高考数学二轮总复习题型专项集训题型练4大题专项（二）文

题型练4大题专项(二)数列的通项、求和问题1.(2022新高考Ⅱ,17)已知{an}为等差数列,{bn}为公比为2的等比数列,且a2b2=a3b3=b4a4.(1)证明:a1=b1;(2)求集合{k|bk=am+a1,1≤m≤500}中元素的个数.2.设数列{an}的前n项和为Sn,已知a1=1,Sn=n+12a(1)求数列{an}的通项公式;(2)设bn=2an,数列{bn}的前n项和为Tn,求证:Tn·Tn+2<3.已知公差不为零的等差数列{an}满足a3=4,且a2,a1,a3成等比数列.(1)求数列{an}的通项公式;(2)若数列{an3n1}的前n项和为Sn,求使Sn≤20成立的最小正整数n.4.(2022山东烟台模拟)已知{an}是公差为2的等差数列,a1>0,且a4是2a2和a52的等比中项.(1)求{an}的通项公式;(2)设数列{bn}满足b1a1+b2a2+…+bnan=25.已知函数f(x)=7x+5x+1,数列{an}满足:2an+12an+an+1an=0,且anan+1≠0.在数列{bn}中,b1=f(0),且bn=f((1)求证:数列1a(2)求数列{|bn|}的前n项和Tn.6.已知等比数列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列{bn}满足b1=1,数列{(bn+1bn)an}的前n项和为2n2+n.(1)求q的值;(2)求数列{bn}的通项公式.答案：1.(1)证明:设等差数列{an}的公差为d,由a2b2=a3b3,等比数列{bn}的公比为2,得a1+d2b1=a1+2d4b1,得d=2b1.由a2b2=b4a4,等比数列{bn}的公比为2,得a1+d2b1=8b1(a1+3d),所以a1+2b12b1=8b1(a1+6b1),所以a1=b1.(2)解:由bk=am+a1,可得b1·2k1=a1+(m1)d+a1,由(1)知d=2b1=2a1,所以b1·2k1=b1+(m1)·2b1+b1,所以2k1=2m.又1≤m≤500,∴2≤2k1≤1000.∴2≤k≤10.又k∈Z,故集合{k|bk=am+a1,1≤m≤500}中元素的个数为9.2.(1)解:∵Sn=n+12an,∴2Sn=(n+1)a∴2Sn+1=(n+2)an+1,∴2Sn+12Sn=(n+2)an+1(n+1)an,即2an+1=(n+2)an+1(n+1)an,∴(n+1)an=nan+1,即an∴ann是以a11=1为首项的常数列,∴(2)证明:由(1)知bn=2n,∴Tn=2(1-2n∴T

n+12Tn·Tn+2=4(2n+11)24(2n1)·(2n+21)=4(22n+22n+2+122n+2+2n+2n+21)=4×2n=2n+2>0,∴Tn·Tn+23.解:(1)设等差数列{an}的公差为d(d≠0).因为a2,a1,a3成等比数列,所以a12=a2a3,即(a32d)2=(a3d)a3,所以(42d)2=4(4d又d≠0,所以d=3.所以a1=a32d=2.所以an=23(n1)=3n+5.(2)由(1)得an3n1=(3n+5)3n1,所以Sn=(230)+(131)+(432)+…+[(3n+5)3n1]=[2+(1)+(4)+…+(3n+5)](30+31+32+…+3n1)=n[所以S1=1,S2=3,S1>S2.由f(x)=7x-3x22在区间[2,+∞)上单调递减,g(x)=3x-12在区间[2,+又S2<S1,所以{Sn}是递减数列.又S3=16,S4=50,所以使Sn≤20成立的最小正整数n为4.4.解:(1)依题意,{an}是公差为2的等差数列,a1>0,且a4是2a2和a52的等比中项,所以a42=2a2·(a52),所以(a1+6)2=2(a1+2)·(a1+6),解得a1=2或a1所以an=2+2(n1)=2n.(2)依题意,b1a1+b2a2+…当n=1时,b1a1=22,b当n≥2时,b1a1+b2a2+①②得bnan=2n,所以bn=2n·2n=n·2所以bn=8所以Tn=8+2×23+3×24+…+n·2n+1,③2Tn=16+2×24+3×25+…+n·2n+2,④③④得Tn=23+24+…+2n+1n·2n+2=8(1-2n-1)1-2n·2所以Tn=(n1)·2n+2+8.5.(1)证明:∵2an+12an+an+1an=0,∴1a故数列1an是以1(2)解:∵b1=f(0)=5,∴7(a1-1)+5a1-1+1=5,7a12=5a1,∴a1=∴an=2n+1,bn=7an-2an当n≤6时,Tn=n2(5+6n)=n当n≥7时,Tn=15+n-62(1+n6)故Tn=n6.解:(1)由a4+2是a3,a5的等差中项,得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1解得q=2或q=12,因为q>1,所以q=2(2)设cn=(bn+1bn)an,数列{cn}前n项和为Sn,由cn=S1,n=1,Sn-由(1)可知an=2n1,所以bn+1bn=(4n1)·12故bnbn1=(4n5)·12n-bnb1=(bnbn1)+(bn1bn2)+…+(b3b2)+(b2b1)=(4n5)·12n-2+(4n9)·12n-3+设Tn=3