1、高考资源网() 您身边的高考专家12015郑州质量预测(二)已知等差数列an的各项均为正数,a11,且a3,a4,a11成等比数列(1)求an的通项公式;(2)设bn,求数列bn的前n项和Tn.解(1)设等差数列an的公差为d,由题意知d0,因为a3,a4,a11成等比数列,所以2a3a11,所以2(12d)(110d),即44d236d450,所以d(d舍去),所以an.(2)bn,所以Tn.22015石家庄一模设数列an的前n项和为Sn,a11,an1Sn1(nN*,1),且a1、2a2、a33为等差数列bn的前三项 (1)求数列an,bn的通项公式;(2)求数列anbn的前n项和解(1)
2、解法一:an1Sn1(nN*),anSn11(n2),an1anan,即an1(1)an(n2),10,又a11,a2S111,数列an是以1为首项,公比为1的等比数列,a3(1)2,4(1)1(1)23,整理得2210,解得1,an2n1,bn13(n1)3n2.解法二:a11,an1Sn1(nN*),a2S111,a3S21(11)1221,4(1)12213,整理得2210,解得1,an1Sn1(nN*),anSn11(n2),an1anan(n2),即an12an(n2),又a11,a22,数列an是以1为首项,公比为2的等比数列,an2n1,bn13(n1)3n2.(2)由(1)知,
3、anbn(3n2)2n1,设Tn为数列anbn的前n项和,Tn11421722(3n2)2n1,2Tn121422723(3n5)2n1(3n2)2n.得,Tn1132132232n1(3n2)2n13(3n2)2n,整理得:Tn(3n5)2n5.32015云南统测在数列an中,a1,an12,设bn,数列bn的前n项和是Sn.(1)证明数列bn是等差数列,并求Sn;(2)比较an与Sn7的大小解(1)证明:bn,an12,bn11bn1,bn1bn1,数列bn是公差为1的等差数列由a1,bn得b1,Sn3n.(2)由(1)知:bnn1n.由bn得an11.anSn73n6.当n4时,y3n6
4、是减函数,y也是减函数,当n4时,anSn7a4S470.又a1S170,a2S270,a3S370,a31,由S37得:a1a2a37,17,6q2q10,q或q(舍)a14,an4n123n.(2)bn23n(3n),TnTn得:Tn111n,Tn1.52015大连一模已知数列an中,a11,其前n项的和为Sn,且满足an(n2) (1)求证:数列是等差数列;(2)证明:当n2时,S1S2S3Sn.证明(1)当n2时,SnSn1,Sn1Sn2SnSn1,2,从而是以1为首项,2为公差的等差数列(2)由(1)可知,(n1)22n1,Sn,当n2时,Sn,从而S1S2S3Sn1Cn1,当n5时,CnCn1,所以当n4时Cn最大,所以C4,所以k.高考资源网版权所有,侵权必究!