1、高考解答题专项三数列中的综合问题1.已知数列an的前n项和为Sn,等比数列an为递增数列,S3=7,且3a2是a1+3和a3+4的等差中项,bn=an+1SnSn+1,设数列bn的前n项和为Tn,是否存在实数k,使得Tnk恒成立?若实数k存在,求出k的最小值;若不存在,请说明理由.2.(2021全国乙,文19)设数列an是首项为1的等比数列,数列bn满足bn=nan3.已知a1,3a2,9a3成等差数列.(1)求an和bn的通项公式;(2)记Sn和Tn分别为an和bn的前n项和.证明:Tn0,数列an为递增数列,所以q=2,所以an=2n-1,Sn=2n-1,所以bn=an+1SnSn+1=2
2、n(2n-1)(2n+1-1)=12n-112n+1-1,所以Tn=1-13+1317+12n-112n+1-1=1-12n+1-11.当k1时,使得Tnk恒成立,故k的最小值为1.2.(1)解设an的公比为q,则an=qn-1.因为a1,3a2,9a3成等差数列,所以1+9q2=23q,解得q=13,故an=13n-1.由bn=nan3,得bn=n313n-1=n13n.(2)证明由(1)可知Sn=1-13n1-13=321-13n.又bn=n3n,则Tn=131+232+333+n-13n-1+n3n,两边同乘13,得13Tn=132+233+334+n-13n+n3n+1,-,得23Tn
3、=13+132+133+134+13nn3n+1,即23Tn=131-13n1-13n3n+1=121-13nn3n+1,整理得Tn=341-13nn23n=342n+343n,则2Tn-Sn=234-2n+343n321-13n=-n3n0.故Tn0,所以数列bn是以1为首项,2为公比的等比数列.(2)解由(1)知,bn=2n-1,Sn=1-2n1-2=2n-1,则cn=|2n-16|,故cn=16-2n(1n4),2n-16(n4),则当1n4时,Tn=(16-21)+(16-22)+(16-2n)=16n-(21+22+2n)=16n-2(1-2n)1-2=16n-2n+1+2.当n4时
4、,Tn=(16-21)+(16-22)+(16-24)+(25-16)+(26-16)+(2n-16)=2T4+(21+22+2n)-16n=234+2(1-2n)1-2-16n=2n+1-16n+66,则Tn=16n-2n+1+2(1n4),2n+1-16n+66(n4).4.解(1)因为2Sn=an+1,所以a1=1.因为2Sn=an+1,所以Sn=(an+1)24.当n2时,Sn-1=(an-1+1)24,-得,2an+2an-1=an2an-12,所以an-an-1=2,所以数列an是首项为1,公差为2的等差数列,所以an=2n-1.(2)由题意可知,b1=a1=S1,b2=a2+a3
5、=S3-S1,b3=a4+a5+a6=S6-S3,b4=a7+a8+a9+a10=S10-S6,所以bn=Sn(n+1)2Sn(n+1)2-n,而Sn=(1+2n-1)n2=n2,所以bn=Sn(n+1)2Sn(n+1)2-n=n(n+1)22-n(n+1)2-n2=n3.由可得(-1)nbnn=(-1)nn2,所以T2n=(-1+22)+(-32+42)+(-52+62)+-(2n-1)2+(2n)2=3+7+(4n-1)=n(3+4n-1)2=n(2n+1).5.解(1)设an的公差为d,由a2=1,S7=14得a1+d=1,7a1+21d=14,解得a1=12,d=12,an=n2.b1
6、b2b3bn=2n2+n2=2n(n+1)2,b1b2b3bn-1=2n(n-1)2(n2),两式相除得bn=2n(n2).当n=1时,b1=2符合上式,bn=2n(nN*).(2)cn=bncos(an)=2ncosn2,T2n=2cos2+22cos+23cos32+24cos(2)+22n-1cos(2n-1)2+22ncos(n)=22cos+24cos(2)+22ncos(n)=-22+24-+(-1)n22n=-41-(-4)n1+4=-4+(-4)n+15.6.解(1)设第一行数的公差为d,各列的公比为q.由题意可知a13a23a33=a233=1,解得a23=1.由a32+a33+a34=3a33=32,解得a33=12,则q=a33a23=12.由a23=a13q=(a11+2d)q=(1+2d)12=1,解得d=12,因此a1n=a11+(n-1)d=1+n-12=n+12.(2)由ann=a1nqn-1=n+1212n-1=n+12n,可得Sn=221+322+423+n+12n,两边同时乘以12可得,12Sn=222+323+n2n+n+12n+1,上述两式相减可得,12Sn=1+122+123+12n-n+12n+1=1+122(1-12n-1)1-12n+12n+1=32n+32n+1,因此Sn=3-n+32n.
