廣西專用2022年高考數(shù)學(xué)一輪復(fù)習(xí)考點(diǎn)規(guī)范練30數(shù)列的概念與表示含解析新人教A版理2

1、PAGE PAGE 6考點(diǎn)規(guī)范練30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+32.已知數(shù)列an的前n項(xiàng)和為Sn=n2-2n+2,則a8=()A.13B.15C.17D.193.(2021廣西河池模擬)數(shù)列an滿足an+1=11-an且a8=2,則a1的值為()A.10B.8C.23D.124.已知數(shù)列an的前n項(xiàng)和為Sn=n2+n+1,bn=(-1)nan(nN*),則數(shù)列bn的前50項(xiàng)和為()A.49B.50C.99D.1005.記Sn為數(shù)列an的前n項(xiàng)和.“任意正整數(shù)n,均有an0”是“Sn是

2、遞增數(shù)列”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件6.設(shè)數(shù)列2,5,22,11,則41是這個(gè)數(shù)列的第項(xiàng).7.已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,Sn+1=2Sn(nN*),則a10=.8.(2021安徽滁州模擬)在數(shù)列an中,a1=3,an+1=an+1n(n+1),則通項(xiàng)公式an=.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=.10.已知數(shù)列an的前n項(xiàng)和為Sn.(1)若Sn=(-1)n+1n,求a5+a6及數(shù)列an的通項(xiàng)公式;(2)若Sn=3n+2n+1,求數(shù)列an的通項(xiàng)公式.能力提升

3、11.(2021云南大理模擬)數(shù)列an滿足a1=2,an=an+1-1an+1+1,其前n項(xiàng)積為Tn,則T10等于()A.16B.-16C.6D.-612.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-113.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=.14.定義函數(shù)f(x)=xx,其中x表示不小于x的最小整數(shù),如1.4=

4、2,-2.3=-2.當(dāng)x(0,n(nN*)時(shí),函數(shù)f(x)的值域?yàn)锳n,記集合An中元素的個(gè)數(shù)為an,則an=.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)求數(shù)列bn的通項(xiàng)公式;(2)若an+1an,求a的取值范圍.高考預(yù)測(cè)16.已知數(shù)列an的通項(xiàng)公式an=a,n=1,4n+(-1)n(8-2a),n2,若對(duì)任意nN*,an0”“數(shù)列Sn是遞增數(shù)列”,“an0”是“數(shù)列Sn是遞增數(shù)列”的充分條件.如數(shù)列an為-1,1,3,5,7,9,顯然數(shù)列Sn是遞增數(shù)列,但是an不一定大于零,還有可能小于零,“數(shù)列Sn是遞增數(shù)列”不能推出“a

5、n0”,“an0”是“數(shù)列Sn是遞增數(shù)列”的不必要條件.“an0”是“數(shù)列Sn是遞增數(shù)列”的充分不必要條件.6.14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).7.256解析因?yàn)閍1=S1=1,Sn+1=2Sn,所以數(shù)列Sn是公比為2的等比數(shù)列,所以Sn=2n-1,所以a10=S10-S9=29-28=28=256.8.4-1n解析因?yàn)閍n+1=an+1n(n+1),即an+1-an=1n-1n+1,則an-an-1=1n-1-1n,an-1-an-2=1n-2-1n-1,an-2-an-3=1n-3-1n-2,a3-a2=12-13,a2-a1=

6、1-12,所以an-an-1+an-1-an-2+an-2-an-3+a3-a2+a2-a1=1n-1-1n+1n-2-1n-1+1n-3-1n-2+12-13+1-12,即an-a1=1-1n.又因?yàn)閍1=3,所以an=1-1n+a1=4-1n.9.1n解析(n+1)an+12-nan2+an+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,故an=anan-1an-1an-2a2a1a1=n-1nn-2n-1121=1n.10.解(1)因?yàn)镾n=(-1)n+1n,所以a5+a6=S6-S4=(-6)-

7、(-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a1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,n2.11.D解析由an=an+1-1an+1+1,得an(an+1+1)=an+1-1,即an+1(an-1)=-(an+1),所以an+1=an+11-an.又因?yàn)閍1=2,則a

8、2=a1+11-a1=2+11-2=-3,a3=a2+11-a2=-3+11-(-3)=-12,a4=a3+11-a3=-12+11-12=13,a5=a4+11-a4=13+11-13=2=a1,所以an是以4為周期的周期數(shù)列,且a5a6a7a8=a1a2a3a4=1,a9a10=a1a2=2(-3)=-6,所以T10=a1a2a3a4a5a6a7a8a9a10=-6.12.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.

9、數(shù)列an是首項(xiàng)為1,公比為32的等比數(shù)列.an=32n-1.13.3n解析a1+3a2+5a3+(2n-3)an-1+(2n-1)an=(n-1)3n+1+3,當(dāng)n2時(shí),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,當(dāng)n=1時(shí),a1=3,符合,所以an=3n.14.n(n+1)2解析由題意得,當(dāng)x(n-1,n時(shí),x=n,所以xx所在的區(qū)間為(n(n-1),n2,區(qū)間長(zhǎng)度為n,xx取到的整數(shù)為n2-n+1,n2-n+2,n2-n+n=n2,共n個(gè),所以,當(dāng)x(0,1時(shí),xx有1個(gè);當(dāng)x(1,2時(shí),xx有2個(gè);當(dāng)x(2,3時(shí),xx有3個(gè);當(dāng)

10、x(n-1,n時(shí),xx有n個(gè).所以x(0,n時(shí),xx共有1+2+3+n=n(n+1)2個(gè)數(shù).故an=n(n+1)2.15.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