可得,12Sn=12+122+123+12nn2n+1=121-(12)n1-12n2n+1=2n+1-n-22n+1,所以Sn=2n+1-n-22n,故选A.3.D解析:因为a1=2,(1-an)an+1=1,所以由(1-a1)a2=1得a2=-1,进而得a3=12,a4=2=a1,可得an+3=an,所以a1+a2+a2021=673(a1+a2+a3)+a1+a2=67332+1=20212.故选D.4.C解析:因为an+an+1=2n,所以an+1+an
5、+2=2n+2,由-得an+2-an=2,所以数列an奇数项与偶数项均成公差为2的等差数列.当n为奇数时,an=a1+n-122=n+a1-1;当n为偶数时,an=a2+n-222=n+a2-2=n+(2-a1)-2=n-a1.又因为a1=a4,所以a1=4-a1,得a1=2,所以an=n+1,n为奇数,n-2,n为偶数,所以S101=(a1+a101)+(a2+a100)=512(2+102)+502(0+98)=5102.故选C.5.A解析:因为an=n(n+1)!=n+1-1(n+1)!=1n!1(n+1)!,所以其前n项和为1-12!+12!13!+1n!1(n+1)!=1-1(n+1
6、)!.故选A.6.4nn+1解析:因为y=2x,所以在点(n,n2)(nN*)处的切线的斜率为k=2n,所以切线方程为y-n2=2n(x-n),即ln的方程为y=2nx-n2,令y=0,得x=n2.ln+1:y=2(n+1)x-(n+1)2,令y=0,得x=n+12.由y=2(n+1)x-(n+1)2,y=2nx-n2得x=2n+12,y=n2+n,直线ln,ln+1的交点坐标为2n+12,n2+n,所以直线ln,ln+1及x轴围成的三角形的面积为Sn=12n+12n2(n2+n)=14n(n+1),所以1Sn=4n(n+1)=41n1n+1,则1S1+1S2+1S3+1Sn=41-12+12
7、13+1314+1n1n+1=41-1n+1=4nn+1.7.证明(1)由题意得an+2+an+1=3an+1+3an,bn0,bn+1=3bn,即bn+1bn=3,且b1=a2+a1=2+1=3,数列bn是以3为首项,3为公比的等比数列,bn=3n.(2)由(1)得,Sn=3(1-3n)1-3=3(3n-1)2,94bnSnSn+1=943n94(3n-1)(3n+1-1)=3n(3n-1)(3n+1-1)=1213n-113n+1-1,Tn=1213-1132-1+132-1133-1+13n-113n+1-1=121213n+1-1=1412(3n+1-1)14.8.解(1)当n=1时,
8、a1=S1=3-12=1,当n2,an=Sn-Sn-1=3n2-n23(n-1)2-(n-1)2=3n-2,所以n=1满足n2时的情况,所以an=3n-2(nN*).因为log3bn=log3b1+(n-1)(-1)=1-n,所以bn=31-n.(2)因为cn=a2n+1+b2n+1=3(2n+1)-2+31-(2n+1)=6n+1+19n,所以Tn=(7+6n+1)n2+191-(19)n1-19,所以Tn=3n2+4n+181-19n.9.B解析:当n2时,an=Sn-Sn-1=(2-3n)-(2-3n-1)=-23n-1.因为当n=1时,a1=-1不满足,所以数列an从第2项开始成等比数
9、列.又a3=-18,则数列an的奇数项构成的数列的前m项和Tm=-18(1-9m-1)1-9-1=549m4.故选B.10.B解析:因为an=1n+1+n=n+1n,所以Sn=21+32+n+1n=n+1-1.S1=1+1-1=0,S2=2+1-1=0,S3=3+1-1=1,S4=4+1-1=1,S7=7+1-1=1,S8=8+1-1=2,S9=9+1-1=2,S14=14+1-1=2,S15=15+1-1=3,S16=16+1-1=3,S23=23+1-1=3,S24=24+1-1=4,S25=25+1-1=4,S34=34+1-1=4,S35=35+1-1=5,S36=36+1-1=5,S
10、40=40+1-1=5.故S1+S2+S40=02+15+27+39+411+56=120.故选B.11.C解析:已知nan+1=(n+1)an+n(n+1),则an+1n+1=ann+1,可得数列ann是首项为1,公差为1的等差数列,即有ann=n,即an=n2,则bn=ancos2n3=n2cos2n3,则S11=-12(12+22+42+52+72+82+102+112)+(32+62+92)=-12(12+22-32-32+42+52-62-62+72+82-92-92+102+112)=-12(5+23+41+59)=-64.故选C.12.803解析:令p=1,q=n,则a1+n=a
11、1an=2an,所以数列an是首项为2,公比为2的等比数列,所以an=2n.当m=1时,b1=0.当2nm3(kN*)时,a2k0,a2k-1=-16(nN*)时,an6(nN*)时,SnSn+1,所以当n=4时,Sn取得最大值,且S4=10.17.(1)解因为等比数列an的前n项和为Sn,公比q0,S2=2a2-2,S3=a4-2,所以S3-S2=a4-2a2=a3,整理得a2q2-2a2=a2q.又a20,所以q2-q-2=0.由于q0,解得q=2.由于a1+a2=2a2-2,解得a1=2,所以an=2n.(2)证明数列an满足a2=4b1,解得b1=1.由于nbn+1-(n+1)bn=n
12、2+n,所以bn+1n+1bnn=1,所以数列bnn是以1为首项,1为公差的等差数列.(3)解因为数列bnn是以1为首项,1为公差的等差数列,所以bnn=1+(n-1)=n,所以bn=n2.因为数列cn的通项公式为cn=-anbn2,n为奇数,anbn4,n为偶数,所以令pn=c2n-1+c2n=-(2n-1)222n-12+(2n)222n4=(4n-1)4n-1,所以T2n=340+741+1142+(4n-1)4n-1,4T2n=341+742+1143+(4n-5)4n-1+(4n-1)4n.-得-3T2n=340+441+44n-1-(4n-1)4n,整理得-3T2n=3+44n-44-1-(4n-1)4n,故T2n=79+12n-794n.