20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：61-數(shù)列的概念與表示-【含解析】-資料下載頁
2025-04-03 03:29本頁面
【正文】 1.例6解(1)由題意得a2=12,a3=14.(2)由an2(2an+11)an2an+1=0,得2an+1(an+1)=an(an+1).因為{an}的各項都為正數(shù),所以an+1an={an}是首項為1,公比為12的等比數(shù)列,因此an=12n1.對點訓(xùn)練3(1)D (2)31n (3)2n2n(4)=13n1 (5)116 (1)因為a=(n,an),b=(an+1,n+1),且a⊥b,所以nan+1+(n+1)an=0,所以an+1an=n+1n,所以a2a1=21,a3a2=32,…,a100a99=,得a100a1=100,因為a1=1,所以a100=.(2)由題意,得an+1an=1n(n+1)=1n1n+1,an=(anan1)+(an1an2)+…+(a2a1)+a1=1n11n+1n21n1+…+1213+112+2=31n.又a1也滿足上式,∴an=31n.(3)由題意,得ann1=n1nan+1n+11,當n≥2時,ann1n1=an+1n+11n,所以數(shù)列ann1n1從第二項開始是常數(shù)列,又a22121=2,所以ann1n1=2,即an=2n2n(n≥2),又n=1也滿足,所以an=2n2n.(4)由題意,得an+1bn+1=bn+1+an2bn+1=12bn+1+12an=1213an+23bn+12an=13(anbn),即an+1bn+1anbn=13,即+1==,b1=,所以c1=a1b1=,故{},公比為13的等比數(shù)列,所以=13n1.(5)由an+1=an2an+1an,得1an+11an=2,∴數(shù)列1an是1a1=2,公差d=2的等差數(shù)列,∴1an=2+2(n1)=2n,a8=116.例70 ∵a1=0,an+1=3+an13an,∴a2=31=3,a3=3+3133=232=3,a4=331+33=0,即數(shù)列{an}為周期數(shù)列,周期為3,且a1+a2+a3=0,則S2020=S3673+1=a1=0.對點訓(xùn)練425 由已知可得,a2=2351=15,a3=215=25,a4=225=45,a5=2451=35,∴{an}為周期數(shù)列,且周期T=4,∴a2019=a5044+3=a3=25.例8(1)(16,+∞) (2)D (1)當n≥7時,數(shù)列{Sn}為遞增數(shù)列,設(shè)Sn+1Sn,即Sn+1Sn=an+10,∴an+1=2(n+1)+λ0,則λ2n2.又n≥7,∴2n2≤16,即λ16.(2)由于{an}是遞增數(shù)列,所以a1,且a2a1,即a22a+3,解得a1或a3,所以a3.對點訓(xùn)練5B ∵anan+1恒成立,∴a滿足12a0,(12a)5+1a,0a1,解得12a712.例9(1)C (2)A (1)∵Sn=n+23an,∴當n≥2時,an=SnSn1=n+23ann+13an1,可化為anan1=n+1n1=1+2n1,由函數(shù)y=2x1在區(qū)間(1,+∞)內(nèi)單調(diào)遞減,可得當n=2時,2n1取得最大值2.∴anan1的最大值為3.(2)因為n∈N*,所以當1≤n≤3時,an=12n150,單調(diào)遞減,當n≥4時,an=12n150,單調(diào)遞減,所以最小項為a3=1815=17,最大項為a4=11615=1,故選A.對點訓(xùn)練6(1)C (2)B (1)an=nn2+90=1n+90n,令f(x)=x+90x(x0),運用基本不等式得f(x)≥610,當且僅當x=310時,=1n+90n,n∈N*,所以1n+90n1610,即當n=9或n=10時,an=119最大.(2)∵Sn=n210n,∴當n≥2時,an=SnSn1=2n11。當n=1時,a1=S1=9也適合上式.∴an=2n11(n∈N*).記f(n)=nan=n(2n11)=2n211n,∴此函數(shù)圖象的對稱軸為直線n=114,又n∈N*,∴當n=3時,f(n)取最小值.∴數(shù)列{nan}中數(shù)值最小的項是第3項.12
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