1.2.2数列递推关系综合应用专题限时训练(小题提速练)(建议用时:45分钟)一、选择题1.设数列满足a1=a,an+1=(n∈N*),若数列是常数列,则a=()A.-2B.-1C.0D.(-1)n解析:因为数列{an}是常数列,所以a=a2==,即a(a+1)=a2-2,解得a=-2.故选A.答案:A2.在数列{an}中,若a1=1,a2=,=+(n∈N*),则该数列的通项为()A.an=B.an=C.an=D.an=解析:由已知=+,可得-=-,所以是首项为=1,公差为-=2-1=1的等差数列,所以=n,即an=.答案:A3.已知等差数列{an}满足a2=3,Sn-Sn-3=51(n>3),若Sn=100,则n的值为()A.8B.9C.10D.11解析:由Sn-Sn-3=51得,an-2+an-1+an=51,所以an-1=17,又a2=3,∴Sn==100,解得n=10.答案:C4.已知数列{an}满足log3an+1=log3an+1(n∈N*),且a2+a4+a6=9,则log(a5+a7+a9)=()A.-5B.-C.5D.解析: log3an+1=log3an+1,∴an+1=3an.∴数列{an}是以3为公比的等比数列. a2+a4+a6=a2(1+q2+q4)=9,∴a5+a7+a9=a5(1+q2+q4)=a2q3(1+q2+q4)=35.∴log35=-5.故选A.答案:A5.已知Sn表示数列{an}的前n项和,若对任意n∈N*满足an+1=an+a2,且a3=2,则S2019=()A.1008×2020B.1008×2019C.1009×2019D.1009×2020解析:在an+1=an+a2中,令n=1,得a2=a1+a2,a1=0;令n=2,得a3=2=2a2,a2=1,于是an+1-an=1,故数列{an}是首项为0,公差为1的等差数列.S2019==1009×2019.答案:C6.在数列{an}中,a1=1,a2=2,当整数n>1时,Sn+1+Sn-1=2(Sn+S1)都成立,则S15=()A.210B.211C.224D.225解析:n>1时,Sn+1-Sn=Sn-Sn-1+2,∴an+1=an+2,∴an+1-an=2.数列{an}从第二项开始组成公差为2的等差数列,所以S15=a1+(a2+…+a15)=1+×14=211.答案:B7.(2019·广东汕头市一模)设Sn是数列{an}的前n项和,且Sn=-an,则an=()A.·n-1B.·n-1C.2·n-D.n解析:由题意,得S1=a1=-a1,所以a1=.又当n≥2时,Sn-Sn-1=an=-an-+an-1,即=,所以数列{an}是首项为,公比为的等比数列,所以an=n.故选D.答案:D8.已知数列{an}满足a1=1,an+1=(n∈N*),则数列{an}的通项公式为()A.an=2n-1B.an=2-C.an=D.an=解析:由题意得=+1,则+1=2,易知+1=2≠0,所以数列是以2为首项,2为公比的等比数列,则+1=2n,则an=.故选C.答案:C9.已知函数f(n)=n2cos(nπ),且an=f(n),则a1+a2+…+a100=()A.0B.100C.5050D.10200解析:a1+a2+a3+…+a100=-12+22-32+42-…-992+1002=(22-12)+(42-32)+…+(1002-992)=3+7+…+199==5050.故选C.答案:C10.已知数列{an}满足a1=0,an+1=an+2+1,则a13=()A.143B.156C.168D.195解析:由an+1=an+2+1,可知an+1+1=an+1+2+1=(+1)2,∴=+1,又=1,故数列{}是首项为1,公差为1的等差数列,所以=n,所以=13,则a13=168.故选C.答案:C11.定义为n个正数p1,p2,…,pn的“均倒数”.若已知数列{an}的前n项的“均倒数”为,且bn=,则++…+=()A.B.C.D.解析:由已知,得=,∴a1+a2+…+an=n(2n+1)=Sn.当n=1时,a1=S1=3;当n≥2时,an=Sn-Sn-1=4n-1.验证知,当n=1时此式也成立,∴an=4n-1.∴bn==n.∴=-,∴++…+=++…+=.故选C.答案:C12.已知正项数列{an}中,a1=1,a2=2,2a=a+a(n≥2),bn=,记数列{bn}的前n项和为Sn,则S33的值是()A.B.C.4D.3解析: 2a=a+a(n≥2),∴数列{a}为等差数列,首项为1,公差为22-1=3.∴a=1+3(n-1)=3n-2.an>0.∴an=.∴bn===(-),故数列{bn}的前n项和为Sn=[(-)+(-)+…+(-)]=(-1).则S33=(-1)=3.故选D.答案:D二、填空题13.已知等比数列{an}的前n项和为Sn,且Sn=m·2n-1-3,则m=.解析:a1=S1=m-3,当n≥2时,an=Sn-Sn-1=m·2n-2,∴a2=m,a3=2m,又a=a1a3,∴m2=(m-3)·2m,整理得m2-6m=0,则m=6或m=0(舍去).答案:614.已知数列{an}的前n项和Sn=n2+2n+1(n∈N*),则an=.解析:当n≥2时,an=Sn-Sn-1=2n+1;当n=1时,a1=S1=4...