1、高考资源网() 您身边的高考专家第五章第4节1数列an中,an,若an的前n项和为,则项数n为( )A2 019B2 016C2 017 D2 018解析:Aan,Sn11,所以n2 019.2.等于()A. B.C. D.解析:B法一:令Sn,则Sn,得Sn.Sn.故选B.法二:取n1时,代入各选项验证可知选B.3已知数列an:,那么数列bn的前n项和为()A4 B4C1 D.解析:A由题意知an,bn4,所以b1b2bn44444.4数列an的通项公式为an(1)n1(4n3),则它的前100项之和S100等于()A200B200C400D400解析:BS100(413)(423)(433
2、)(41003)4(12)(34)(99100)4(50)200.5数列an满足a11,且对任意的nN*都有an1a1ann,则的前100项和为()A. B. C. D.解析:D数列an满足a11,且对任意的nN*都有an1a1ann,an1an1n,anan1n,an(anan1)(an1an2)(a2a1)a1n(n1)21,2,的前100项和22,故选D.6(2019聊城市一模)已知数列an的前n项和公式为Snn2,若bn2an,则数列bn的前n项和Tn_.解析:Snn2,当n1时,S1a11,当n2时,Sn1(n1)2,由可得an2n1,当n1时也成立,an2n1,bn2an24n1,
3、Tn(4n1)答案:(4n1)7数列an的前n项和Snn24n2,则|a1|a2|a10|_.解析:当n1时,a1S11.当n2时,anSnSn12n5.an令2n50,得n,当n2时,an0,|a1|a2|a10|(a1a2)(a3a4a10)S102S266.答案:668数列an的前n项和Sn2n1,则aaa_.解析:当n1时,a1S11,当n2时,anSnSn12n1(2n11)2n1,又a11适合上式an2n1,a4n1.数列a是以a1为首项,以4为公比的等比数列aaa(4n1)答案:(4n1)9. (2017全国卷)记Sn为等比数列an的前n项和已知S22,S36.(1)求an的通项
4、公式;(2)求Sn,并判断Sn1,Sn,Sn2是否成等差数列解:(1)设an的公比为q,由题设可得,解得故an的通项公式为an(2)n.(2)由(1)可得Sn(1)n.由于Sn2Sn1(1)n22Sn,故Sn1,Sn,Sn2成等差数列10(2018天津卷)设an是等差数列,其前n项和为Sn(nN*);bn是等比数列,公比大于0,其前n项和为Tn(nN*)已知b11,b3b22,b4a3a5,b5a42a6.(1)求Sn和Tn;(2)若Sn(T1T2Tn)an4bn,求正整数n的值解:(1)设等比数列bn的公比为q,由b11,b3b22,可得q2q20,因为q0,可得q2,故bn2n1.所以,Tn2n1.设等差数列an的公差为d,由b4a3a5,可得a13d4,由b5a42a6,可得3a113d16,从而a11,d1,故ann.所以Sn.(2)由(1),有T1T2Tn(21222n)nn2n1n2.由Sn(T1T2Tn)an4bn可得2n1n2n2n1,整理得n23n40,解得n1(舍),或n4.所以n的值为4.高考资源网版权所有,侵权必究!