課時(shí)規(guī)范練46數(shù)列中的構(gòu)造問(wèn)題一、基礎(chǔ)鞏固練1.已知在數(shù)列{an}中,a1=4,an+1=4an-6,則an=()A.22n+1+2 B.22n+1-2C.22n-1+2 D.22n-1-22.在數(shù)列{an}中,a1=14,an+12nA.{an2n+3}是等比數(shù)列 B.{C.{an2n+32}是等比數(shù)列3.(2024·山東菏澤模擬)已知在數(shù)列{an}中,a1=1且an+1=3anan+3(n∈N*),則A.16 B.14 C.13 4.已知數(shù)列{an}滿足a1=1,nan+1=2(n+1)an,設(shè)bn=ann,則數(shù)列{bn}的前6項(xiàng)和為(A.127 B.255 C.31 D.635.已知數(shù)列{an}滿足an+1=an2an+1,n∈N*,若a4=19,則{6.已知數(shù)列{an}滿足a1=1,ann+1=an-1n(7.(2024·四川樂(lè)山模擬)已知數(shù)列{an}滿足an+1=2an+2,a1=1,則an=.

8.(2024·江西景德鎮(zhèn)模擬)已知在數(shù)列{an}中,a1=1,當(dāng)n≥2時(shí),(n-1)an=2nan-1,則數(shù)列{an}的通項(xiàng)公式為.

9.已知數(shù)列{an}滿足a1=1,an+1=2an+3n,則數(shù)列{an}的通項(xiàng)公式為.

10.設(shè)數(shù)列{an}滿足a1=4,an=3an-1+2n-1(n≥2),則數(shù)列{an}的通項(xiàng)公式為.

二、綜合提升練11.(2024·江西臨川模擬)已知數(shù)列{an}滿足a1=3,an+1=an+2an+1+1,則a10=(A.80 B.100 C.120 D.14312.(多選題)(2024·黑龍江伊春模擬)已知數(shù)列{an}的首項(xiàng)為4,且滿足2(n+1)an-nan+1=0(n∈N*),則()A.{ann}為等比數(shù)列 B.{C.{an}為遞增數(shù)列 D.{an}為遞減數(shù)列13.已知在數(shù)列{an}中,nan+1=2(n+1)an+n(n+1)且a1=1,則數(shù)列{an}的通項(xiàng)公式為.

課時(shí)規(guī)范練46數(shù)列中的構(gòu)造問(wèn)題1.C解析因?yàn)閍n+1=4an-6,所以an+1-2=4(an-2).又a1-2=2,所以數(shù)列{an-2}是一個(gè)以2為首項(xiàng),4為公比的等比數(shù)列,所以an-2=2×4n-1,所以an=22n-1+2.2.B解析由題知an+12n+1=an2n-1-3,所以an+12所以{an2n-3}是等比數(shù)列,且首項(xiàng)為4,3.A解析易知an≠0,由an+1=3anan+3,得∴數(shù)列{1an}是以1為首項(xiàng),13為公差的等差數(shù)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項(xiàng)和為S6=1-265.an=12n+1解析由題得an≠0,則等式兩邊同取倒數(shù)得1an+1=2a則數(shù)列{1an}為公差為2的等差數(shù)列,則1an=1a4+2(n-4)=6.50解析根據(jù)題意,令bn=ann+1,則由ann+1=an-1n(n≥2),得bn=bn-所以{bn}是首項(xiàng)為b1=12的常數(shù)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解析當(dāng)n≥2時(shí),(n-1)an=2nan-1,即ann=2·an-因此數(shù)列{ann}是以1為首項(xiàng),2為公比的等比數(shù)列,則ann=1×2n-1=2n-1,所以數(shù)列{an}的通項(xiàng)公式為an=n·9.an=3n-2n解析an+1=2an+3n兩邊同除以3n+1得an令bn=an3n,則bn+1=23bn+13.設(shè)bn+1+λ=23(bn則bn+1-1=23(bn-1).又b1-1=-2∴數(shù)列{bn-1}是以-23為首項(xiàng),23為公比的等比數(shù)列,bn-1=-(23)n,得an=3n-10.an=2·3n-n-1解析設(shè)an+pn+q=3[an-1+p(n-1)+q],化簡(jiǎn)后得an=3an-1+2pn+(2q-3p),與原遞推式比較,對(duì)應(yīng)項(xiàng)的系數(shù)相等,得2p=2即an+n+1=3(an-1+n-1+1).令bn=an+n+1,則bn=3bn-1.又b1=6,故bn=6·3n-1=2·3n,得an=2·3n-n-1.11.C解析因?yàn)閍n+1=an+2an+1所以an+1+1=(an+1)2+2an+1+1,即an+1+1=(a等式兩邊開方可得an+1+1=an+1+1,即an+1+1?an+1=1,所以數(shù)列{an+1}是首項(xiàng)為a1+1=2,公差為1的等差數(shù)列,所以an+1=2+(n-1)×1=n+1,12.ABC解析由題意,2(n+1)an-nan+1=0?an+1n+1=2·ann.又a11=4,故{ann}是首項(xiàng)為4,公比為2的等比數(shù)列,故A正確;ann=4·2n-1=2n+1,顯然{ann}是遞增數(shù)列,故B正確;an=n·2n+1,由an+1an=(n+113.an=n·(2n-1)解析∵nan+1=2(n+1)an+n(n+1),等式兩邊同除以n(n+1),可得an+1n+1