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1、專題升級訓(xùn)練27 解答題專項訓(xùn)練(數(shù)列)1(2020云南昆明質(zhì)檢,17)已知等差數(shù)列an的前n項和為Sn,a23,S10100.(1)求數(shù)列an的通項公式;(2)設(shè)bnnan,求數(shù)列bn的前n項和Tn.2(2020山東濟(jì)南二模,18)已知等比數(shù)列an的前n項和為Sn,且滿足Sn3nk,(1)求k的值及數(shù)列an的通項公式;(2)若數(shù)列bn滿足,求數(shù)列bn的前n項和Tn.3(2020河南豫東、豫北十校段測,18)已知數(shù)列an的前n項和為Sn,a11,Snnann(n1)(nN*)(1)求數(shù)列an的通項公式;(2)設(shè)bn,求數(shù)列bn的前n項和Tn.4(2020河北石家莊二模,17)已知Sn是等比數(shù)列
2、an的前n項和,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項的積5.(2020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項和為Sn.(1)求an及Sn;(2)令bn(nN*),求數(shù)列bn的前n項和Tn.6(2020廣西南寧三測,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;(2)設(shè)cn,求數(shù)列cn3n1的前n項和Tn.7(2020安徽蕪湖一中,理21)已知數(shù)列an的相鄰兩項an,an1是關(guān)于x的方程x22nxbn0(nN*)的兩根,且a11.(1
3、)證明:數(shù)列是等比數(shù)列(2)求數(shù)列an的前n項和Sn.(3)是否存在常數(shù),使得bnSn0對于任意的正整數(shù)n都成立,若存在,求出的取值范圍;若不存在,請說明理由8(2020北京石景山統(tǒng)測,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(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項之積為Tn,即Tn(2a11)(2a21)(2an1),求數(shù)列an的通項及Tn關(guān)于n的表達(dá)式;(3)記bnlog2an1Tn,求數(shù)列bn的前n項和S
4、n,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,ana1(n1)d2n1.(2)Tn3253(2n1)n,Tn23354(2n1)n1,相減,得Tn22232n(2n1)n1n.Tn1.2解:(1)當(dāng)n2時,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,Sn1(n1)an1(n1)(n2),anSnSn1nann(n1)(n1)an1(n1)(n2)anan12.數(shù)列an是首項a11,公差d2的等差數(shù)列故an1(n1)
5、22n1,nN*.(2)由(1)知bn,Tnb1b2bn1.4解:(1)當(dāng)q1時,2S10S4S7,q1.由2S10S4S7,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項的積為Tn,Tna13a23a33an313q3(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的首項為a1,公差為d.由于a37,a5a726,所以a12d7,2a110d26.解得a13,d2.由于ana1(
6、n1)d,Sn,所以an2n1,Snn(n2)(2)因為an2n1,所以a14n(n1)因此bn,故Tnb1b2bn,所以數(shù)列bn的前n項和Tn(n1)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(1)證明:由題知anan12n,an12n1,故數(shù)列是首項為a1,公比為1的等比數(shù)列(2)解:an2n(1)n1,即an2n(1)n由題知Sna1a2
7、a3an(222232n)(1)(1)2(1)n.(3)解:bnanan12n(1)n2n1(1)n122n1(2)n1要使bnSn0對任意nN*都成立,即22n1(2)n10(*)對任意nN*都成立當(dāng)n為正奇數(shù)時,由(*)式得(2n12n1)(2n11)0,即(2n11)(2n1)(2n11)0.2n110,(2n1)對任意正奇數(shù)n都成立當(dāng)且僅當(dāng)n1時,(2n1)有最小值1,0,即(2n11)(2n1)(2n1)0.2n10,(2n11)對任意正整數(shù)n都成立當(dāng)且僅當(dāng)n2時,(2n11)有最小值,0對任意nN*都成立,且的取值范圍是(,1)8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)為首項是lg 5,公比為2的等比數(shù)列(2)lg(2an1)lg(2a11)2n12n1lg 5lg 52n1,2an1,an(1)lg Tnlg(2a11)lg(2an1)(2n1)lg 5,Tn52n1.(3)bn2,Sn2n2.Sn2 012,2n22 012.n1 007.nmin1 007.