第一节数列的概念及简单表示法A组基础题组1.数列1,,,,,…的一个通项公式是()A.an=B.an=C.an=D.an=2.已知数列{an}的前n项和Sn=n2-2n,则a2+a18=()A.36B.35C.34D.333.设数列{an}的前n项和为Sn,且Sn=2(an-1),则an=()A.2nB.2n-1C.2nD.2n-14.已知数列{an}满足a1=1,an+1=-2an+1(n∈N*),则a2017=()A.1B.0C.2017D.-20175.数列{an}中,a1=1,对于所有的n≥2,n∈N*,都有a1·a2·a3·…·an=n2,则a3+a5=()A.B.C.D.6.已知数列{an}的前n项和Sn=2n-3,则数列{an}的通项公式是.7.数列{an}的通项公式是an=(n+1)·,则此数列的最大项是第项.8.已知数列{an}满足a1=2,an+1=(n∈N*),则该数列的前2018项的乘积a1·a2·a3·…·a2018=.9.已知各项都为正数的数列{an}满足a1=1,-(2an+1-1)·an-2an+1=0.(1)求a2,a3;(2)求{an}的通项公式.110.(2018河北石家庄调研)已知数列{an}中,a1=1,前n项和Sn=an.(1)求a2,a3;(2)求{an}的通项公式.B组提升题组1.已知数列{an}满足a1=1,an+1an=2n(n∈N*),则a10=()A.64B.32C.16D.82.设数列{an}的前n项和为Sn,且a1=1,{Sn+nan}为常数列,则an=()A.(n∈N*)B.(n∈N*)C.(n∈N*)D.(n∈N*)3.设Sn是数列{an}的前n项和,已知a1=3,an+1=2Sn+3(n∈N*).(1)求数列{an}的通项公式;(2)令bn=(2n-1)an,求数列{bn}的前n项和Tn.2答案精解精析A组基础题组1.B数列可写成,,,…,故通项公式可写为an=.故选B.2.C当n≥2时,an=Sn-Sn-1=2n-3;当n=1时,a1=S1=-1,适合上式,所以an=2n-3(n∈N*),所以a2+a18=34.3.C当n=1时,a1=S1=2(a1-1),可得a1=2,当n≥2时,an=Sn-Sn-1=2an-2an-1,∴an=2an-1,∴数列{an}为等比数列,公比为2,首项为2,所以an=2n.4.A∵a1=1,∴a2=(a1-1)2=0,a3=(a2-1)2=1,a4=(a3-1)2=0,…,可知数列{an}是以2为周期的数列,∴a2017=a1=1.5.A当n≥2时,a1·a2·a3·…·an=n2.当n≥3时,a1·a2·a3·…·an-1=(n-1)2.两式相除得an=(n≥2,n∈N*),∴a3=,a5=,∴a3+a5=.6.答案an=解析当n=1时,a1=S1=2-3=-1,当n≥2时,an=Sn-Sn-1=(2n-3)-(2n-1-3)=2n-2n-1=2n-1.又a1=-1不适合上式,故an=7.答案9或10解析∵an+1-an3=(n+2)-(n+1)=×,当n<9时,an+1-an>0,即an+1>an;当n=9时,an+1-an=0,即an+1=an;当n>9时,an+1-an<0,即an+1
1时,有an=Sn-Sn-1=an-·an-1,整理得=,因此··…··=··…··,化简得an=·a1=,n=1时,a1=1满足上式,所以{an}的通项公式为an=.B组提升题组1.B因为an+1an=2n,所以an+2an+1=2n+1,两式相除得=2.又a1a2=2,a1=1,所以a2=2.所以···=24,即a10=25=32.2.B由题意知,Sn+nan=2①,当n≥2时,Sn-1+(n-1)·an-1=2②,∴由①-②可得(n+1)an=(n-1)an-1,n≥2,从而···…·=××…×,n≥2,则an=,n≥2,当n=1时上式成立,所以an=(n∈N*),故选B.3.解析(1)当n≥2时,由an+1=2Sn+3,得an=2Sn-1+3,两式相减,得an+1-an=2Sn-2Sn-1=2an,∴an+1=3an,∴=3.当n=1时,a1=3,a2=2S1+3=2a1+3=9,则=3,∴数列{an}是以3为首项,3为公比的等比数列,∴an=3×3n-1=3n.(2)由(1)得bn=(2n-1)an=(2n-1)×3n.∴Tn=1×3+3×32+5×33+…+(2n-1)×3n,①3Tn=1×32+3×33+5×34+…+(2n-1)×3n+1,②①-②得-2Tn=1×3+2×32+2×33+…+2×3n-(2n-1)×3n+15=3+2×(32+33+…+3n)-(2n-1)×3n+1=3+2×-(2n-1)×3n+1=-6-(2n-2)×3n+1.∴Tn=(n-1)×3n+1+3.6
