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1、專題升級(jí)訓(xùn)練28解答題專項(xiàng)訓(xùn)練(數(shù)列)1(2012云南昆明質(zhì)檢,17)已知等差數(shù)列an的前n項(xiàng)和為Sn,a23,S10100.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè),求數(shù)列bn的前n項(xiàng)和Tn.2(2012山東濟(jì)南二模,18)已知等比數(shù)列an的前n項(xiàng)和為Sn,且滿足Sn3nk,(1)求k的值及數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足(4k)anbn,求數(shù)列bn的前n項(xiàng)和Tn.3(2012河南豫東、豫北十校段測(cè),18)已知數(shù)列an的前n項(xiàng)和為Sn,a11,Snnann(n1)(nN*)(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Tn.4(2012河北石家莊二模,17)已知Sn是等比
2、數(shù)列an的前n項(xiàng)和,S4,S10,S7成等差數(shù)列(1)求證a3,a9,a6成等差數(shù)列;(2)若a11,求數(shù)列a的前n項(xiàng)的積5(2012陜西西安三質(zhì)檢,19)已知等差數(shù)列an滿足a27,a5a726,an的前n項(xiàng)和為Sn.(1)求a4及Sn;(2)令bn(nN*),求數(shù)列bn的前n項(xiàng)和Tn.6(2012廣西南寧三測(cè),20)已知數(shù)列an滿足a12,nan1(n1)an2n(n1)(1)證明:數(shù)列為等差數(shù)列,并求數(shù)列an的通項(xiàng);(2)設(shè)cn,求數(shù)列cn3n1的前n項(xiàng)和Tn.7(2012山東濟(jì)寧模擬,20)已知等差數(shù)列an滿足:a25,a4a622.數(shù)列bn滿足b12b22n1bnnan.設(shè)數(shù)列bn的
3、前n項(xiàng)和為Sn.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)求滿足13Sn14的n的集合8(2012北京石景山統(tǒng)測(cè),20)若數(shù)列An滿足An1A,則稱數(shù)列An為“平方遞推數(shù)列”已知數(shù)列an中,a12,點(diǎn)(an,an1)在函數(shù)f(x)2x22x的圖象上,其中n為正整數(shù)(1)證明數(shù)列2an1是“平方遞推數(shù)列”,且數(shù)列l(wèi)g(2an1)為等比數(shù)列;(2)設(shè)(1)中“平方遞推數(shù)列”的前n項(xiàng)之積為Tn,即Tn(2a11)(2a21)(2an1),求數(shù)列an的通項(xiàng)及Tn關(guān)于n的表達(dá)式;(3)記,求數(shù)列bn的前n項(xiàng)和Sn,并求使Sn2 012的n的最小值參考答案1解:(1)設(shè)an的公差為d,有解得a11,d2,a
4、na1(n1)d2n1.(2)Tn35(2n1),Tn35(2n1),相減,得Tn222(2n1).Tn1.2解:(1)當(dāng)n2時(shí),由anSnSn13nk3n1k23n1,a1S13k,所以k1.(2)由(4k)anbn,可得bn,bn,Tn,Tn,所以Tn,Tn.3解:(1)Snnann(n1),當(dāng)n2時(shí),Sn1(n1)an1(n1)(n2),anSnSn1nann(n1)(n1)an1(n1)(n2)anan12.數(shù)列an是首項(xiàng)a11,公差d2的等差數(shù)列故an1(n1)22n1,nN*.(2)由(1)知bn,Tnb1b2bn1.4解:(1)當(dāng)q1時(shí),2S10S4S7,q1.由2S10S4S7
5、,得.a10,q1,2q10q4q7.則2a1q8a1q2a1q5.2a9a3a6.a3,a9,a6成等差數(shù)列(2)依題意設(shè)數(shù)列a的前n項(xiàng)的積為Tn,Tn13q3(q2)3(qn1)3q3(q3)2(q3)n1(q3)123(n1).又由(1)得2q10q4q7,2q6q310,解得q31(舍),q3.Tn.5解:(1)設(shè)等差數(shù)列an的首項(xiàng)為a1,公差為d.由于a37,a5a726,所以a12d7,2a110d26.解得a13,d2.由于ana1(n1)d,Sn,所以an2n1,Snn(n2)(2)因?yàn)閍n2n1,所以a14n(n1)因此bn,故Tnb1b2bn,所以數(shù)列bn的前n項(xiàng)和Tn(n
6、1)6解:(1)nan1(n1)an2n(n1),2.數(shù)列為等差數(shù)列不妨設(shè)bn,則bn1bn2,從而有b2b12,b3b22,bnbn12,累加得bnb12(n1),即bn2n.an2n2.(2)cnn,Tn130231332n3n1,3Tn13232333n3n,兩式相減,得Tn3n,Tn3n.7解:(1)設(shè)等差數(shù)列an的公差為d.a25,a4a622,a1d5,(a13d)(a15d)22.解得a13,d2,an2n1.在b12b22n1bnnan中,令n1得b1a13.又b12b22nbn1(n1)an1,2nbn1(n1)an1nan.2nbn1(n1)(2n3)n(2n1)4n3.b
7、n1.bn(n2)經(jīng)檢驗(yàn),b13也符合上式,所以數(shù)列bn的通項(xiàng)公式為bn.(2)Sn37(4n1),Sn37(4n5)(4n1),兩式相減得:Sn34(4n1),Sn34(4n1).Sn14.nN*,Sn14.數(shù)列bn的各項(xiàng)為正,Sn單調(diào)遞增又計(jì)算得S51413,S61413,滿足13Sn14的n的集合為n|n6,nN8解:(1)an12an22an,2an112(2an22an)1(2an1)2,數(shù)列2an1是“平方遞推數(shù)列”由以上結(jié)論lg(2an11)lg(2an1)22lg(2an1),數(shù)列l(wèi)g(2an1)為首項(xiàng)是lg 5,公比為2的等比數(shù)列(2)lg(2an1)lg(2a11)2n12n1lg 5,2an1,an(1)lg Tnlg(2a11)lg(2an1)(2n1)lg 5,Tn.(3)bn2,Sn2n2.Sn2 012,2n22 012.n1 007.nmin1 007.