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1、2022高考數(shù)學(xué)二輪復(fù)習(xí)”一本“培養(yǎng)優(yōu)選練 中檔大題分類練2 數(shù)列 理1(2018海南省二模)已知數(shù)列an是公差為1的等差數(shù)列,且a4,a6,a9成等比數(shù)列(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn(2)an(1)nan,求數(shù)列bn的前2n項(xiàng)和解(1)因?yàn)閍4,a6,a9成等比數(shù)列,所以aa4a9,又因?yàn)閿?shù)列an是公差為1的等差數(shù)列,a6a15,a4a13,a9a18,所以(a15)2(a13)(a18),解得a11,所以ana1(n1)dn.(2)由(1)可知ann,因?yàn)閎n(2)an(1)nan,所以bn(2)n(1)nn.所以S2n2(2)2(2)2n(123452n)nn.【教師備選】(
2、2017全國(guó)卷)設(shè)數(shù)列an滿足a13a2(2n1)an2n.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和解(1)因?yàn)閍13a2(2n1)an2n,故當(dāng)n2時(shí),a13a2(2n3)an12(n1),兩式相減得(2n1)an2,所以an(n2)又由題設(shè)可得a12,滿足上式,所以an的通項(xiàng)公式為an.(2)記的前n項(xiàng)和為Sn.由(1)知,則Sn.2設(shè)Sn為數(shù)列an的前n項(xiàng)和,已知a37,an2an1a22(n2)(1)證明:an1為等比數(shù)列;(2)求an的通項(xiàng)公式,并判斷n,an,Sn是否成等差數(shù)列?解(1)證明:a37,a33a22,a23,an2an11,a11,2(n2),an1是首項(xiàng)為2,
3、公比為2的等比數(shù)列(2)由(1)知,an12n,an2n1,Snn2n1n2,nSn2ann2n1n22(2n1)0,nSn2an,即n,an,Sn成等差數(shù)列【教師備選】已知數(shù)列an滿足:a11,an1an.(1)設(shè)bn,求數(shù)列bn的通項(xiàng)公式;(2)求數(shù)列an的前n項(xiàng)和Sn.解(1)由an1an可得.又bn,bn1bn,由a11,得b11,累加法可得:(b2b1)(b3b2)(bnbn1),化簡(jiǎn)并代入b11得:bn2.(2)由(1)可知an2n,設(shè)數(shù)列的前n項(xiàng)和為Tn,則Tn,Tn,Tn2,Tn4.Snn(n1)4.3設(shè)各項(xiàng)均為正數(shù)的數(shù)列an的前n項(xiàng)和為Sn,滿足4Sna4n1,nN*,且a2
4、,a5,a14構(gòu)成等比數(shù)列(1)求數(shù)列an的通項(xiàng)公式;(2)證明:對(duì)一切正整數(shù)n,有.解(1)當(dāng)n2時(shí),4Sn1a4(n1)1,4an4Sn4Sn1aa4,aa4an4(an2)2,an0,an1an2,當(dāng)n2時(shí),an是公差d2的等差數(shù)列a2,a5,a14構(gòu)成等比數(shù)列,aa2a14,(a26)2a2(a224),解得a23,當(dāng)n1時(shí),4a1a54,a11,a2a1312,an是首項(xiàng)a11,公差d2的等差數(shù)列數(shù)列an的通項(xiàng)公式為an2n1.(2).4已知數(shù)列an為等比數(shù)列,其前n項(xiàng)和為Sn,且Sn4n31(R)(1)求an的通項(xiàng)公式;(2)設(shè)bnlog21,求數(shù)列的前n項(xiàng)和Tn.解(1)由Sn4n31,得Sn14n131(x2)anSnSn134n1當(dāng)n1時(shí),a1S11,4.an是以1為首項(xiàng),4為公比的等比數(shù)列16,.an4n1,當(dāng)n1時(shí),a1,符合上式an4n1.(2)由(1)知bnlog21log212n.Tn,Tn,得:Tn11,Tn.