高考数学一轮复习 课时规范练28 数列的概念与表示 理 新人教B版.doc

课时规范练28数列的概念与表示基础巩固组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.(2017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.(2017吉林模拟改编)若数列an满足a1=12,an=1-1an-1(n2,且nn+),则a2 018等于()a.-1b.12c.1d.26.已知数列an的首项a1=1,其前n项和sn=n2an(nn+),则a9=()a.136b.145c.155d.1667.(2017宁夏银川二模)已知数列an满足a1=2,且a12+a23+a34+an-1n=an-2(n2),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.(2017广东江门一模)已知正项数列an的前n项和为sn,sn=12an(an+1),nn+.(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综合提升组11.(2017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(n2),a1=m,a2=n,sn为数列an的前n项和,则s2 017的值为()a.2 017n-mb.n-2 017mc.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)(nn+),则an等于()a.2n-1b.nc.2n-1d.32n-113.(2017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(nn+),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.(2017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=.创新应用组16.(2017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)(a2 015a2 017-a2 0162)=()a.1b.-1c.2 017d.-2 017导学号2150073117.已知数列an中,a1=-1,an+1=2an+3n-1(nn+),求数列an的通项公式.参考答案课时规范练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.aa1=12,an=1-1an-1(n2,且nn*),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,a2 018=a6723+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=a9a8a8a7a2a1a1=810796824131=290=145.7.an=n+1a12+a23+a34+an-1n=an-2(n2),a12+a23+a34+ann+1=an+1-2(n2),-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anan-1,anan+1,(n+2)78n(n+1)78n-1,(n+2)78n(n+3)78n+1,解得n6,n5.n=5或n=6.9.2nan+12-an+1an-2an2=0,(an+1+an)(an+1-2an)=0.数列an的各项均为正数,an+1+an0,an+1-2an=0,即an+1=2an(nn+),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),a10,解得a1=1.nn+,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+an0,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.can+1=an-an-1(n2),a1=m,a2=n,a3=n-m,a4=-m,a5=-n,a6=m-n,a7=m,a8=n,an+6=an.则s2 017=s3366+1=336(a1+a2+a6)+a1=3360+m=m.12.d由题意知f(sn+2)=f(an)+f(3)=f(3an)(nn+),sn+2=3an,sn-1+2=3an-1(n2),两式相减,得2an=3an-1(n2),则anan-1=32(n2).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n2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=22n-1=2n,an=2n-1.16.ba1a3-a22=12-12=1,a2a4-a32=13-22=-1,a3a5-a42=25-32=1,a2 015a2 017-a2 0162=1.(a1a3-a22)(a2a4-a32)(a3a5-a42)(a2 015a2 017-a2 0162)=11 008(-1)1 007=-1.17.解 an+1=2an+3n-1(nn+),a1=-1,a2=0.当n2时,an=2an-1+3n-4,由-可得an+1-an=2a