2022年高考理科《數(shù)學(xué)》人教A版總復(fù)習(xí)練習(xí)題-考點規(guī)范練29　數(shù)列的概念與表示

1、考點規(guī)范練29數(shù)列的概念與表示考點規(guī)范練A冊第19頁基礎(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a

2、4-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.若數(shù)列an滿足a1=12,an=1-1an-1(n2,且nN*),則a2 018等于()A.-1B.12C.1D.2答案:A解析:a1=12

3、,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,依此類推,可得an+3=an,a2 018=a6723+2=a2=-1,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+

4、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+1an=0,則它的通項公式an=.答案:1n解析:(n+1)an+12-nan2+an+1an=0,(n+1)an+1-nanan+1+an=0.an是首項為1的正項數(shù)列

5、,(n+1)an+1=nan,即an+1an=nn+1,所以an=anan-1an-1an-2a2a1a1=n-1nn-2n-1121=1n.10.已知數(shù)列an的前n項和為Sn.(1)若Sn=(-1)n+1n,求a5+a6及an;(2)若Sn=3n+2n+1,求an.解:(1)因為Sn=(-1)n+1n,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+1n-(-1)n(n-1)=(-1)n+1n+(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=

6、S1=6;當(dāng)n2時,an=Sn-Sn-1=(3n+2n+1)-3n-1+2(n-1)+1=23n-1+2.因為a1不適合式,所以an=6,n=1,23n-1+2,n2.能力提升11.已知數(shù)列an滿足an+1=an-an-1(n2),a1=m,a2=n,Sn為數(shù)列an的前n項和,則S2 017的值為()A.2 017n-mB.n-2 017mC.mD.n答案:C解析:an+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.則S2 017=S3366+1=336(a1+a2+a6)+a1=3360+m=m.12.

7、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an滿足an+1=an+2n,且a1=32,則ann的

8、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)+21+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)在區(qū)間1,42)內(nèi)單調(diào)遞減;在區(qū)間42,+內(nèi)單調(diào)遞增.又f(5)=5+325-1=525,f(6)=6+326-1=313a1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=23n-1+(a-3)2n-2,故an+1-an=43n-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,+).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