考點(diǎn)規(guī)范練27數(shù)列的概念與簡單表示法基礎(chǔ)鞏固組1.數(shù)列1,3,5,7,9,…的一個通項(xiàng)公式為()A.an=2n1B.an=B(1)n(2n1)C.an=(1)n+1(2n1)D.an=D(1)n(2n+1)2.數(shù)列{an}中,a1=1,對所有n∈N*都有a1a2…an=n2,則a3+a5等于()A.6116 B.259 C.25163.(2017浙江溫州測試)設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,且Sn=32(an1)(n∈N*),則an=(A.3(3n2n) B.3n+2C.3n D.3·2n14.(2017廣西南寧測試)已知數(shù)列{an}滿足:an+1an+1+1=12,且a2A.12 B.23 C.12 D.5.數(shù)列{an}滿足an+1+an=2n3,若a1=2,則a8a4=()A.7 B.6 C.5 D.46.已知數(shù)列{an}中,首項(xiàng)a1=1,an=an1·3n1(n≥2,n∈N*),則數(shù)列{bn}的通項(xiàng)公式為.

7.數(shù)列{an}滿足:a1+3a2+5a3+…+(2n1)·an=(n1)·3n+1+3(n∈N*),則數(shù)列{an}的通項(xiàng)公式an=.

8.若數(shù)列{an}滿足a1=2,an+1=1+an1-an(n∈N*),則該數(shù)列的前2018項(xiàng)的乘積a1·a2·a3·…·能力提升組9.(2017浙江嘉興模擬)已知數(shù)列{an}中的任意一項(xiàng)都為正實(shí)數(shù),且對任意m,n∈N*,有am·an=am+n,如果a10=32,則a1的值為()A.2 B.2 C.2D.D210.已知函數(shù)f(x)是定義在(0,+∞)上的單調(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(xiàng)和為Sn,且滿足f(Sn+2)f(an)=f(3)(n∈N*),則an等于()A.2n1B.nB C.2n1 D.311.已知數(shù)列{an}滿足:a1=1,an+1=anan+2(n∈N*).若bn+1=(nλ)·1an+1,b1=λ,且數(shù)列{bn}是單調(diào)遞增數(shù)列A.λ>2 B.λ>3 C.λ<2 D.λ<312.(2017遼寧沈陽期末)若數(shù)列{an}滿足1an+1-2an=0,則稱{an}為“夢想數(shù)列”,已知正項(xiàng)數(shù)列1bn為“夢想數(shù)列”,且b1+b2+b3=2,則bA.4 B.16 C.32 D.6413.已知數(shù)列{an}滿足a1=43,an+11=an2an(n∈N*),則m=1a1+1aA.1 B.2 C.3 D.414.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,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個五角形數(shù)an=.

15.(2017浙江溫州瑞安模擬)已知數(shù)列{an}中,an=1+1a+2(n-1)(n∈N*,a(1)若a=7,求數(shù)列{an}中的最大項(xiàng)和最小項(xiàng)的值;(2)若對任意的n∈N*,都有an≤a6成立,求a的取值范圍.16.在數(shù)列{an}中,a1=1,2anan+1+an+1an=0(n∈N*).(1)求證:數(shù)列1an為等差數(shù)列,并求{an}(2)若tan+1(an1)+1≥0對任意n≥2的整數(shù)恒成立,求實(shí)數(shù)t的取值范圍.答案：1.C由數(shù)列{an}中1,3,5,7,9,…可以看出:符號正負(fù)相間,通項(xiàng)的絕對值為1,3,5,7,9…為等差數(shù)列{bn},其通項(xiàng)公式bn=2n1.∴數(shù)列1,3,5,7,9,…的一個通項(xiàng)公式為an=(1)n+1(2n1).故選C.2.A∵當(dāng)n≥2時,a1a2a3…an=n2,當(dāng)n≥3時,a1a2a3…an1=(n1)2,兩式相除,得an=nn∴a3=94,a5=2516.∴a3+a5=613.C當(dāng)n≥2時,an=SnSn1=32(an1)32(an11),整理,得an=3an1.由a1=32(a11),得a1=3,∴anan-1=3,∴數(shù)列{an}是以3為首項(xiàng),3為公比的等比數(shù)列,∴4.D由已知得an+1+1an+1=2,則{an+1}是公比為2的等比數(shù)列,所以a4+1=(a2+1)·22=12,則a45.D依題意得(an+2+an+1)(an+1+an)=[2(n+1)3](2n3),即an+2an=2,所以a8a4=(a8a6)+(a6a4)=2+2=4.6.an=3n(n-1)2∵an=anan-1·an-1an-2·…·a2a1·7.3na1+3a2+5a3+…+(2n3)·an1+(2n1)·an=(n1)·3n+1+3,把n替換成n1得,a1+3a2+5a3+…+(2n3)·an1=(n2)·3n+3,兩項(xiàng)相減得an8.6經(jīng)計(jì)算,得a1=2,a2=3,a3=12,a4=13,a5=則{an}是以4為周期的一個周期數(shù)列.∴a1a2a3a4=1.∴a1·a2·…·a2013·a2014·a2018=2×(3)=6.9.C令m=1,則an+1an=a1,所以數(shù)列{an}是以a1為首項(xiàng),公比為a1的等比數(shù)列,從而an=a1n,因?yàn)閍10=10.D由題意知f(Sn+2)=f(an)+f(3)=f(3an)(n∈N*),∴Sn+2=3an,Sn1+2=3an1(n≥2),兩式相減得,2an=3an1(n≥2).又n=1時,S1+2=3a1=a1+2,∴a1=1.∴數(shù)列{an}是首項(xiàng)為1,公比為32的等比數(shù)列.∴an=11.C由已知可得1an+1=2an+又1a1+1=2≠0,則1an+1=2n,bn+1=2nbn=2n1(n1λ)(n≥2).b1=λ也適合上式,故bn=2n1(n1λ)(n∈N*).由bn+1>bn,得2n(nλ)>2n1(n1λ),即λ<n+1恒成立.而n+1的最小值為2,故λ的取值范圍為λ<2.12.D因?yàn)檎?xiàng)數(shù)列1bn為“夢想數(shù)列”,所以11bn+1-21bn=0,即bn+1=2bn,所以{bn}是以2為公比的等比數(shù)列,所以b6+b7+b8=(b1+b2+b3)×213.B∵a1=43,an+11=an2an(n∈∴an+1an=(an1)2>0,∴an+1>an,∴數(shù)列{an}是單調(diào)遞增數(shù)列,由an+11=an2an=an(a∴1∴1an=1an-1-1an+1-1由a1=43>1,則an+1an=(an1)2>∴a2=1+49,a3=1+5281,a4=1+6…,a2018>2,∴0<1a2018-1<1,∴2<m<3,14.32n212n觀察圖象,發(fā)現(xiàn)a1=1,a2=a1+4,a3=a2+7,a4=a猜測當(dāng)n≥2時,an=an1+3n2,∴anan1=3n2.∴an=(anan1)+(an1an2)+…+(a2a1)+a1=(3n2)+[3(n1)2]+…+(3×22)+1=32n21215.解(1)∵an=1+1a+2(n-1)(n∈N*,a∵a=7,∴an=1+12n-9(n∈結(jié)合函數(shù)f(x)=1+12x-9的單調(diào)性,可知1>a1>a2>aa5>a6>a7>…>an>1(n∈N*).∴數(shù)列{an}中的最大項(xiàng)為a5=2,最小項(xiàng)為a4=0.(2)an=1+1a+2(n-已知對任意的n∈N*,都有an≤a6成立,結(jié)合函數(shù)f(x)=1+12x可知5<2-a2<6,即10即a的取值范圍是(10,8).16.解(1)由題意得2anan