2021高考数学大一轮复习考点规范练29数列的概念与表示理新人教A版.docx

6考点规范练29数列的概念与表示考点规范练A册基础巩固1.数列123354759…的一个通项公式an=()A.n2n+1B.n2n-1C.n2n-3D.n2n+3答案:B2.若Sn为数列{an}的前n项和且Sn=nn+1则1a5等于()A.56B.65C.130D.30答案:D解析:当n≥2时an=Sn-Sn-1=nn+1-n-1n=1n(n+1)则1a5=5×(5+1)=30.3.已知数列{an}满足an+1+an=n若a1=2则a4-a2=()A.4B.3C.2D.1答案:D解析:由an+1+an=n得an+2+an+1=n+1两式相减得an+2-an=1令n=2得a4-a2=1.4.(2019广东六校第一次联考)已知数列{an}的前n项和为Sn=n2+n+1bn=(-1)nan(n∈N*)则数列{bn}的前50项和为()A.49B.50C.99D.100答案:A解析:由题意得当n≥2时an=Sn-Sn-1=2n当n=1时a1=S1=3所以数列{bn}的前50项和为-3+4-6+8-10+…+96-98+100=1+48=49故选A.5.若数列{an}满足a1=12an=1-1an-1(n≥2且n∈N*)则a2018等于()A.-1B.12C.1D.2答案:A解析:∵a1=12an=1-1an-1(n≥2且n∈N*)∴a2=1-1a1=1-112=-1∴a3=1-1a2=1-1-1=2∴a4=1-1a3=1-12=12……依此类推可得an+3=an∴a2018=a672×3+2=a2=-1故选A.6.设数列252211…则41是这个数列的第项.答案:14解析:由已知得数列的通项公式为an=3n-1.令3n-1=41解得n=14即为第14项.7.(2019安徽合肥高三调研)已知数列{an}的前n项和为Sna1=1Sn+1=2Sn(n∈N*)则a10=.答案:256解析:因为a1=S1=1Sn+1=2Sn所以数列{Sn}是公比为2的等比数列所以Sn=2n-1所以a10=S10-S9=29-28=28=256.8.(2019河北衡水中学摸底联考)已知数列{an}若数列{3n-1an}的前n项和Tn=15×6n-15则a5=.答案:16解析:根据题意得a1+3a2+32a3+…+3n-1an=15×6n-15故当n≥2时a1+3a2+32a3+…+3n-2an-1=15×6n-1-15两式相减得3n-1an=15×6n-15×6n-1=6n-1.即当n≥2时an=2n-1故a5=16.9.设数列{an}是首项为1的正项数列且(n+1)an+12-nan2+an+1·an=0则它的通项公式an=.答案:1n解析:∵(n+1)an+12-nan2+an+1·an=0∴(n+1)an+1-nanan+1+an=0.∵{an}是首项为1的正项数列∴(n+1)an+1=nan即an+1an=nn+1故an=anan-1·an-1an-2·…·a2a1·a1=n-1n·n-2n-1·…·12·1=1n.10.已知数列{an}的前n项和为Sn.(1)若Sn=(-1)n+1·n求a5+a6及an;(2)若Sn=3n+2n+1求an.解:(1)因为Sn=(-1)n+1·n所以a5+a6=S6-S4=(-6)-(-4)=-2.当n=1时a1=S1=1;当n≥2时an=Sn-Sn-1=(-1)n+1·n-(-1)n·(n-1)=(-1)n+1·[n+(n-1)]=(-1)n+1·(2n-1).又a1也适合于此式所以an=(-1)n+1·(2n-1).(2)当n=1时a1=S1=6;当n≥2时an=Sn-Sn-1=(3n+2n+1)-[3n-1+2(n-1)+1]=2·3n-1+2.①因为a1不适合①式所以an=6n=12·3n-1+2n≥2.能力提升11.已知数列{an}满足an+1=an-an-1(n≥2)a1=ma2=nSn为数列{an}的前n项和则S2017的值为()A.2017n-mB.n-2017mC.mD.n答案:C解析:∵an+1=an-an-1(n≥2)a1=ma2=n∴a3=n-ma4=-ma5=-na6=m-na7=ma8=n…∴an+6=an.则S2017=S336×6+1=336×(a1+a2+…+a6)+a1=336×0+m=m.12.已知函数f(x)是定义在区间(0+∞)内的单调函数且对任意的正数xy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.32n-1答案:D解析:由题意知f(Sn+2)=f(an)+f(3