新课标2022届高考数学二轮复习第三部分题型指导考前提分题型练4大题专项二数列的通项求和问题理

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a≠0,a≠1.(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:Tn<13.-5-\n4.已知等差数列{an}的前n项和为Sn,公比为q的等比数列{bn}的首项是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列{an},{bn}的通项公式an,bn;(2)求数列1anan+1+1bnbn+1的前n项和Tn.5.已知数列{an}满足a1=12,且an+1=an-an2(n∈N*).(1)证明:1≤anan+1≤2(n∈N*);(2)设数列{an2}的前n项和为Sn,证明:12(n+2)≤Snn≤12(n+1)(n∈N*).-5-\n6.已知数列{an}的首项为1,Sn为数列{an}的前n项和,Sn+1=qSn+1,其中q>0,n∈N*.(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+…+en>4n-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n≥2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=211×2+12×3+13×4+…+-5-\n1n(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n≥2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=a·an-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+1<13n,13n+1-1>13n+1,所以bn=13n+1-13n+1-1<13n-13n+1.所以Tn=b1+b2+…+bn<13-132+132-133+…+13n-13n+1=13-13n+1.因为-13n+1<0,所以13-13n+1<13,即Tn<13.4.解(1)设{an}公差为d,由题意得a1+2d=8,a1+2q=3,a1+d+2q=6,解得a1=2,d=3,q=12,故an=3n-1,bn=12n.(2)∵1anan+1+1bnbn+1=131an-1an+1+1bnbn+1=131an-1an+1+22n+1,∴Tn=1312-15+15-18+…+13n-1-13n+2+8(1-4n)1-4=1312-13n+2+13(22n+3-8)=1322n+3-13n+2-52.5.证明(1)由题意得an+1-an=-an2≤0,即an+1≤an,故an≤12.由an=(1-an-1)an-1,得an=(1-an-1)(1-an-2)…(1-a1)a1>0.由0<an≤12,得anan+1=anan-an2=11-an∈[1,2],即1≤anan+1≤2.(2)由题意得an2=an-an+1,所以Sn=a1-an+1.①由1an+1-1an=anan+1和1≤anan+1≤2,得1≤1an+1-1an≤2,所以n≤1an+1-1a1≤2n,因此12(n+1)≤an+1≤1n+2(n∈N*).②由①②得12(n+2)≤Snn≤12(n+1)(n∈N*).6.解(1)由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,-5-\n两式相减得到an+2=qan+1,n≥1.又由S2=qS1+1得到a2=qa1,故an+1=qan对所有n≥1都成立.所以,数列{an}是首项为1,公比为q的等比数列.从而an=qn-1.由2a2,a3,a2+2成等差数列,可得2a3=3a2+2,即2q2=3q+2,则(2q+1)(q-2)=0,由已知,q>0,故q=2.所以an=2n-1(n∈N*).(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(k∈N*).于是e1+e2+…+en>1+q+…+qn-1=qn-1q-1,故e1+e2+…+en>4n-3n3n-1.-5-