2025优化设计一轮课时规范练46　数列中的构造问题

课时规范练46数列中的构造问题一、基础巩固练1.已知在数列{an}中,a1=4,an+1=4an-6,则an=()A.22n+1+2 B.22n+1-2C.22n-1+2 D.22n-1-22.在数列{an}中,a1=14,an+12nA.{an2n+3}是等比数列 B.{C.{an2n+32}是等比数列3.(2024·山东菏泽模拟)已知在数列{an}中,a1=1且an+1=3anan+3(n∈N*),则A.16 B.14 C.13 4.已知数列{an}满足a1=1,nan+1=2(n+1)an,设bn=ann,则数列{bn}的前6项和为(A.127 B.255 C.31 D.635.已知数列{an}满足an+1=an2an+1,n∈N*,若a4=19,则{6.已知数列{an}满足a1=1,ann+1=an-1n(7.(2024·四川乐山模拟)已知数列{an}满足an+1=2an+2,a1=1,则an=.

8.(2024·江西景德镇模拟)已知在数列{an}中,a1=1,当n≥2时,(n-1)an=2nan-1,则数列{an}的通项公式为.

9.已知数列{an}满足a1=1,an+1=2an+3n,则数列{an}的通项公式为.

10.设数列{an}满足a1=4,an=3an-1+2n-1(n≥2),则数列{an}的通项公式为.

二、综合提升练11.(2024·江西临川模拟)已知数列{an}满足a1=3,an+1=an+2an+1+1,则a10=(A.80 B.100 C.120 D.14312.(多选题)(2024·黑龙江伊春模拟)已知数列{an}的首项为4,且满足2(n+1)an-nan+1=0(n∈N*),则()A.{ann}为等比数列 B.{C.{an}为递增数列 D.{an}为递减数列13.已知在数列{an}中,nan+1=2(n+1)an+n(n+1)且a1=1,则数列{an}的通项公式为.

课时规范练46数列中的构造问题1.C解析因为an+1=4an-6,所以an+1-2=4(an-2).又a1-2=2,所以数列{an-2}是一个以2为首项,4为公比的等比数列,所以an-2=2×4n-1,所以an=22n-1+2.2.B解析由题知an+12n+1=an2n-1-3,所以an+12所以{an2n-3}是等比数列,且首项为4,3.A解析易知an≠0,由an+1=3anan+3,得∴数列{1an}是以1为首项,13为公差的等差数列,∴1an=1+1∴an=3n+2,∴a164.D解析由an+1n+1=2×ann及bn=ann,得bn+1=2bn.又b1=a于是{bn}的前6项和为S6=1-265.an=12n+1解析由题得an≠0,则等式两边同取倒数得1an+1=2a则数列{1an}为公差为2的等差数列,则1an=1a4+2(n-4)=6.50解析根据题意,令bn=ann+1,则由ann+1=an-1n(n≥2),得bn=bn-所以{bn}是首项为b1=12的常数列,故bn=12,即ann+1=12,所以a99=12×(99+1)=7.3×2n-1-2解析由an+1=2an+2得an+1+2=2(an+2).又a1+2=3,所以an+1+2an+2=2,即{所以an+2=3×2n-1,即an=3×2n-1-2.8.an=n·2n-1解析当n≥2时,(n-1)an=2nan-1,即ann=2·an-因此数列{ann}是以1为首项,2为公比的等比数列,则ann=1×2n-1=2n-1,所以数列{an}的通项公式为an=n·9.an=3n-2n解析an+1=2an+3n两边同除以3n+1得an令bn=an3n,则bn+1=23bn+13.设bn+1+λ=23(bn则bn+1-1=23(bn-1).又b1-1=-2∴数列{bn-1}是以-23为首项,23为公比的等比数列,bn-1=-(23)n,得an=3n-10.an=2·3n-n-1解析设an+pn+q=3[an-1+p(n-1)+q],化简后得an=3an-1+2pn+(2q-3p),与原递推式比较,对应项的系数相等,得2p=2即an+n+1=3(an-1+n-1+1).令bn=an+n+1,则bn=3bn-1.又b1=6,故bn=6·3n-1=2·3n,得an=2·3n-n-1.11.C解析因为an+1=an+2an+1所以an+1+1=(an+1)2+2an+1+1,即an+1+1=(a等式两边开方可得an+1+1=an+1+1,即an+1+1−an+1=1,所以数列{an+1}是首项为a1+1=2,公差为1的等差数列,所以an+1=2+(n-1)×1=n+1,12.ABC解析由题意,2(n+1)an-nan+1=0⇔an+1n+1=2·ann.又a11=4,故{ann}是首项为4,公比为2的等比数列,故A正确;ann=4·2n-1=2n+1,显然{ann}是递增数列,故B正确;an=n·2n+1,由an+1an=(n+113.an=n·(2n-1)解析∵nan+1=2(n+1)an+n(n+1),等式两边同除以n(n+1),可得an+1n+1