《(江蘇專用)2013年高考數(shù)學(xué)總復(fù)習(xí) 第五章第4課時(shí) 數(shù)列通項(xiàng)與求和隨堂檢測(含解析)》由會(huì)員分享,可在線閱讀,更多相關(guān)《(江蘇專用)2013年高考數(shù)學(xué)總復(fù)習(xí) 第五章第4課時(shí) 數(shù)列通項(xiàng)與求和隨堂檢測(含解析)(2頁珍藏版)》請?jiān)谘b配圖網(wǎng)上搜索。
1、(江蘇專用)2013年高考數(shù)學(xué)總復(fù)習(xí) 第五章第4課時(shí) 數(shù)列通項(xiàng)與求和 課時(shí)闖關(guān)(含解析)1設(shè)an是正項(xiàng)數(shù)列,其前n項(xiàng)和Sn滿足條件4Sn(an1)(an3),則數(shù)列an的通項(xiàng)公式an_.解析:4Sn(an1)(an3)a2an3,當(dāng)n1時(shí),4S14a1a2a13,a13(a11舍去);當(dāng)n2時(shí),4Sn1a2an13,得4an(aa)2(anan1)(anan1)(anan12)0,an0,anan10,anan12.故an為首項(xiàng)為3,公差為2的等差數(shù)列,an3(n1)22n1.答案:2n12已知數(shù)列an滿足a11,anan11(n2)則an_.解析:兩邊同時(shí)加上2,則有an2(an12),即.
2、數(shù)列an2成等比數(shù)列,公比為,首項(xiàng)為 a121,an2()n1,故an2()n1.答案:2()n13已知函數(shù)f(n)且anf(n)f(n1),則a1a2a3a100等于_解析:當(dāng)n為奇數(shù)時(shí),ann2(n1)2(2n1),當(dāng)n為偶函數(shù)時(shí),ann2(n1)22n1,則an(1)n(2n1)a1a2a3a1003579199201250100.答案:1004設(shè)數(shù)列an是一個(gè)公差不為0的等差數(shù)列,它的前10項(xiàng)的和S10110,且a1,a2,a4成等比數(shù)列(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bnn2an,求數(shù)列bn 的前n項(xiàng)的和Tn.解:(1)設(shè)數(shù)列an的公差為d(d0),則an2(n1)22n(nN*
3、)(2)由bnn2ann4n,則Tn14242343(n1)4n1n4n,4Tn142243(n2)4n1(n1)4nn4n1兩式相減得:3Tn4424nn4n1(4n1)n4n1,Tn1(3n1)4n5函數(shù)f(x)(m0),x1,x2R,當(dāng)x1x21時(shí),f(x1)f(x2).(1)求m的值;(2)已知數(shù)列an滿足anf(0)f()f()f()f(1),求an.(3)若Sna1a2an,求Sn.解:(1)令x1x2,則2f()即.則m2.當(dāng)x1x21時(shí),f(x1)f(x2).m2合題意(2)anf(0)f()f()f()f(1),anf(1)f()f()f()f(0)整理得:2an,即an.(3)Sna1a2an(12n)n.6(2011高考課標(biāo)全國卷)等比數(shù)列an的各項(xiàng)均為正數(shù),且2a13a21,a9a2a6.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bnlog3a1log3a2log3an,求數(shù)列的前n項(xiàng)和解:(1)設(shè)數(shù)列an的公比為q,由a9a2a6得a9a,q2,由條件可知q0故q.由 2a13a21,q知a1.故數(shù)列an的通項(xiàng)公式為an.(2)bnlog3a1log3a2log3an(123n).故2(),2(1)()().所以數(shù)列的前n項(xiàng)和為.