2022年新教材高考数学一轮复习 考点规范练25 数列的概念（含解析）新人教版.docx

1、考点规范练25数列的概念一、基础巩固1.已知数列5,11,17,23,29,则55是它的()A.第19项B.第20项C.第21项D.第22项2.若Sn为数列an的前n项和,且Sn=nn+1,则1a5等于()A.56B.65C.130D.303.已知数列an满足an+1+an=n,若a1=2,则a4-a2等于()A.4B.3C.2D.14.已知数列an的前n项和Sn=n2,若bn=(n-10)an,则数列bn的最小项为()A.第10项B.第11项C.第6项D.第5项5.(多选)已知数列an满足a1=-12,an+1=11-an,则下列各数是数列an的项的有()A.-2B.23C.32D.36.已

2、知数列an满足m,nN*,都有anam=an+m,且a1=12,则a5等于()A.132B.116C.14D.127.已知数列an满足a1+3a2+5a3+(2n-1)an=(n-1)3n+1+3,则数列an的通项公式an=.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.已知数列an的前n项和为Sn,Sn=2an-n,则an=.10.已知数列an的前n项和为Sn.(1)若Sn=(-1)n+1n,求a5+a6及an;(2)若Sn=3n+2n+1,求an.二、综合应用11.已知数列an满足an+1=an-an-1(n2),a1=m,a2=n,Sn为数列an的前

3、n项和,则S2 023的值为()A.2 023n-mB.n-2 023mC.mD.n12.设数列an的首项为1且各项均为正数,若(n+1)an+12-nan2+an+1an=0,则它的通项公式an=.13.已知数列an满足an+1=an+2n,且a1=32,则ann的最小值为.14.已知数列an的通项公式是an=-n2+12n-32,其前n项和是Sn,则对任意的nm(其中m,nN*),Sn-Sm的最大值是.15.设数列an的前n项和为Sn.已知a1=a(a3),an+1=Sn+3n,bn=Sn-3n.(1)求数列bn的通项公式;(2)若an+1an,求a的取值范围.三、探究创新16.已知函数f

4、(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-1考点规范练25数列的概念1.C数列5,11,17,23,29,中的各项可变形为5,5+6,5+26,5+36,5+46,即该数列的通项公式an=5+6(n-1)=6n-1,令6n-1=55,得n=21.2.D当n2时,an=Sn-Sn-1=nn+1-n-1n=1n(n+1),则1a5=5(5+1)=30.3.D由an+1+an=n,得an+2+an+1=n+1

5、,两式相减得an+2-an=1,令n=2,得a4-a2=1.4.D由Sn=n2,得当n=1时,a1=1,当n2时,an=Sn-Sn-1=n2-(n-1)2=2n-1,当n=1时显然适合上式,所以an=2n-1,所以bn=(n-10)an=(n-10)(2n-1).令f(x)=(x-10)(2x-1),易知其图象的对称轴为直线x=214,所以数列bn的最小项为第5项.5.BD因为数列an满足a1=-12,an+1=11-an,所以a2=11-12=23,a3=11-a2=3,a4=11-a3=-12=a1.故数列an是周期为3的数列,且前3项依次为-12,23,3.6.A数列an满足m,nN*,

6、都有anam=an+m,且a1=12,a2=a1a1=14,a3=a1a2=18,a5=a3a2=132.7.3n当n=1时a1=3.当n2时,a1+3a2+5a3+(2n-3)an-1+(2n-1)an=(n-1)3n+1+3,把n换成n-1,得a1+3a2+5a3+(2n-3)an-1=(n-2)3n+3,两式相减得an=3n.又a1=3符合上式,故an=3n.8.5或6由题意令anan-1,anan+1,得(n+2)78n(n+1)78n-1,(n+2)78n(n+3)78n+1,解得n6,n5.故n=5或n=6.9.2n-1当n2时,an=Sn-Sn-1=2an-n-2an-1+(n-

7、1),即an=2an-1+1,an+1=2(an-1+1).又S1=2a1-1,a1=1.数列an+1是首项为a1+1=2,公比为2的等比数列,an+1=22n-1=2n,an=2n-1.10.解(1)因为Sn=(-1)n+1n,所以a5+a6=S6-S4=(-6)-(-4)=-2.当n=1时,a1=S1=1;当n2时,an=Sn-Sn-1=(-1)n+1n-(-1)n(n-1)=(-1)n+1n+(n-1)=(-1)n+1(2n-1).又a1=1也适合此式,所以an=(-1)n+1(2n-1).(2)当n=1时,a1=S1=6;当n2时,an=Sn-Sn-1=(3n+2n+1)-3n-1+2

8、(n-1)+1=23n-1+2.因为a1=6不适合式,所以an=6,n=1,23n-1+2,n2.11.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S2023=S3376+1=337(a1+a2+a6)+a1=3370+m=m.12.1n(n+1)an+12-nan2+an+1an=0,(n+1)an+1-nanan+1+an=0.数列an的首项为1,且各项均为正数,(n+1)an+1=nan,即an+1an=nn+1,则an=anan-1an-1an-2a2a1a1=n-1nn-2n-112

9、1=1n.13.313an+1=an+2n,即an+1-an=2n,an=an-an-1+(an-1-an-2)+a2-a1+a1=2(n-1)+2(n-2)+21+32=2(1+n-1)(n-1)2+32=n2-n+32.ann=n+32n-1.令f(x)=x+32x-1(x1),则f(x)=1-32x2=x2-32x2.f(x)在区间1,42)内单调递减,在区间(42,+)内单调递增.又f(5)=5+325-1=525,f(6)=6+326-1=3130,得4na1对任意的a都成立.由(1)知Sn=3n+(a-3)2n-1.于是,当n2时,an=Sn-Sn-1=3n+(a-3)2n-1-3n-1-(a-3)2n-2=23n-1+(a-3)2n-2,故an+1-an=43n-1+(a-3)2n-2=2n-21232n-2+a-3.当n2时,由an+1an,可知1232n-2+a-30,即a-9.又a3,故所求的a的取值范围是-9,3)(3,+).16.D由题意知f(Sn+2)=f(an)+f(3)=f(3an)(nN*),则Sn+2=3an,Sn-1+2=3an-1(n2),两式相减,得2an=3an-1(n2),又an0,所以anan-1=32(n2).又当n=1时,S1+2=3a1=a1+2,所以a1=1.即数列an是首项为1,公比为32的等比数列.故an=32n-1.