2018届高三数学复习数列第四节数列求和夯基提能作业本文.docx

第四节数列求和A组基础题组1.数列an的通项公式是an=1n+n+1,前n项和为9,则n等于()A.9B.99C.10D.1002.已知数列an满足an+1=12+an-an2,且a1=12,则该数列的前2 016项的和等于()A.1 509B.3 018C.1 512D.2 0163.在数列an中,a1=1,a2=2,an+2-an=1+(-1)n,那么S100的值为()A.2 500B.2 600C.2 700D.2 8004.已知数列an的前n项和Sn=n2-6n,则|an|的前n项和Tn=()A.6n-n2 B.n2-6n+18C.D.5.设数列an的前n项和为Sn.若S2=4,an+1=2Sn+1,nN*,则a1=,S5=.6.(2015课标,16,5分)设Sn是数列an的前n项和,且a1=-1,an+1=SnSn+1,则Sn=.7.对于数列an,定义数列an+1-an为数列an的“差数列”,若a1=2,an的“差数列”的通项为2n,则数列an的前n项和Sn=.8.已知等比数列an的前n项和为Sn,且满足Sn=3n+k.(1)求k的值及数列an的通项公式;(2)若数列bn满足n=anbn,求数列bn的前n项和Tn.9.正项数列an的前n项和Sn满足:Sn2-(n2+n-1)Sn-(n2+n)=0.(1)求数列an的通项公式an;(2)令bn=n+1(n+2)2an2,数列bn的前n项和为Tn.证明:对于任意的nN*,都有Tn564.B组提升题组10.在数列an中,an=2n-12n,若an的前n项和Sn=32164,则n=()A.3B.4C.5D.611.数列an的通项公式为an=ncos ,其前n项和为Sn,则S2 015等于()A.1 002B.-1 004C.1 006D.-1 00812.已知数列an的前n项和为Sn,a1=1,当n2时,an+2Sn-1=n,则S2 015的值为()A.2 015B.2 013C.1 008D.1 00713.(2016江西八校联考)在数列an中,已知a1=1,an+1+(-1)nan=cos(n+1),记Sn为数列an的前n项和,则S2 015=.14.(2017安徽师大附中模拟)用x表示不超过x的最大整数,例如3=3,1.2=1,-1.3=-2.已知数列an满足a1=1,an+1=an2+an,则=.15.在数列an中,a2=4,a3=15,若Sn为an的前n项和,且数列an+n是等比数列,则Sn=.16.(2015安徽,18,12分)已知数列an是递增的等比数列,且a1+a4=9,a2a3=8.(1)求数列an的通项公式;(2)设Sn为数列an的前n项和,bn=an+1SnSn+1,求数列bn的前n项和Tn.17.已知an是递增的等差数列,a2,a4是方程x2-5x+6=0的根.(1)求an的通项公式;(2)求数列an2n的前n项和.答案全解全析A组基础题组1.Ban=1n+n+1=n+1-n,Sn=a1+a2+an=(n+1-n)+(n-n-1)+(3-2)+(2-1)=n+1-1,令n+1-1=9,得n=99,故选B.2.C因为a1=12,an+1=12+an-an2,所以a2=1,从而a3=12,a4=1,可得an=故数列的前2 016项的和S2 016=1 0081+12=1 512.3.B当n为奇数时,an+2-an=0an=1,当n为偶数时,an+2-an=2an=n,故an=于是S100=50+=2 600.4.C由Sn=n2-6n知an是等差数列,且首项为-5,公差为2.an=-5+(n-1)2=2n-7,n3时,an3时,an0,易得Tn=5.答案1;121解析由an+1=2Sn+1,得a2=2S1+1,即S2-a1=2a1+1,又S2=4,4-a1=2a1+1,解得a1=1.又an+1=Sn+1-Sn,Sn+1-Sn=2Sn+1,即Sn+1=3Sn+1,则Sn+1+12=3Sn+12,又S1+12=32,Sn+12是首项为32,公比为3的等比数列,Sn+12=323n-1,即Sn=3n-12,S5=35-12=121.6.答案-1n解析an+1=Sn+1-Sn,an+1=SnSn+1,Sn+1-Sn=Sn+1Sn,又由a1=-1,知Sn0,1Sn-1Sn+1=1,1Sn是等差数列,且公差为-1,而1S1=1a1=-1,1Sn=-1+(n-1)(-1)=-n,Sn=-1n.7.答案2n+1-2解析由题意知an+1-an=2n,当n2时,an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2n-1+2n-2+22+2+2=2-2n1-2+2=2n-2+2=2n,又a1=2满足上式,an=2n(nN*),Sn=2-2n+11-2=2n+1-2.8.解析(1)当n2时,由an=Sn-Sn-1=3n+k-3n-1-k=23n-1,得等比数列an的公比q=3,首项为2.a1=S1=3+k=2,数列an的通项公式为an=23n-1,k=-1.(2)由n=anbn,可得bn=,即bn=32n3n.Tn=32,13Tn=32,23Tn=32,Tn=94.9.解析(1)由Sn2-(n2+n-1)Sn-(n2+n)=0,得Sn-(n2+n)(Sn+1)=0.由于an是正项数列,所以Sn0,Sn=n2+n.于是a1=S1=2,n2时,an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n.a1=2适合上式,故数列an的通项公式为an=2n.(2)证明:由于an=2n,bn=n+1(n+2)2an2,所以bn=n+14n2(n+2)2=1161n2-1(n+2)2.所以Tn=1161-132+122-142+132-152+1(n-1)2-1(n+1)2+1n2-1(n+2)2=1161,1an+1=1an(an+1)=1an-1an+1,1an+1=1an-1an+1,1a1+1+1a2+1+=1a1-1a2+1a2-1a3+1a2 016-1a2 017=1-1a2 017(0,1).又anan+1=1-1an+1,a1a1+1+a2 016a2 016+1=2 016-1-1a2 017,=2 015.15.答案3n-n2+n2-1解析an+n是等比数列,数列an+n的公比q=a3+3a2+2=15+34+2=186=3,则an+n的通项为an+n=(a2+2)3n-2=63n-2=23n-1,则an=23n-1-n,Sn=2(1-3n)1-3-n(1+n)2=3n-n2+n2-1.16.解析(1)由题设知a1a4=a2a3=8,又a1+a4=9,可解得a1=1,a4=8或a1=8,a4=1(舍去).由a4=a1q3得q=2,故an=a1qn-1=2n-1.(2)Sn=a1(1-qn)1-q=2n-1,又bn=an+1SnSn+1=Sn+1-SnSnSn+1=1Sn-1Sn+1,所以Tn=b1+b2+bn=1S1-1S2+1S2-1S3+1Sn-1Sn+1=1S1-=1-12n+1-1.17.解析(1)方程x2-5x+6=0的两根分别为2,3,由题意得a2=2,a4=3.设数列an的公差为d,则a4-a2=2d,故d=12,从而a1=32.所以