2024-2025學年新教材高中數(shù)學第四章數(shù)列4.1第2課時數(shù)列的遞推公式課后提升訓練含解析新人教A版選擇性必修第二冊

第四章數(shù)列4.1數(shù)列的概念第2課時數(shù)列的遞推公式課后篇鞏固提升基礎(chǔ)達標練1.數(shù)列12,14,A.an=12n+1(n∈N*) B.an=12n(n∈C.an+1=12an(n∈N*) D.an+1=2an(n∈N*解析數(shù)列從第2項起,后一項是前一項的12,故遞推公式為an+1=12an(n∈N*答案C2.(多選)符合遞推關(guān)系式an=2an-1的數(shù)列是()A.1,2,3,4,… B.1,2,2,22,…C.2,2,22,4,… D.0,2,2,22,…解析B與C中從第2項起,后一項是前一項的2倍,符合遞推公式an=2an-1.答案BC3.在數(shù)列{an}中,an+1=an+2-an,a1=2,a2=5,則a5=()A.-3 B.-11 C.-5 D.19解析由an+1=an+2-an,得an+2=an+an+1,則a3=a1+a2=7,a4=a2+a3=12,a5=a3+a4=19.答案D4.已知數(shù)列{an}的前n項和為Sn,且Sn=2an-2,則a2=()A.4 B.2 C.1 D.-2解析Sn=2an-2?a1=S1=2a1-2?a1=2?a1+a2=S2=2a2-2?a2=4.故選A.答案A5.已知a1=1,an=n(an+1-an)(n∈N*),則數(shù)列{an}的通項公式是()A.2n-1 B.n+1nn-1 C.解析法一:構(gòu)造法.由已知整理,得(n+1)an=nan+1,∴an+1n+1=a且ann=a11=法二:累乘法.當n≥2時,an…a3兩邊分別相乘,得ana1=n.∵a1=1,∴a答案D6.在數(shù)列{an}中,若a1=2,an+1=an+n-1,則a4=.

解析a2=a1+1-1=2,a3=a2+2-1=3,a4=a3+3-1=5.答案57.(2024濰坊高三檢測)已知數(shù)列{an}的前n項和Sn=n2-9n,第k項滿意5<ak<8,則k=.

解析∵an=S∴an=2n-10.由5<2k-10<8,得7.5<k<9,∴k=8.答案88.若數(shù)列{an}滿意an+1=2an-1,且a8=16,則a6=.

解析∵an+1=2an-1,∴a8=2a7-1=16,解得a7=172,又a7=2a6-1=172,解得a6=答案199.在數(shù)列{an}中,a1=2,an+1=an+ln1+1n,求a解由題意,得an+1-an=lnn+1∴an-an-1=lnnn-1(an-1-an-2=lnn-…a2-a1=ln21∴當n≥2時,an-a1=lnnn-1·n-1n-2·…·21=lnn,∴an=當n=1時,a1=2+ln1=2,符合上式,∴an=2+lnn(n∈N*).10.已知各項均不為0的數(shù)列{an}滿意a1=12,anan-1=an-1-an(n≥2,n∈N*),求數(shù)列{an}的通項公式解∵anan-1=an-1-an,且各項均不為0,∴1an-1an-1=1.∴當n≥=2+1+1+…+1(∴1an∴當n≥2時,an=1n+1∵a1=12也符合上式,∴an=1實力提升練1.已知數(shù)列{an},a1=2,a2=1,an+2=3an+1-an,則a6+a4-3a5的值為()A.3 B.-2 C.-1 D.0解析∵an+2=3an+1-an,∴an+2+an=3an+1.令n=4,得a6+a4=3a5,∴a6+a4-3a5=0.答案D2.已知數(shù)列{an},an+1=11-an,a1=3,則a2019A.23 B.3 C.-12 D解析由題意,可知:a1=3,a2=11-aa3=11a4=11-a5=11-a…∴數(shù)列{an}是一個以3為最小正周期的周期數(shù)列.∵2024÷3=673,∴a2024=a3=23答案A3.已知數(shù)列{an},a1=1,lnan+1-lnan=1,則數(shù)列{an}的通項公式是()A.an=n B.an=1C.an=en-1 D.an=1解析∵lnan+1-lnan=1,∴l(xiāng)nan+1an=1.∴由累乘法可得an=en-1.答案C4.在數(shù)列{an}中,a1=1,an+1=an2-1,則此數(shù)列的前4項和為解析∵a1=1,an+1=an2-1,∴a2=12-1=0,a3=02-1=-1,a4=(-1)2-1=0,故前4項和a1+a2+a3+a4=答案05.(2024浙江期末)已知數(shù)列{an}滿意:a1=2,an=an+1an-1(n∈N*).設(shè)Sn為數(shù)列{an}的前n項和,則a2=,解析由題意知a2=a1+1a1-1=2+12-1=3,又a3=3+13-1=2=a1,…,∴數(shù)列{an}是周期數(shù)列,周期為2,答案350476.已知在數(shù)列{an}中,an+1=2anan+2對隨意正自然數(shù)n都成立,且a7=12,則解析由已知a7=2a6a6+2=1又因為a6=2a5a5+2=2答案17.在數(shù)列{an}中,已知a1=2,a2=3,an+2=3an+1-2an(n≥1),寫出此數(shù)列的前6項.解a1=2,a2=3,a3=3a2-2a1=3×3-2×2=5,a4=3a3-2a2=3×5-2×3=9,a5=3a4-2a3=3×9-2×5=17,a6=3a5-2a4=3×17-2×9=33.8.已知數(shù)列{an}滿意an+1=2若a1=67,試求a2018+a2019解∵a1=67∴a2=2a1-1=57∴a3=2a2-1=37∈0,12,∴a4=∴數(shù)列{an}是周期數(shù)列,且周期為3.∴a2024+a2024=a672×3+2+a673×3=a2+a3=57素養(yǎng)培優(yōu)練已知數(shù)列a1=1,a2,a3,…,an(n∈N*)的法則如下:若an為自然數(shù),則an+1=an-2,否則an+1=an+3,則a6=.

解析∵a1=1是自然數(shù)