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1、(全國(guó)通用版)2022高考數(shù)學(xué)二輪復(fù)習(xí) 中檔大題規(guī)范練(二)數(shù)列 文1(2018濰坊模擬)已知數(shù)列an的前n項(xiàng)和為Sn,且1,an,Sn成等差數(shù)列(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足anbn12nan,求數(shù)列bn的前n項(xiàng)和Tn.解(1)由已知1,an,Sn成等差數(shù)列,得2an1Sn,當(dāng)n1時(shí),2a11S11a1,a11.當(dāng)n2時(shí),2an11Sn1,得2an2an1an,2,數(shù)列an是以1為首項(xiàng),2為公比的等比數(shù)列,ana1qn112n12n1(nN*)(2)由anbn12nan,得bn2n,Tnb1b2bn242n(242n)n2n2(nN*)2(2018四川成都市第七中學(xué)三診)已
2、知公差不為零的等差數(shù)列an中,a37,且a1,a4,a13成等比數(shù)列(1)求數(shù)列an的通項(xiàng)公式;(2)記數(shù)列an2n的前n項(xiàng)和為Sn,求Sn.解(1)設(shè)等差數(shù)列an 的公差為d(d0),則a3a12d7.又a1,a4,a13成等比數(shù)列,aa1a13,即(a13d)2a1(a112d),整理得2a13da10,由解得an32(n1)2n1(nN*)(2)由(1)得an2n(2n1)2n,Sn32522(2n1)2n1(2n1)2n,2Sn322523(2n1)2n(2n1)2n1,得Sn623242n1(2n1)2n122223242n1(2n1)2n1(2n1)2n12(12n)2n1.Sn2
3、(2n1)2n1(nN*)3(2018廈門質(zhì)檢)已知等差數(shù)列an滿足(n1)an2n2nk,kR.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn,求數(shù)列bn的前n項(xiàng)和Sn.解(1)方法一由(n1)an2n2nk,令n1,2,3,得到a1,a2,a3,an是等差數(shù)列,2a2a1a3,即,解得k1.由于(n1)an2n2n1(2n1)(n1),又n10,an2n1(nN*)方法二an是等差數(shù)列,設(shè)公差為d,則ana1d(n1)dn(a1d),(n1)an(n1)(dna1d)dn2a1na1d,dn2a1na1d2n2nk對(duì)于nN*均成立,則解得k1,an2n1(nN*)(2)由bn111,得Snb1b
4、2b3bn1111nnn(nN*)4(2018安徽省江南十校模擬)數(shù)列an滿足a12a23a3nan2.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn,求bn的前n項(xiàng)和Tn.解(1)當(dāng)n1時(shí),a12;當(dāng)n2時(shí),由a12a23a3nan2,a12a23a3(n1)an12,得nan2 ,可得an,又當(dāng)n1時(shí)也成立,an(nN*)(2)bn 2,Tn22(nN*)5(2018宿州模擬)已知數(shù)列an的前n項(xiàng)和為Sn,數(shù)列Sn的前n項(xiàng)和為Tn,滿足Tn2Snn2.(1)證明數(shù)列an2是等比數(shù)列,并求出數(shù)列an的通項(xiàng)公式;(2)設(shè)bnnan,求數(shù)列bn的前n項(xiàng)和Kn.解(1)由Tn2Snn2,得a1S1T12S11,解得a1S11,由S1S22S24,解得a24.當(dāng)n2時(shí),SnTnTn1 2Snn22Sn1(n1)2,即Sn2Sn12n1,Sn12Sn2n1,由得an12an2,an122(an2),又a222(a12),數(shù)列an2是以a123為首項(xiàng),2為公比的等比數(shù)列,an232n1,即an32n12(nN*)(2)bn3n2n12n,Kn3(120221n2n1)2(12n)3(120221n2n1)n2n.記Rn120221n2n1,2Rn121222(n1)2n1n2n,由,得Rn2021222n1n2nn2n (1n)2n1,Rn(n1)2n1.Kn3(n1)2nn2n3(nN*)