高考數(shù)學(xué)專題復(fù)習(xí)導(dǎo)練測 第六章 數(shù)列階段測試(八)課件 理 新人教A版.ppt
數(shù)學(xué) A(理),45分鐘階段測試(八),第六章 數(shù) 列,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,1.(2014遼寧)設(shè)等差數(shù)列an的公差為d.若數(shù)列 為遞減數(shù)列,則( ) A.d0 C.a1d0 解析 設(shè)bn ,則bn1 ,由于 是遞減數(shù)列,則bnbn1,即 .y2x是單調(diào)增函數(shù),a1ana1an1,a1ana1(and)0,a1(anand)0,即a1(d)0,a1d0.,C,3,4,5,6,7,8,9,10,1,2,2.已知等比數(shù)列an的首項為1,若4a1,2a2,a3成等差數(shù)列,則數(shù)列 的前5項和為( ),解析 設(shè)公比為q,4a24a1a3, 4q4q2,q2.數(shù)列 是首項為1,,3,4,5,6,7,8,9,10,1,2,答案 A,2,4,5,6,7,8,9,10,1,3,3.若數(shù)列an滿足:a119,an1an3(nN*),則數(shù)列an的前n項和數(shù)值最大時,n的值是( ) A.6 B.7 C.8 D.9 解析 an1an3,anan13, an是以19為首項,以3為公差的等差數(shù)列, an19(n1)(3)223n.,2,4,5,6,7,8,9,10,1,3,nN*,n7,故答案為B.,答案 B,2,3,5,6,7,8,9,10,1,4,4.已知數(shù)列an的前n項和Sn2an1,則滿足 2的正整數(shù)n的集合為( ) A.1,2 B.1,2,3,4 C.1,2,3 D.1,2,4 解析 因為Sn2an1, 所以當(dāng)n2時,Sn12an11, 兩式相減得an2an2an1,整理得an2an1,,2,3,5,6,7,8,9,10,1,4,所以an是公比為2的等比數(shù)列, 又因為a12a11,解得a11, 故an的通項公式為an2n1. 而 2,即2n12n,所以有n1,2,3,4. 答案 B,2,3,4,6,7,8,9,10,1,5,2,3,4,6,7,8,9,10,1,5,答案 C,2,3,4,5,7,8,9,10,1,6,6.已知公差不為零的等差數(shù)列an的前n項和為Sn,若a10S4,則 _. 解析 由a10S4,,二、填空題,即a1d0.,2,3,4,5,7,8,9,10,1,6,答案 4,2,3,4,5,6,8,9,10,1,7,7.設(shè)Sn是等比數(shù)列an的前n項和,若2S1,3S2,4S3成等差數(shù)列,則等比數(shù)列an的公比q_. 解析 由2S1,3S2,4S3成等差數(shù)列, 得6S22S14S3, 即3S2S12S3,2(S2S3)S2S10,2,3,4,5,6,9,10,1,7,8,8.設(shè)數(shù)列an,若an1anan2(nN*),則稱數(shù)列an為“凸數(shù)列”,已知數(shù)列bn為“凸數(shù)列”,且b11,b22,則數(shù)列bn前2 013項的和為_. 解析 由“凸數(shù)列”的定義,可寫出數(shù)列的前幾項, 即b11,b22,b33,b41,b52,b63, b71,b82,,2,3,4,5,6,9,10,1,7,8,故數(shù)列bn為周期為6的周期數(shù)列. 又b1b2b3b4b5b60, 故S2 013S33563 b1b2b31234.故填4. 答案 4,2,3,4,5,6,7,8,10,1,9,9.設(shè)數(shù)列an的前n項和為Sn,已知a11,an13Sn1,nN*. (1)求數(shù)列an的通項公式; 解 由題意,an13Sn1, 則當(dāng)n2時,an3Sn11. 兩式相減,得an14an(n2).,三、解答題,2,3,4,5,6,7,8,10,1,9,所以數(shù)列an是以首項為1,公比為4的等比數(shù)列, 所以數(shù)列an的通項公式是an4n1(nN*).,2,3,4,5,6,7,8,10,1,9,(2)記Tn為數(shù)列nan的前n項和,求Tn.,解 因為Tn(1a1)(2a2)(3a3)(nan) (12n)(14424n1),2,3,4,5,6,7,8,9,1,10,(1)求數(shù)列an的通項公式; 解 當(dāng)n1時,a1S16, 當(dāng)n2時,anSnSn1,而當(dāng)n1時,n56,符合上式, ann5(nN*).,2,3,4,5,6,7,8,9,1,10,2,3,4,5,6,7,8,9,1,10,Tnc1c2cn,2,3,4,5,6,7,8,9,1,10,所以kmax671.,