新版高三理科数学新课标二轮习题：第三部分 题型指导考前提分 题型练4 Word版含答案

1 1题型练4大题专项(二)数列的通项、求和问题1.设数列an的前n项和为Sn,满足(1-q)Sn+qan=1,且q(q-1)0.(1)求an的通项公式;(2)若S3,S9,S6成等差数列,求证:a2,a8,a5成等差数列.2.已知等差数列an的首项a1=1,公差d=1,前n项和为Sn,bn=1Sn.(1)求数列bn的通项公式;(2)设数列bn前n项和为Tn,求Tn.3.已知数列an的前n项和Sn满足:Sn=aa-1(an-1),a为常数,且a0,a1.(1)求数列an的通项公式;(2)若a=13,设bn=an1+an-an+11-an+1,且数列bn的前n项和为Tn,求证:Tn0,nN*.(1)若2a2,a3,a2+2成等差数列,求数列an的通项公式;(2)设双曲线x2-y2an2=1的离心率为en,且e2=53,证明:e1+e2+en4n-3n3n-1.参考答案题型练4大题专项(二)数列的通项、求和问题1.(1)解当n=1时,由(1-q)S1+qa1=1,a1=1.当n2时,由(1-q)Sn+qan=1,得(1-q)Sn-1+qan-1=1,两式相减,得an=qan-1.又q(q-1)0,所以an是以1为首项,q为公比的等比数列,故an=qn-1.(2)证明由(1)可知Sn=1-anq1-q,又S3+S6=2S9,所以1-a3q1-q+1-a6q1-q=2(1-a9q)1-q,化简,得a3+a6=2a9,两边同除以q,得a2+a5=2a8.故a2,a8,a5成等差数列.2.解(1)在等差数列an中,a1=1,公差d=1,Sn=na1+n(n-1)2d=n2+n2,bn=2n2+n.(2)bn=2n2+n=2n(n+1)=21n-1n+1,Tn=b1+b2+b3+bn=2112+123+134+1n(n+1)=21-12+12-13+13-14+1n-1n+1=21-1n+1=2nn+1.故Tn=2nn+1.3.(1)解因为a1=S1=aa-1(a1-1),所以a1=a.当n2时,an=Sn-Sn-1=aa-1an-aa-1an-1,得anan-1=a,所以数列an是首项为a,公比也为a的等比数列.所以an=aan-1=an.(2)证明当a=13时,an=13n,所以bn=an1+an-an+11-an+1=13n1+13n-13n+11-13n+1=13n+1-13n+1-1.因为13n+113n+1,所以bn=13n+1-13n+1-113n-13n+1.所以Tn=b1+b2+bn13-132+132-133+13n-13n+1=13-13n+1.因为-13n+10,所以13-13n+113,即Tn0.由00,故q=2.所以an=2n-1(nN*).(2)由(1)可知,an=qn-1.所以双曲线x2-y2an2=1的离心率en=1+an2=1+q2(n-1).由e2=1+q2=53,解得q=43.因为1+q2(k-1)q2(k-1),所以1+q2(k-1)qk-1(kN*).于是e1+e2+en1+q+qn-1=qn-1q-1,故e1+e2+en4n-3n3n-1.