2022年高考數(shù)學(xué)一輪復(fù)習(xí)考點(diǎn)規(guī)范練28數(shù)列的概念與表示含解析新人教A版2

1、PAGE PAGE 6考點(diǎn)規(guī)范練28數(shù)列的概念與表示基礎(chǔ)鞏固1.數(shù)列1,23,35,47,59,的一個(gè)通項(xiàng)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(xiàng)和,且Sn=nn+1,則1a5等于()A.56B.65C.130D.30答案:D解析:當(dāng)n2時(shí),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(xiàng)和為Sn=n2,若bn=(n-10)an,則數(shù)列bn的最小項(xiàng)為()A.第10項(xiàng)B.第11項(xiàng)C.第6項(xiàng)D.第5項(xiàng)答案:D解析:由Sn=n2,得當(dāng)n=1時(shí),a1=1,當(dāng)n2時(shí),an=Sn-Sn-1=n2-(n-1)2=2n-1,當(dāng)n=1時(shí)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(duì)稱軸為x=514,所以數(shù)列bn的最小項(xiàng)為第5項(xiàng).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(xiàng)和,則S2 022的值為()A.0B.2C.5D.6答

3、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2022=6337,S2022=337(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(gè)數(shù)列的第項(xiàng).答案:14解析:由已知,得數(shù)列的通項(xiàng)公式為an=3n-1.令3n-1=41,解得n=14,即為第14項(xiàng).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(xiàng)公式an=.答案:3n解

4、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(xiàng)公式為an=(n+2)78n,則當(dāng)an取得最大值時(shí),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(xiàng)為1的正項(xiàng)數(shù)列,且(n+1)an+12-nan2+an+1an=0,則它的通項(xiàng)公式an=.答案:1n解析:(n+1)an+12-nan2+an+

5、1an=0,(n+1)an+1-nanan+1+an=0.an是首項(xiàng)為1的正項(xiàng)數(shù)列,(n+1)an+1=nan,即an+1an=nn+1,an=anan-1an-1an-2a2a1a1=n-1nn-2n-1121=1n.10.已知數(shù)列an的前n項(xiàng)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?yàn)镾n=(-1)n+1n,所以a5+a6=S6-S4=(-6)-(-4)=-2.當(dāng)n=1時(shí),a1=S1=1;當(dāng)n2時(shí),an=Sn-Sn-1=(-1)n+1n-(-1)n(n-1)=(-1)n+1n+(n-1)=(-1)n+1(2n-1).又a

6、1也適合于此式,所以an=(-1)n+1(2n-1).(2)當(dāng)n=1時(shí),a1=S1=6;當(dāng)n2時(shí),an=Sn-Sn-1=(3n+2n+1)-3n-1+2(n-1)+1=23n-1+2.因?yàn)閍1不適合式,所以an=6,n=1,23n-1+2,n2.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

7、=1+(n-1)5=5n-4.所以b20=20a20=520-4=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(duì)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(xiàng)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(shí),S1+2=3a1=a1+2,a1=1.數(shù)列an是首項(xiàng)為

8、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(xiàng)和為Sn,Sn=2an-n,則an=.答案:2n-1解析:當(dāng)n2時(shí),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(xiàng),公比為2的等比數(shù)列,an+1=22n-1=2n,an=2n-1.14.數(shù)列an滿足an+2+(-1)nan=3n-1,前16項(xiàng)和為540,則a1=.答案:7解析:當(dāng)n為偶數(shù)時(shí),有an+2+an=3n-1,則(a2+a4)+(a6+a8)+(a10+a12)+(a14+a16)=5+17

9、+29+41=92,因?yàn)榍?6項(xiàng)和為540,所以a1+a3+a5+a7+a9+a11+a13+a15=448.當(dāng)n為奇數(shù)時(shí),有an+2-an=3n-1,由累加法得an+2-a1=3(1+3+5+n)-1+n2=34n2+n+14,所以an+2=34n2+n+14+a1,所以a1+3412+1+14+a1+3432+3+14+a1+3452+5+14+a1+3472+7+14+a1+3492+9+14+a1+34112+11+14+a1+34132+13+14+a1=448,解得a1=7.15.設(shè)數(shù)列an的前n項(xiàng)和為Sn.已知a1=a(a3),an+1=Sn+3n,nN*,bn=Sn-3n.(1

10、)求數(shù)列bn的通項(xiàng)公式;(2)若an+1an,求a的取值范圍.解:(1)因?yàn)閍n+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(xiàng)公式為bn=(a-3)2n-1.(2)由題意可知,a2a1對(duì)任意的a都成立.由(1)知Sn=3n+(a-3)2n-1.于是,當(dāng)n2時(shí),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(shí),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