广西2020版高考数学一轮复习 考点规范练28 数列的概念与表示 文

考点规范练28数列的概念与表示一、基础巩固1.数列1,23,35,47,59,的一个通项公式an=()A.n2n+1B.n2n-1C.n2n-3D.n2n+3答案B2.若Sn为数列an的前n项和,且Sn=nn+1,则1a5等于()A.56B.65C.130D.30答案D解析当n2时,an=Sn-Sn-1=nn+1-n-1n=1n(n+1),则1a5=5(5+1)=30.3.已知数列an满足an+1+an=n,若a1=2,则a4-a2=()A.4B.3C.2D.1答案D解析由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.数列an的前n项和为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项答案D解析由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x=514,所以数列bn的最小项为第5项.5.已知数列an满足an+2=an+1-an,且a1=2,a2=3,Sn为数列an的前n项和,则S2 016的值为()A.0B.2C.5D.6答案A解析an+2=an+1-an,a1=2,a2=3,a3=a2-a1=1,a4=a3-a2=-2,a5=a4-a3=-3,a6=a5-a4=-1,a7=a6-a5=2,a8=a7-a6=3.数列an是周期为6的周期数列.又2016=6336,S2016=336(2+3+1-2-3-1)=0,故选A.6.设数列2,5,22,11,则41是这个数列的第项.答案14解析由已知,得数列的通项公式为an=3n-1.令3n-1=41,解得n=14,即为第14项.7.已知数列an满足:a1+3a2+5a3+(2n-1)an=(n-1)3n+1+3(nN*),则数列an的通项公式an=.答案3n解析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.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=.答案5或6解析由题意令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.设数列an是首项为1的正项数列,且(n+1)an+12-nan2+an+1an=0,则它的通项公式an=.答案1n解析(n+1)an+12-nan2+an+1an=0,(n+1)an+1-nanan+1+an=0.an是首项为1的正项数列,(n+1)an+1=nan,即an+1an=nn+1,an=anan-1an-1an-2a2a1a1=n-1nn-2n-1121=1n.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.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也适合于此式,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(n-1)+1=23n-1+2.因为a1不适合式,所以an=6,n=1,23n-1+2,n2.二、能力提升11.设数列an满足a1=1,a2=3,且2nan=(n-1)an-1+(n+1)an+1,则a20的值是()A.415B.425C.435D.445答案D解析由2nan=(n-1)an-1+(n+1)an+1,得nan-(n-1)an-1=(n+1)an+1-nan=2a2-a1=5.令bn=nan,则数列bn是公差为5的等差数列,故bn=1+(n-1)5=5n-4.所以b20=20a20=520-4=96,所以a20=9620=445.12.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-1答案D解析由题意知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).又当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.已知数列an的前n项和为Sn,Sn=2an-n,则an=.答案2n-1解析当n2时,an=Sn-Sn-1=2an-n-2an-1+(n-1),即an=2an-1+1an+1=2(an-1+1).又S1=2a1-1,a1=1.数列an+1是以a1+1=2为首项,公比为2的等比数列,an+1=22n-1=2n,an=2n-1.14.已知an满足an+1=an+2n,且a1=32,则ann的最小值为.答案313解析an+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1,42内单调递减,在42,+内单调递增.又f(5)=5+325-1=525,f(6)=6+326-1=313a1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,+).三、高考预测16.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.答案10解析由an=-n2+12n-32=-(n-4)(n-8)0得4n8,即在数列an中,前三项以及从第9项起后的各项均为负且a4=a8=0,因此Sn-Sm的最大值是a5+a6+a7=3+4+3=10.6