课时规范练28 数列的概念与表示一、基础巩固组1.数列1,23,35,47,59,…的一个通项公式an=( )A.n2n+1B.n2n-1C.n2n-3D.n2n+32.已知数列{an}的前n项和为Sn,且Sn=2(an-1),则a2等于( )A.4B.2C.1D.-23.(2022江西上饶模拟)已知数列{an}满足an+1+an=n,若a1=2,则a4-a2=( )A.4B.3C.2D.14.已知数列{an}满足a1=0,an+1=an+2n-1,则数列{an}的一个通项公式为( )A.an=n-1B.an=(n-1)2C.an=(n-1)3D.an=(n-1)45.(2022吉林市模拟改编)若数列{an}满足a1=12,an=1-1an-1(n≥2,且n∈N*),则a2018等于( )A.-1B.12C.1D.26.已知数列{an}的首项a1=1,其前n项和Sn=n2an(n∈N*),则a9=( )A.136B.145C.155D.1667.(2022宁夏银川二模)已知数列{an}满足a1=2,且a12+a23+a34+…+an-1n=an-2(n≥2),则{an}的通项公式为 . 8.已知数列{an}的通项公式为an=(n+2)78n,则当an取得最大值时,n= . 9.已知各项都为正数的数列{an}满足an+12-an+1an-2an2=0,且a1=2,则an= . 10.(2022广东江门一模)已知正项数列{an}的前n项和为Sn,Sn=12an(an+1),n∈N*.(1)求数列{an}的通项公式;(2)若bn=1Sn,求数列{bn}的前n项和Tn.〚导学号21500730〛二、综合提升组11.(2022河南郑州、平顶山、濮阳二模,理7)已知数列{an}满足an+1=an-an-1(n≥2),a1=m,a2=n,Sn为数列{an}的前n项和,则S2017的值为( )A.2017n-mB.n-2017mC.mD.n12.已知函数f(x)是定义在区间(0,+∞)内的单调函数,且对任意的正数x,y都有f(xy)=f(x)+f(y).若数列{an}的前n项和为Sn,且满足f(Sn+2)-f(an)=f(3)(n∈N*),则an等于( )A.2n-1B.nC.2n-1D.32n-113.(2022山西晋中二模,理15)我们可以利用数列{an}的递推公式an=n,n为奇数,an2,n为偶数(n∈N*),求出这个数列各项的值,使得这个数列中的每一项都是奇数,则a64+a65= . 14.(2022山西吕梁二模,理16)在数列{an}中,已知a2n=a2n-1+(-1)n,a2n+1=a2n+n,a1=1,则a20= .15.已知数列{an}的前n项和为Sn,Sn=2an-n,则an= . 三、创新应用组16.(2022河南洛阳一模)意大利著名数学家斐波那契在研究兔子繁殖问题时,发现有这样一列数:1,1,2,3,5,8,13,….该数列的特点是:前两个数都是1,从第三个数起,每一个数都等于它前面两个数的和,人们把这样的一列数所组成的数列{an}称为“斐波那契数列”,则(a1a3-a22)(a2a4-a32)(a3a5-a42)·…·(a2015a2017-a20162)=( )A.1B.-1C.2017D.-2017〚导学号21500731〛17.已知数列{an}中,a1=-1,an+1=2an+3n-1(n∈N*),求数列{an}的通项公式.3\n课时规范练28 数列的概念与表示1.B 由已知得,数列可写成11,23,35,…,故通项为n2n-1.2.A 由Sn=2(an-1),得a1=2(a1-1),即a1=2,又a1+a2=2(a2-1),所以a2=4.3.D 由an+1+an=n,得an+2+an+1=n+1,两式相减得an+2-an=1,令n=2,得a4-a2=1.4.B 因为a1=0,an+1=an+2n-1,所以a2=0+1=1,a3=1+3=4,a4=4+5=9,故数列{an}的一个通项公式为an=(n-1)2.5.A ∵a1=12,an=1-1an-1(n≥2,且n∈N*),∴a2=1-1a1=1-112=-1,∴a3=1-1a2=1-1-1=2,∴a4=1-1a3=1-12=12,……依此类推,可得an+3=an,∴a2018=a672×3+2=a2=-1,故选A.6.B 由Sn=n2an,得Sn+1=(n+1)2an+1,所以an+1=(n+1)2an+1-n2an,化简得(n+2)an+1=nan,即an+1an=nn+2,所以a9=a9a8·a8a7·…·a2a1·a1=810×79×68×…×24×13×1=290=145.7.an=n+1 ∵a12+a23+a34+…+an-1n=an-2(n≥2),①a12+a23+a34+…+ann+1=an+1-2(n≥2),②②-①得ann+1=an+1-an,整理得an+1an=n+2n+1,∴an+1n+2ann+1=1,又a11+1=1,∴数列ann+1是以1为首项,1为公比的等比数列,即常数列1,∴an=n+1.8.5或6 由题意令an≥an-1,an≥an+1,∴(n+2)78n≥(n+1)78n-1,(n+2)78n≥(n+3)78n+1,解得n≤6,n≥5.∴n=5或n=6.9.2n ∵an+12-an+1an-2an2=0,∴(an+1+an)(an+1-2an)=0.∵数列{an}的各项均为正数,∴an+1+an>0,∴an+1-2an=0,即an+1=2an(n∈N*),∴数列{an}是以2为公比的等比数列.∵a1=2,∴an=2n.10.解(1)a1=S1=12a1(a1+1),a1>0,解得a1=1.∀n∈N*,an+1=Sn+1-Sn=12an+1(an+1+1)-12an(an+1),移项整理并因式分解得(an+1-an-1)(an+1+an)=0,因为{an}是正项数列,所以an+1+an>0,所以an+1-an-1=0,an+1-an=1.所以{an}是首项a1=1、公差为1的等差数列,所以an=n.(2)由(1)得Sn=12an(an+1)=12n(n+1),bn=1Sn=2n(n+1)=2n-2n+1,Tn=b1+b2+…+bn=21-22+22-23+…+2n-2n+1=21-2n+1=2nn+1.11.C ∵an+1=an-an-1(n≥2),a1=m,a2=n,3\n∴a3=n-m,a4=-m,a5=-n,a6=m-n,a7=m,a8=n,…,∴an+6=an.则S2017=S336×6+1=336×(a1+a2+…+a6)+a1=336×0+m=m.12.D 由题意知f(Sn+2)=f(an)+f(3)=f(3an)(n∈N*),∴Sn+2=3an,Sn-1+2=3an-1(n≥2),两式相减,得2an=3an-1(n≥2),则anan-1=32(n≥2).又n=1时,S1+2=3a1=a1+2,∴a1=1.∴数列{an}是首项为1,公比为32的等比数列.∴an=32n-1.13.66 由题得,这个数列各项的值分别为1,1,3,1,5,3,7,1,9,5,11,3,…∴a64+a65=a32+65=a16+65=a8+65=a4+65=1+65=66.14.46 由a2n=a2n-1+(-1)n,得a2n-a2n-1=(-1)n,由a2n+1=a2n+n,得a2n+1-a2n=n,∴a2-a1=-1,a4-a3=1,a6-a5=-1,…,a20-a19=1,10个式子之和为0,a3-a2=1,a5-a4=2,a7-a6=3,…,a19-a18=9,9个式子之和为9(1+9)2=45.累加得a20-a1=45.又a1=1,故a20=46,故答案为46.15.2n-1 当n≥2时,an=Sn-Sn-1=2an-n-2an-1+(n-1),即an=2an-1+1,∴an+1=2(an-1+1).又a1=S1=2a1-1,∴a1=1.∴数列{an+1}是以首项为a1+1=2,公比为2的等比数列,∴an+1=2·2n-1=2n,∴an=2n-1.16.B ∵a1a3-a22=1×2-12=1,a2a4-a32=1×3-22=-1,a3a5-a42=2×5-32=1,…,a2015a2017-a20162=1.∴(a1a3-a22)(a2a4-a32)(a3a5-a42)·…·(a2015a2017-a20162)=11008×(-1)1007=-1.17.解∵an+1=2an+3n-1(n∈N*),①a1=-1,∴a2=0.当n≥2时,an=2an-1+3n-4,②由①-②可得an+1-an=2an-2an-1+3,即an+1-an+3=2(an-an-1+3),∴数列{an-an-1+3}为等比数列,首项为4,公比为2.∴an-an-1+3=4×2n-2,∴an-an-1=2n-3.∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n-3+2n-1-3+…+22-3-1=4(2n-1-1)2-1-3(n-1)-1=2n+1-3n-2.3
