考点规范练27数列的概念与表示考点规范练A册第20页基础巩固组1.数列0,,…的一个通项公式为()A.an=B.an=C.an=D.an=答案:C解析:将0写成,观察数列中每一项的分子、分母可知,分子为偶数列,可表示为2(n-1),n∈N+;分母为奇数列,可表示为2n-1,n∈N+,故选C.2.若Sn为数列{an}的前n项和,且Sn=,则等于()A.B.C.D.30答案:D解析:当n≥2时,an=Sn-Sn-1=,∴=5×(5+1)=30.3.数列{an}的前n项积为n2,则当n≥2时,an=()A.2n-1B.n2C.D.答案:D解析:设数列{an}的前n项积为Tn,则Tn=n2,当n≥2时,an=.4.设数列{an}满足:a1=2,an+1=1-,记数列{an}的前n项和为Sn,则S2016的值为()A.1006B.1007C.1008D.1008.5导学号〚32470473〛答案:B解析:由a2=,a3=-1,a4=2可知,数列{an}是周期为3的周期数列,从而S2016=(a1+a2+a3)×672=×672=1008.5.已知数列{an}的前n项和为Sn,a1=1,Sn=2an+1,则Sn等于()A.2n-1B.C.D.答案:B解析:∵Sn=2an+1,∴当n≥2时,Sn-1=2an.∴an=Sn-Sn-1=2an+1-2an(n≥2),即(n≥2).又a2=,∴an=(n≥2).当n=1时,a1=1≠,∴an=∴Sn=2an+1=2×.6.(2015大连双基测试)数列{an}满足:a1+3a2+5a3+…+(2n-1)·an=(n-1)·3n+1+3(n∈N+),则数列{an}的通项公式an=.答案:3n解析:a1+3a2+5a3+…+(2n-3)·an-1+(2n-1)·an=(n-1)·3n+1+3,把n换成n-1得,a1+3a2+5a3+…+(2n-3)·an-1=(n-2)·3n+3,两式相减得an=3n.7.若数列{an}的前n项和Sn=an+,则{an}的通项公式是an=.答案:(-2)n-1解析:∵Sn=an+,①∴当n≥2时,Sn-1=an-1+.②①-②,得an=an-an-1,即=-2(n≥2).∵a1=S1=a1+,∴a1=1.∴{an}是以1为首项,-2为公比的等比数列,故an=(-2)n-1.8.已知数列{an}的通项公式为an=(n+2),则当an取得最大值时,n=.导学号〚32470475〛答案:5或6解析:由题意令∴解得∴n=5或6.9.设{an}是首项为1的正项数列,且(n+1)-n+an+1·an=0(n=1,2,3,…),则它的通项公式an=.答案:解析:∵(n+1)+an+1·an-n=0,∴(an+1+an)[(n+1)an+1-nan]=0.又an+1+an>0,∴(n+1)an+1-nan=0,即,∴·…·×…×.又a1=1,∴an=.10.(2015重庆模拟)设数列{an}的前n项和为Sn,且Sn=2n-1.数列{bn}满足b1=2,bn+1-2bn=8an.(1)求数列{an}的通项公式;(2)证明:数列为等差数列,并求{bn}的通项公式.解:(1)当n=1时,a1=S1=21-1=1;当n≥2时,an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1.∵a1=1适合通项公式an=2n-1,∴an=2n-1.(2)∵bn+1-2bn=8an,∴bn+1-2bn=2n+2,即=2,又=1,∴是首项为1,公差为2的等差数列.∴=1+2(n-1)=2n-1,∴bn=(2n-1)×2n.11.(2015陕西五校模拟)设数列{an}的前n项和为Sn,且Sn=4an-p,其中p是不为零的常数.(1)证明:数列{an}是等比数列;(2)当p=3时,数列{bn}满足bn+1=bn+an(n∈N+),b1=2,求数列{bn}的通项公式.(1)证明:∵Sn=4an-p,∴Sn-1=4an-1-p(n≥2),∴当n≥2时,an=Sn-Sn-1=4an-4an-1,整理得(n≥2).由Sn=4an-p,令n=1,得a1=4a1-p,解得a1=.∴{an}是首项为,公比为的等比数列.(2)解:当p=3时,由(1)知,an=.由bn+1=bn+an,得bn+1-bn=.当n≥2时,可得bn=b1+(b2-b1)+(b3-b2)+…+(bn-bn-1)=2+=3-1,当n=1时,上式也成立.∴数列{bn}的通项公式为bn=3-1.能力提升组12.已知函数f(x)是定义在(0,+∞)上的单调函数,且对任意的正数x,y都有f(xy)=f(x)+f(y).若数列{an}的前n项和为Sn,且满足f(Sn+2)-f(an)=f(3)(n∈N+),则an等于()A.2n-1B.nC.2n-1D.导学号〚32470476〛答案:D解析:由题意知f(Sn+2)=f(an)+f(3)=f(3an)(n∈N+),∴Sn+2=3an,Sn-1+2=3an-1(n≥2),两式相减得,2an=3an-1(n≥2).又n=1时,S1+2=3a1=a1+2,∴a1=1.∴数列{an}是首项为1,公比为的等比数列.∴an=.13.已知数列{an}满足:a1=1,an+1=(n∈N+).若bn+1=(n-λ)·,b1=-λ,且数列{bn}是递增数列,则实数λ的取值范围为()A.λ>2B.λ>3C.λ<2D.λ<3导学号〚32470477〛答案:C解析:由已知可得+1,+1=2.又+1=2≠0,则+1=2n,bn+1=2n(n-λ),bn=2n-1(n-1-λ)(n≥2).b1=-λ也适合上式,故bn=2n-1(n-1-λ).由bn+1>bn,得2n(n-λ)>2n-1(n-1-λ),即λ
