廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 考點規(guī)范練28 數(shù)列的概念與表示 文

考點規(guī)范練28數(shù)列的概念與表示一、基礎(chǔ)鞏固1.數(shù)列1,23,35,47,59,的一個通項公式an=()A.n2n+1B.n2n-1C.n2n-3D.n2n+3答案B2.若Sn為數(shù)列an的前n項和,且Sn=nn+1,則1a5等于()A.56B.65C.130D.30答案D解析當(dāng)n2時,an=Sn-Sn-1=nn+1-n-1n=1n(n+1),則1a5=5×(5+1)=30.3.已知數(shù)列an滿足an+1+an=n,若a1=2,則a4-a2=()A.4B.3C.2D.1答案D解析由an+1+an=n,得an+2+an+1=n+1,兩式相減得an+2-an=1,令n=2,得a4-a2=1.4.數(shù)列an的前n項和為Sn=n2,若bn=(n-10)an,則數(shù)列bn的最小項為()A.第10項B.第11項C.第6項D.第5項答案D解析由Sn=n2,得當(dāng)n=1時,a1=1,當(dāng)n2時,an=Sn-Sn-1=n2-(n-1)2=2n-1,當(dāng)n=1時顯然適合上式,所以an=2n-1,所以bn=(n-10)an=(n-10)(2n-1).令f(x)=(x-10)(2x-1),易知其圖象的對稱軸為x=514,所以數(shù)列bn的最小項為第5項.5.已知數(shù)列an滿足an+2=an+1-an,且a1=2,a2=3,Sn為數(shù)列an的前n項和,則S2 016的值為()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.數(shù)列an是周期為6的周期數(shù)列.又2016=6×336,S2016=336×(2+3+1-2-3-1)=0,故選A.6.設(shè)數(shù)列2,5,22,11,則41是這個數(shù)列的第項. 答案14解析由已知,得數(shù)列的通項公式為an=3n-1.令3n-1=41,解得n=14,即為第14項.7.已知數(shù)列an滿足:a1+3a2+5a3+(2n-1)·an=(n-1)·3n+1+3(nN*),則數(shù)列an的通項公式an=. 答案3n解析a1+3a2+5a3+(2n-3)·an-1+(2n-1)·an=(n-1)·3n+1+3,把n換成n-1,得a1+3a2+5a3+(2n-3)·an-1=(n-2)·3n+3,兩式相減得an=3n.8.已知數(shù)列an的通項公式為an=(n+2)78n,則當(dāng)an取得最大值時,n=. 答案5或6解析由題意令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.設(shè)數(shù)列an是首項為1的正項數(shù)列,且(n+1)an+12-nan2+an+1·an=0,則它的通項公式an=. 答案1n解析(n+1)an+12-nan2+an+1·an=0,(n+1)an+1-nanan+1+an=0.an是首項為1的正項數(shù)列,(n+1)an+1=nan,即an+1an=nn+1,an=anan-1·an-1an-2··a2a1·a1=n-1n·n-2n-1··12·1=1n.10.已知數(shù)列an的前n項和為Sn.(1)若Sn=(-1)n+1·n,求a5+a6及an;(2)若Sn=3n+2n+1,求an.解(1)因為Sn=(-1)n+1·n,所以a5+a6=S6-S4=(-6)-(-4)=-2.當(dāng)n=1時,a1=S1=1;當(dāng)n2時,an=Sn-Sn-1=(-1)n+1·n-(-1)n·(n-1)=(-1)n+1·n+(n-1)=(-1)n+1·(2n-1).又a1也適合于此式,所以an=(-1)n+1·(2n-1).(2)當(dāng)n=1時,a1=S1=6;當(dāng)n2時,an=Sn-Sn-1=(3n+2n+1)-3n-1+2(n-1)+1=2·3n-1+2.因為a1不適合式,所以an=6,n=1,2·3n-1+2,n2.二、能力提升11.設(shè)數(shù)列an滿足a1=1,a2=3,且2nan=(n-1)an-1+(n+1)an+1,則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,則數(shù)列bn是公差為5的等差數(shù)列,故bn=1+(n-1)×5=5n-4.所以b20=20a20=5×20-4=96,所以a20=9620=445.12.已知函數(shù)f(x)是定義在區(qū)間(0,+)內(nèi)的單調(diào)函數(shù),且對任意的正數(shù)x,y都有f(xy)=f(x)+f(y).若數(shù)列an的前n項和為Sn,且滿足f(Sn+2)-f(an)=f(3)(nN*),則an等于()A.2n-1B.nC.2n-1D.32n-1答案D解析由題意知f(Sn+2)=f(an)+f(3)=f(3an)(nN*),Sn+2=3an,Sn-1+2=3an-1(n2),兩式相減,得2an=3an-1(n2).又當(dāng)n=1時,S1+2=3a1=a1+2,a1=1.數(shù)列an是首項為1,公比為32的等比數(shù)列.an=32n-1.13.已知數(shù)列an的前n項和為Sn,Sn=2an-n,則an=. 答案2n-1解析當(dāng)n2時,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.數(shù)列an+1是以a1+1=2為首項,公比為2的等比數(shù)列,an+1=2·2n-1=2n,an=2n-1.14.已知an滿足an+1=an+2n,且a1=32,則ann的最小值為. 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)+2×1+32=2×(1+n-1)(n-1)2+32=n2-n+32.ann=n+32n-1.令f(x)=x+32x-1(x1),則f'(x)=1-32x2=x2-32x2.f(x)在1,42內(nèi)單調(diào)遞減,在42,+內(nèi)單調(diào)遞增.又f(5)=5+325-1=525,f(6)=6+326-1=313<f(5),當(dāng)n=6時,ann取最小值為313.15.設(shè)數(shù)列an的前n項和為Sn.已知a1=a(a3),an+1=Sn+3n,nN*,bn=Sn-3n.(1)求數(shù)列bn的通項公式;(2)若an+1an,求a的取值范圍.解(1)因為an+1=Sn+3n,所以Sn+1-Sn=an+1=Sn+3n,即Sn+1=2Sn+3n,由此得Sn+1-3n+1=2(Sn-3n),即bn+1=2bn.又b1=S1-3=a-3,故bn的通項公式為bn=(a-3)2n-1.(2)由題意可知,a2>a1對任意的a都成立.由(1)知Sn=3n+(a-3)2n-1.于是,當(dāng)n2時,an=Sn-Sn-1=3n+(a-3)2n-1-3n-1-(a-3)2n-2=2×3n-1+(a-3)2n-2,故an+1-an=4×3n-1+(a-3)2n-2=2n-21232n-2+a-3.當(dāng)n2時,由an+1an,可知1232n-2+a-30,即a-9.又a3,故所求的a的取值范圍是-9,3)(3,+).三、高考預(yù)測16.已知數(shù)列an的通項公式是an=-n2+12n-32,其前n項和是Sn,則對任意的n>m(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4<n<8,即在數(shù)列an中,前三項以及從第9項起后的各項均為負(fù)且a4=a8=0,因此Sn-Sm的最大值是a5+a6+a7=3+4+3=10.6