一輪復(fù)習(xí)精品資料(高中)PAGE1-課時(shí)作業(yè)梯級(jí)練三十七數(shù)列的概念及其表示一、選擇題(每小題5分,共35分)1.不能作為數(shù)列2,0,2,0,…的通項(xiàng)公式的是 ()A.an=1+(-1)n+1 B.an=1-(-1)nC.an=1+(-1)n D.an=1-cosnπ〖解析〗選C.驗(yàn)證易知,只有C選項(xiàng)中的式子不能作為已知數(shù)列的通項(xiàng)公式.2.在數(shù)列{an}中,a1=-2,an+1=QUOTE,則a2014= ()A.-2 B.-QUOTE C.QUOTE D.3〖解析〗選B.因?yàn)閍1=-2,an+1=QUOTE,所以a2=-QUOTE,a3=QUOTE,a4=3,a5=-2.可知數(shù)列{an}是以4為周期的數(shù)列,所以a2014=a2=-QUOTE.3.數(shù)列{an}的前n項(xiàng)和Sn=2n2-3n,則{an}的通項(xiàng)公式為 ()A.4n-5 B.4n-3C.2n-3 D.2n-1〖解析〗選A.因?yàn)镾n=2n2-3n,所以當(dāng)n≥2時(shí),Sn-1=2(n-1)2-3(n-1),兩式相減可得an=Sn-Sn-1=4n-5,又當(dāng)n=1時(shí),a1=S1=-1,滿足上式.4.如圖,給出的3個(gè)三角形圖案中圓的個(gè)數(shù)依次構(gòu)成一個(gè)數(shù)列的前3項(xiàng),則這個(gè)數(shù)列的一個(gè)通項(xiàng)公式是 ()A.2n+1 B.3nC.QUOTE D.QUOTE〖解析〗選D.由an-an-1=n+1,再根據(jù)累加法得an=a1+(a2-a1)+…+(an-an-1)=3+3+4+5+…+n+1=QUOTE.5.已知數(shù)列QUOTE滿足a1=28,QUOTE=2,則an= ()A.n2 B.n2+nC.n2-n+28 D.n2-n〖解析〗選C.由an+1-an=2n知,a2-a1=2×1,a3-a2=2×2,…,an-an-1=2QUOTE,相加得:an-a1=n2-n,所以an=n2-n+28.6.已知數(shù)列QUOTE滿足an+2=an+1-an,若a1=1,a3=3,則a17= ()A.-4 B.-3 C.3 D.4〖解析〗選A.因?yàn)閿?shù)列QUOTE滿足an+2=an+1-an,故有an+3=an+2-an+1=QUOTE-an+1=-an,所以an+6=-an+3=an,故數(shù)列QUOTE是以6為周期的周期數(shù)列,所以a17=a5=-a2,又因?yàn)閍1=1,a3=3,a3=a2-a1得a2=4,故a17=a5=-a2=-4.7.已知數(shù)列QUOTE的通項(xiàng)公式為an=QUOTE(n∈N*),則下列說法正確的是()A.這個(gè)數(shù)列的第10項(xiàng)為QUOTEB.QUOTE是該數(shù)列中的項(xiàng)C.數(shù)列中的各項(xiàng)都在區(qū)間QUOTE內(nèi)D.數(shù)列QUOTE是單調(diào)遞減數(shù)列〖解析〗選C.an=QUOTE=QUOTE=QUOTE.令n=10,得a10=QUOTE,故選項(xiàng)A不正確;令QUOTE=QUOTE,得9n=300,此方程無正整數(shù)解,故QUOTE不是該數(shù)列中的項(xiàng),故選項(xiàng)B不正確;因?yàn)閍n=QUOTE=QUOTE=1-QUOTE,又n∈N*,所以數(shù)列QUOTE是單調(diào)遞增數(shù)列,所以QUOTE≤an<1,所以數(shù)列中的各項(xiàng)都在區(qū)間QUOTE內(nèi),故選項(xiàng)C正確;選項(xiàng)D由上面分析知不正確.二、填空題(每小題5分,共15分)8.數(shù)列{an}的前n項(xiàng)和為Sn,若Sn+Sn-1=2n-1(n≥2,n∈N*),且S2=3,則a1的值為________.

〖解析〗因?yàn)镾n+Sn-1=2n-1(n≥2,n∈N*),令n=2,得S2+S1=3,由S2=3得a1=S1=0.〖答案〗:09.古希臘人常用小石子在沙灘上擺成各種形狀來研究數(shù),如圖所示.他們研究過圖中的1,5,12,22,…,由于這些數(shù)能夠表示成五角形,將其稱為五角形數(shù),若按此規(guī)律繼續(xù)下去,第n個(gè)五角形數(shù)an=________.

〖解析〗觀察圖形,發(fā)現(xiàn)a1=1,a2=a1+4,a3=a2+7,a4=a3+10,猜測(cè)當(dāng)n≥2時(shí),an=an-1+3n-2,所以an-an-1=3n-2,所以an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=(3n-2)+〖3(n-1)-2〗+…+(3×2-2)+1=QUOTEn2-QUOTEn.〖答案〗:QUOTEn2-QUOTEn10.設(shè)數(shù)列QUOTE滿足nan=n2+λ,若數(shù)列QUOTE是單調(diào)遞增數(shù)列,則實(shí)數(shù)λ的取值范圍是________.

〖解析〗因?yàn)閚an=n2+λ,所以an=n+QUOTE,由于數(shù)列QUOTE是單調(diào)遞增數(shù)列,則an+1>an,即n+1+QUOTE>n+QUOTE,整理得λ<n2+n,令bn=n2+n,所以bn+1-bn=QUOTE-QUOTE=2n+2>0,所以,數(shù)列QUOTE單調(diào)遞增,則數(shù)列QUOTE的最小項(xiàng)為b1=12+1=2,所以λ<2.因此,實(shí)數(shù)λ的取值范圍是QUOTE.〖答案〗:QUOTE〖加練備選·拔高〗(2020·石家莊模擬)設(shè)數(shù)列QUOTE的前n項(xiàng)和為Sn,若a1=QUOTE且當(dāng)n≥2時(shí),an=-Sn·Sn-1,則QUOTE的通項(xiàng)公式an=________.

〖解析〗當(dāng)n≥2時(shí),an=-Sn·Sn-1,則Sn-Sn-1=-Sn·Sn-1,所以QUOTE-QUOTE=1,因?yàn)閍1=QUOTE,所以S1=QUOTE,即QUOTE=2,所以QUOTE=2+QUOTE×1=n+1,所以Sn=QUOTE,所以當(dāng)n≥2時(shí),an=Sn-Sn-1=QUOTE-QUOTE=QUOTE,當(dāng)n=1時(shí)a1=QUOTE,不滿足上式,故an=QUOTE〖答案〗:QUOTE1.(5分)已知數(shù)列QUOTE滿足an=QUOTE(n∈N*),則數(shù)列QUOTE的最小項(xiàng)是第________項(xiàng) ()

A.3 B.4 C.5 D.6〖解析〗選C.因?yàn)閍n=QUOTE,-6=a5<a4=-QUOTE,-6=a5<a6=QUOTE,所以數(shù)列QUOTE的最小項(xiàng)是第5項(xiàng).2.(5分)在數(shù)列{an}中,其前n項(xiàng)和為Sn,且Sn=QUOTEan,則QUOTE的最大值為()A.-3 B.-1 C.3 D.1〖解析〗選C.當(dāng)n≥2時(shí),Sn=QUOTEan,Sn-1=QUOTEan-1.兩式作差可得an=Sn-Sn-1=QUOTEan-QUOTEan-1,則QUOTE=QUOTE=1+QUOTE,據(jù)此可得,當(dāng)n=2時(shí),QUOTE取到最大值3.3.(5分)在數(shù)列{an}中,a1=1,an+1-an=sinQUOTE,記Sn為數(shù)列{an}的前n項(xiàng)和,則S2021= ()A.0 B.2020C.1011 D.2021〖解析〗選C.由a1=1及an+1-an=sinQUOTE,得an+1=an+sinQUOTE,所以a2=a1+sinπ=1,a3=a2+sinQUOTE=0,a4=a3+sinQUOTE=0,a5=a4+sinQUOTE=1,a6=a5+sinQUOTE=1,a7=a6+sinQUOTE=0,a8=a7+sinQUOTE=0,…,可見數(shù)列{an}為周期數(shù)列,周期T=4,所以S2021=505(a1+a2+a3+a4)+a1=1011.4.(5分)已知數(shù)列QUOTE中,a1=1,an+1=an+(-1)n·n2(n∈N*),則a101= ()A.-5150 B.-5151C.5050 D.5051〖解析〗選D.由題意,數(shù)列QUOTE中,a1=1,an+1=an+(-1)n·n2QUOTE,則a2-a1=-12,a3-a2=22,a4-a3=-32,…,a101-a100=1002,各式相加,可得a101-a1=-12+22-32+42-…-992+1002=(-1+2)·(1+2)+(-3+4)·(3+4)+…+(-99+100)·(99+100)=1+2+3+…+100=QUOTE=5050,所以a101=a1+5050=5051.〖加練備選·拔高〗數(shù)列QUOTE滿足QUOTEa1+QUOTEa2+QUOTEa3+…+QUOTEan=2n+1,則QUOTE的通項(xiàng)公式為an=________.

〖解析〗因?yàn)镼UOTEa1+QUOTEa2+QUOTEa3+…+QUOTEan=2n+1,所以QUOTEa1+QUOTEa2+QUOTEa3+…+QUOTEan+QUOTEan+1=2QUOTE+1,兩式相減得QUOTEan+1=2,即an=2n+1,n≥2,又QUOTEa1=3,所以a1=6,不滿足上式,因此an=QUOTE〖答案〗:QUOTE5.(10分)已知函數(shù)f(x)=QUOTE,設(shè)an=f(n)(n∈N*),(1)求證:an<1;(2){an}是遞增數(shù)列還是遞減數(shù)列?為什么?〖解析〗(1)an=f(n)=QUOTE=1-QUOTE<1.(2)因?yàn)閍n+1-an=QUOTE-QUOTE=QUOTE-QUOTE=QUOTE>0,所以an+1>an,所以{an}是遞增數(shù)列.6.(10分)已知數(shù)列QUOTE的前n項(xiàng)和為Sn,且a1=1,Sn=2an+1.(1)求數(shù)列QUOTE的通項(xiàng)公式;(2)當(dāng)bn=QUOTE(3an+1)時(shí),求數(shù)列QUOTE的前n項(xiàng)和Tn.〖解析〗(1)因?yàn)镾n=2an+1,所以Sn-1=2an(n≥2),得Sn-Sn-1=2an+1-2an,所以QUOTE=QUOTE(n≥2),又a1=1,a2=QUOTE,所以QUOTE≠Q(mào)UOTE,所以an=QUOTE(2)bn=QUOTE(3an+1)=n,所以QUOTE=QUOTE=QUOTE-QUOTE,所以Tn=QUOTE+QUOTE+…+QUOTE=1-QUOTE=QUOTE.課時(shí)作業(yè)梯級(jí)練三十七數(shù)列的概念及其表示一、選擇題(每小題5分,共35分)1.不能作為數(shù)列2,0,2,0,…的通項(xiàng)公式的是 ()A.an=1+(-1)n+1 B.an=1-(-1)nC.an=1+(-1)n D.an=1-cosnπ〖解析〗選C.驗(yàn)證易知,只有C選項(xiàng)中的式子不能作為已知數(shù)列的通項(xiàng)公式.2.在數(shù)列{an}中,a1=-2,an+1=QUOTE,則a2014= ()A.-2 B.-QUOTE C.QUOTE D.3〖解析〗選B.因?yàn)閍1=-2,an+1=QUOTE,所以a2=-QUOTE,a3=QUOTE,a4=3,a5=-2.可知數(shù)列{an}是以4為周期的數(shù)列,所以a2014=a2=-QUOTE.3.數(shù)列{an}的前n項(xiàng)和Sn=2n2-3n,則{an}的通項(xiàng)公式為 ()A.4n-5 B.4n-3C.2n-3 D.2n-1〖解析〗選A.因?yàn)镾n=2n2-3n,所以當(dāng)n≥2時(shí),Sn-1=2(n-1)2-3(n-1),兩式相減可得an=Sn-Sn-1=4n-5,又當(dāng)n=1時(shí),a1=S1=-1,滿足上式.4.如圖,給出的3個(gè)三角形圖案中圓的個(gè)數(shù)依次構(gòu)成一個(gè)數(shù)列的前3項(xiàng),則這個(gè)數(shù)列的一個(gè)通項(xiàng)公式是 ()A.2n+1 B.3nC.QUOTE D.QUOTE〖解析〗選D.由an-an-1=n+1,再根據(jù)累加法得an=a1+(a2-a1)+…+(an-an-1)=3+3+4+5+…+n+1=QUOTE.5.已知數(shù)列QUOTE滿足a1=28,QUOTE=2,則an= ()A.n2 B.n2+nC.n2-n+28 D.n2-n〖解析〗選C.由an+1-an=2n知,a2-a1=2×1,a3-a2=2×2,…,an-an-1=2QUOTE,相加得:an-a1=n2-n,所以an=n2-n+28.6.已知數(shù)列QUOTE滿足an+2=an+1-an,若a1=1,a3=3,則a17= ()A.-4 B.-3 C.3 D.4〖解析〗選A.因?yàn)閿?shù)列QUOTE滿足an+2=an+1-an,故有an+3=an+2-an+1=QUOTE-an+1=-an,所以an+6=-an+3=an,故數(shù)列QUOTE是以6為周期的周期數(shù)列,所以a17=a5=-a2,又因?yàn)閍1=1,a3=3,a3=a2-a1得a2=4,故a17=a5=-a2=-4.7.已知數(shù)列QUOTE的通項(xiàng)公式為an=QUOTE(n∈N*),則下列說法正確的是()A.這個(gè)數(shù)列的第10項(xiàng)為QUOTEB.QUOTE是該數(shù)列中的項(xiàng)C.數(shù)列中的各項(xiàng)都在區(qū)間QUOTE內(nèi)D.數(shù)列QUOTE是單調(diào)遞減數(shù)列〖解析〗選C.an=QUOTE=QUOTE=QUOTE.令n=10,得a10=QUOTE,故選項(xiàng)A不正確;令QUOTE=QUOTE,得9n=300,此方程無正整數(shù)解,故QUOTE不是該數(shù)列中的項(xiàng),故選項(xiàng)B不正確;因?yàn)閍n=QUOTE=QUOTE=1-QUOTE,又n∈N*,所以數(shù)列QUOTE是單調(diào)遞增數(shù)列,所以QUOTE≤an<1,所以數(shù)列中的各項(xiàng)都在區(qū)間QUOTE內(nèi),故選項(xiàng)C正確;選項(xiàng)D由上面分析知不正確.二、填空題(每小題5分,共15分)8.數(shù)列{an}的前n項(xiàng)和為Sn,若Sn+Sn-1=2n-1(n≥2,n∈N*),且S2=3,則a1的值為________.

〖解析〗因?yàn)镾n+Sn-1=2n-1(n≥2,n∈N*),令n=2,得S2+S1=3,由S2=3得a1=S1=0.〖答案〗:09.古希臘人常用小石子在沙灘上擺成各種形狀來研究數(shù),如圖所示.他們研究過圖中的1,5,12,22,…,由于這些數(shù)能夠表示成五角形,將其稱為五角形數(shù),若按此規(guī)律繼續(xù)下去,第n個(gè)五角形數(shù)an=________.

〖解析〗觀察圖形,發(fā)現(xiàn)a1=1,a2=a1+4,a3=a2+7,a4=a3+10,猜測(cè)當(dāng)n≥2時(shí),an=an-1+3n-2,所以an-an-1=3n-2,所以an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=(3n-2)+〖3(n-1)-2〗+…+(3×2-2)+1=QUOTEn2-QUOTEn.〖答案〗:QUOTEn2-QUOTEn10.設(shè)數(shù)列QUOTE滿足nan=n2+λ,若數(shù)列QUOTE是單調(diào)遞增數(shù)列,則實(shí)數(shù)λ的取值范圍是________.

〖解析〗因?yàn)閚an=n2+λ,所以an=n+QUOTE,由于數(shù)列QUOTE是單調(diào)遞增數(shù)列,則an+1>an,即n+1+QUOTE>n+QUOTE,整理得λ<n2+n,令bn=n2+n,所以bn+1-bn=QUOTE-QUOTE=2n+2>0,所以,數(shù)列QUOTE單調(diào)遞增,則數(shù)列QUOTE的最小項(xiàng)為b1=12+1=2,所以λ<2.因此,實(shí)數(shù)λ的取值范圍是QUOTE.〖答案〗:QUOTE〖加練備選·拔高〗(2020·石家莊模擬)設(shè)數(shù)列QUOTE的前n項(xiàng)和為Sn,若a1=QUOTE且當(dāng)n≥2時(shí),an=-Sn·Sn-1,則QUOTE的通項(xiàng)公式an=________.

〖解析〗當(dāng)n≥2時(shí),an=-Sn·Sn-1,則Sn-Sn-1=-Sn·Sn-1,所以QUOTE-QUOTE=1,因?yàn)閍1=QUOTE,所以S1=QUOTE,即QUOTE=2,所以QUOTE=2+QUOTE×1=n+1,所以Sn=QUOTE,所以當(dāng)n≥2時(shí),an=Sn-Sn-1=QUOTE-QUOTE=QUOTE,當(dāng)n=1時(shí)a1=QUOTE,不滿足上式,故an=QUOTE〖答案〗:QUOTE1.(5分)已知數(shù)列QUOTE滿足an=QUOTE(n∈N*),則數(shù)列QUOTE的最小項(xiàng)是第________項(xiàng) ()

A.3 B.4 C.5 D.6〖解析〗選C.因?yàn)閍n=QUOTE,-6=a5<a4=-QUOTE,-6=a5<a6=QUOTE,所以數(shù)列QUOTE的最小項(xiàng)是第5項(xiàng).2.(5分)在數(shù)列{an}中,其前n項(xiàng)和為Sn,且Sn=QUOTEan,則QUOTE的最大值為()A.-3 B.-1 C.3 D.1〖解析〗選C.當(dāng)n≥2時(shí),Sn=QUOTEan,Sn-1=QUOTEan-1.兩式作差可得an=Sn-Sn-1=QUOTEan-QUOTEan-1,則QUOTE=QUOTE=1+QUOTE,據(jù)此可得,當(dāng)n=2時(shí),QUOTE取到最大值3.3.(5分)在數(shù)列{an}中,a1=1,an+1-an=sinQUOTE,記Sn為數(shù)列{an}的前n項(xiàng)和,則S2021= ()A.0 B.2020C.1011 D.2021〖解析〗選C.由a1=1及an+1-an=sinQUOTE,得an+1=an+sinQUOTE,所以a2=a1+sinπ=1,a3=a2+sinQUOTE=0,a4=a3+sinQUOTE=0,a5=a4+sinQUOTE=1,a6=a5+sinQUOTE=1,a7=a6+sinQUOTE=0,a8=a7+sinQUOTE=0,…,可見數(shù)列{an}為周期數(shù)列,周期T=4,所以S2021=505(a1+a2+a3+a4)+a1=1011.4.(5分)已知數(shù)列QUOTE中,a1=1,an+1=an+(-1)n·n2(n∈N*),則a101= ()A.-5150 B.-5151C.5050 D.5051〖解析〗選D.由題意,數(shù)列QUOTE中,a1=1,an+1=an+(-1)n·n2QUOTE,則a2-a1=-12,a3-a2=22,a4-a3=-32,…,a101-a100=1002,各式相加,可得a101-a1=-12+22-32+42-…-992+1002=(-1+2)·(1+2)+(-3+4)·(3+4)+…+(-99+100)·(99+100)=1+2+3+…+100=QUOTE=5