全國(guó)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.doc

(二)數(shù)列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(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Snb1b2b3bn1111nnn(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Rn120221n2n1，2Rn12